Laws Of Motion
Class 11th Physics Part I CBSE Solution
Exercise
Question 1.Give the magnitude and direction of the net force acting on
A. a drop of rain falling down with a constant speed,
B. a cork of mass 10 g floating on water,
C. a kite skill fully held stationary in the sky,
D. a car moving with a constant velocity of 30 km/h on a rough road,
E. a high-speed electron in space far from all material objects, and free of electric and magnetic fields.
Answer:
A. Since, the drop of rain is falling down with constant speed, the rate of change of velocity i.e. acceleration is zero. Thus, according to the Newton’s second law of motion, the force is zero.
Hence, the net force on a drop of rain falling down with a constant speed is zero.
B. There are two equal and opposite forces acting the cork floating on water. The force due to gravity i.e. weight of the cork is acting in the downward direction which is balanced by the buoyant force acting in the upward direction in the water.
Hence, the net force acting on the floating cork is zero.
C. No net force is acting on the kite held stationary in the sky. It is not executing any kind of motion at all. Thus, according to the first law of motion, it will continue to be in rest until any external force starts acting on it.
Hence, the net force acting on the kite is zero.
D. The car is moving with a constant velocity of 30km/h, the rate of change of velocity i.e. acceleration is zero. Thus, according to the Newton’s second law of motion, the force is zero.
Hence, the net force on the car moving with a constant velocity is zero.
E. There is no electric, magnetic field acting on the high-speed electron and since it is far away in space from all the material objects the gravitational field is also zero.
Thus, according to the Newton’s first law of motion, the net force acting on the high- speed electron is zero.
Question 2.
A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble,
A. during its upward motion
B. during its downward motion
C. at the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of 45° with the horizontal direction?
Ignore air resistance.
Answer:
According to the Newton’s second law of motion:
F = m × a
Where,
‘F’ is the net force acting on an object.
‘m’ is the mass of the object.
‘a’ is the acceleration of the object.
Given,
Mass of the pebble ‘m’= 0.05 kg
a = acceleration due to gravity = 10 m/s2
In all the three cases A, B and C the only force acting on the pebble is that of gravitational force in the downward direction. Irrespective of the motion of an object, the acceleration due to gravity always acts downwards.
The magnitude of the force due to gravity will be,
F = 0.05 kg × 10 m/s2
= 0.5 N
Thus, the net force acting on the pebble ignoring the air resistance in upward or downward motion is 0.5 N and this gravitational force always acts downwards towards the centre of the earth.
The only difference if a pebble is thrown at an angle of 45° with the horizontal is the change in its velocity as it travels all the way up where the vertical velocity becomes zero at the highest peak but the horizontal velocity remains. The acceleration due to gravity is still acting constantly downwards in this case too and ignoring the air resistance. Thus, the net force still remains the same.
Question 3.
Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg,
A. just after it is dropped from the window of a stationary train,
B. just after it is dropped from the window of a train running at a constant velocity of 36 km/h,
C. just after it is dropped from the window of a train accelerating with 1 m s-2,
D. lying on the floor of a train which is accelerating with 1 m s-2, the stone being at rest relative to the train.
Neglect air resistance throughout.
Answer:
According to the Newton’s second law of motion:
F = m × a
Where,
‘F’ is the magnitude of the force acting on an object.
‘m’ is the mass of the object.
‘a’ is the acceleration of the object.
Given,
Mass of the stone, m = 0.1 kg
Acceleration due to gravity is taken as ‘g’ = 10 m/s2
A. When the train is stationary the net force acting on it is only due to gravitational force. Thus, the magnitude of the force is
F= m×a = m×g
=0.1 kg×10 m/s2
F = 1 N
Therefore, the net force acting on the stone will be due to gravitation which is always in the downward direction.
B. When the train is moving with a constant velocity. The rate of change of velocity i.e. acceleration is zero. Thus, there is no net force acting on the stone in the horizontal direction.
The only force that acts is in vertical direction i.e. gravitational force with the magnitude,
F= m×a= m×g
= 0.1 kg × 10 m/s2
F = 1 N
Therefore, the net force acting on the stone is of the magnitude 1 N acting in the vertically downward direction.
C. When the stone leaves the train, the net force acting on it just after it has left does not consist of horizontal force as the force due to train stops acting on it the instant it leaves the train. According to Newton’s first law of motion, the force acting on a body at an instant depends on that instant.
The net force acting on the stone is given only by acceleration due to gravity.
F= m×a= m×g
= 0.1 kg × 10 m/s2
F = 1 N acting in the downward direction.
Therefore, the net force acting on the stone is of the magnitude 1 N acting in the vertically downward direction.
D. Given,
Acceleration of the train,‘a’ = 1 m/s2
When the stone is lying on the floor of the train the weight of the stone is balanced by the normal reaction of the floor. The vertical forces vanish and only the horizontal force remains which is due to the acceleration of the train.
Thus the net force acting on the stone will be in the same direction as the train,
The magnitude of the net force is,
F= ma
⇒ F=0.1 m/s2 × 1 kg = 0.1 N
Therefore, the net force acting on the stone is of the magnitude 1 N acting in the same direction as the train.
Question 4.
One end of a string of length l is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v the net force on the particle (directed towards the centre) is:
T is the tension in the string. [Choose the correct alternative].
A . T,
B. 
C.
D. 0
Answer:
The centripetal force experienced by a particle connected to a string revolving in a circular path around a centre is given by,
Where,
F is the centripetal force acting on the particle
m is the mass of the particle.
v is the speed of the particle and
l is the length of the string.
Here, the tension produces in the spring provides the centripetal force required by the body to stay in a circular motion.
Thus, the net force on the particle is given as,
Where,
T is the tension in the string.
Question 5.
A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 m s-1. How long does the body take to stop?
Answer:
The Newton’s second law of motion states that,
F = m× a
⇒ a = F / m
Where,
F is the force on the body,
m is the mass of the body and
a is the acceleration of the body.
Given,
F = -50 N
The negative sign is because the force applied to the body is retarding and acts in the opposite direction to the motion of the body.
m= 20 kg
Thus, the acceleration is
The first equation of motion at time t, can be written as,
v = u + at
where,
v is the final velocity,
u is the initial velocity,
a is the acceleration of the body.
For the body to stop the final velocity must be equal to zero, thus the equation becomes,
Ans: the total time taken by the body to come to rest is 6 seconds.
Question 6.
A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m s-1 to 3.5 m s-1 in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?
Answer:
The first equation of motion is given as :
v = u + at …….(i)
where,
v is the final speed,
u is the initial speed,
a is the acceleration of the body.
t is the time taken by the body.
Given,
Mass of the body, m = 3 kg
Initial speed of the body, u = 2 m/s
Final speed of the body, v = 3.5 m/s
Time taken by the body, t = 25 sec
From equation (i), 
⇒ 
From Newton’s second law of motion, the magnitude of the force is given as,
F= ma
= 3 kg × 0.06 m/s2
⇒ F = 0.18 N
The magnitude of the force is 0.18 N and since the acceleration does not change the direction of the motion. The direction of the force is in the same direction of the motion.
Question 7.
A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the acceleration of the body.
Answer:
The magnitude of the resultant, Z of the two vectors X and Y at an angle θ with each other is given as,
Z
and the direction of the resultant can be calculated as θ , where from the fig.
Given,
Mass of the body, m = 5 kg.
Force F1 along AB = 8 N.
Force F2 along AC = 6 N
The resultant force F along AD is given by formula for resultant, Z,
Since, the forces are perpendicular, cos θ =0
⇒
The direction of the resultant thus is,
⇒ θ = -36.87°
The negative sign shows the configuration of the θ with respect to F2.
The Newton’s second law of motion, the force can be defined as:
F = m× a
⇒ 
Where,
m is the mass of the body
a is the acceleration of the body.
Thus, the acceleration of the body is,
Ans: The magnitude of the acceleration is 2 m/s2 and is 36.87° in the clockwise direction from the force taken on the horizontal axis i.e. F2.
Question 8.
The driver of a three-wheeler moving with a speed of 36 km/h sees a child standing in the middle of the road and brings his vehicle to rest in 4.0 s just in time to save the child. What is the average retarding force on the vehicle? The mass of the three-wheeler is 400 kg and the mass of the driver is 65 kg.
Answer:
The first equation of motion is given as:
v = u + at …….(i)
where,
v is the final speed of the body,
u is the initial speed of the body,
a is the acceleration of the body.
t is the time taken by the body.
Given,
Mass of the three-wheeler, M = 400 kg
Mass of the driver, m = 65 kg
Initial speed of the three-wheeler, u = 36 km/h
Final speed of the three-wheeler, v = 0 km/h
Time taken by the vehicle to completely stop, t = 4 sec
Total mass of the system= M + m= 400+65 = 465 kg
From equation (i),
⇒ 
Here, the negative sign indicates that the body is retarding and the velocity of the system is decreasing with the time.
From Newton’s second law of motion, the magnitude of the force is given as,
F= (M + m)× a
= 465 kg × (-2.5 m/s2) = -1162.5 N
The negative sign indicates that the force is acting in the opposite direction of the motion of the three-wheeler.
The average retarding force on the vehicle is 1162.5 N
Question 9.
A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 m s-2. Calculate the initial thrust (force) of the blast.
Answer:
According to the Newton’s second law of motion, the net force on the rocket can be given as,
F – mg = ma (the thrust should overcome the force of gravity to travel with acceleration a)
⇒ F = m (g + a)
Where,
‘m’ is the mass of the rocket,
a is the acceleration of the rocket
g is the acceleration due to gravity= 10 m/s2
Given,
Mass of the rocket, m = 20,000 kg
Initial acceleration of the rocket, a = 5 m/s2
Thus, the initial thrust can be calculated as,
F= 20000 kg×(10+5) m/s2
⇒ F = 20000 kg × 15 m/s2 = 3 × 105 N
The initial thrust of the blasted rocket is 3 × 105 N.
Question 10.
A body of mass 0.40 kg moving initially with a constant speed of 10 m s-1 to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0, the position of the body at that time to be x = 0, and predict its position at t = –5 s, 25 s, 100 s.
Answer:
Given,
Mass of the body, m = 0.40 kg
Initial speed of the body, u = 10 m/s
Force acting on the body, F = -8 N
The force is negative because it acts in the opposite direction i.e. south to the motion of the body i.e. north.
Time for which the force acts on the body, t =30 s
According to the second law of motion, the acceleration of the body is:
The third equation of motion is given as,
…..(i)
Where,
‘s’ is the position of the body,
‘u’ is the initial speed of the body,
‘a’ is the acceleration of the body and
‘t’ is the time.
Given,
1. At t=-5 s
The acceleration of the body is zero since the force was applied starting t=0
u= 10 m/s
using equation (i) we get,
s=10 m/s × (-5 sec)
⇒ s= -50 m
2. At t= 25s
Acceleration, a’ = -20 m/s2 and
u = 10 m/s
using equation (i) we get,
⇒ s = 250 m- 6250 m = -6000 m
3. At t=100 s
We divide this time period in two parts with displacements s’ and s”,
The first part is time 0<t≤30s,
Given,
Acceleration of the body, a = -20 m/s2
u = 10 m/s
The displacement travelled by the body using equation (i) during this time t’ is,
⇒ s’= 300 m – 9000 m = -8700 m
The displacement travelled by the body till t=30 s , s’=-8700 m
The second part is 30s<t≤100s,
Given,
Acceleration of the body=0
The final velocity for the first part is the initial velocity for the second part,
From the first equation of motion, The final velocity after t=30 s is,
v= u + at
⇒ v=10 m/s+ (-20) m/s2× 30 s = -590 m/s
Thus, velocity of the body after 30 sec= -590 m/s
For, motion between 30 s to 100 s, i.e. t = 70 s, we use equation (i):
s”=v t + 0
⇒ s”= -590 m/s × 70 s = -41300 m
Thus, s = s’+ s”= -8700 m – 41300 m
⇒ s = -50000 m
The position of the body at t=-5 s, 25 s, 100 s is -50 m, -6000 m, -50000 m respectively.
Question 11.
A truck starts from rest and accelerates uniformly at 2.0 m s-2. At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the (a) velocity, and (b) acceleration of the stone at t = 11s? (Neglect air resistance.)
Answer:
Given,
The initial velocity of the truck, u= 0
Acceleration of the truck,‘a’ = 2 m/s2
Time, t=10 s
According to the first equation of motion, the final velocity, ‘v’ is
v= u+ at
⇒ v= 0 + 2 m/s2× 10 s =20 m/s
The final velocity after t =10s is 20 m/s
(a) At t = 11 s,
The horizontal component of the velocity remains the same, in the absence of air resistance,
Thus, vx = 20 m/s
According to the first equation of motion, The vertical component of velocity of the stone is given by,
vy = u + ayδt
where,
δt = 11 s -10 s = 1 s and
since, the direction is vertical the acceleration acting on it is due to the gravity.
Thus ay= g =10 m/s2
⇒ vy = 0+ 10 m/s2 × 1 s = 10 m/s
The final resultant velocity of the stone is given as,
vres = (vx2 + vy2)1/2
⇒ vres = (202 + 102)1/2 =
→ vres =22.36 m/s
We suppose that the angle made by the resultant velocity with the horizontal velocity, vx is θ,
Thus,
tan θ = (vy/vx)
⇒ θ = tan-1(10/20)
⇒ θ = tan-1(0.5) = 26.57°
The velocity of the stone at t=11 s is 22.36 m/s and is at angle 26.57° with the horizontal
(b) When the stone is dropped from the truck, the horizontal force provided by the truck acting on the stone becomes zero. The only force and thus, the acceleration, that remains is that in the vertical direction i.e. acceleration due to gravity.
Hence, the acceleration of the stone is 10 m/s2 and it is in the downward direction.
Question 12.
A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is 1 m s-1. What is the trajectory of the bob if the string is cut when the bob is (a) at one of its extreme positions, (b) at its mean position.
Answer:
(a) The velocity of the bob at the extreme positions becomes zero because the kinetic energy becomes zero. Thus, there is no horizontal velocity.
The net force just after it has been cut at extreme positions is just gravitational force which always acts vertically downward. Thus, the bob will fall vertically downwards to the ground if the string is cut at the extreme positions.
(b) Given,
The speed of the bob at the mean position is 1 m/s.
The kinetic energy is thus maximum here and the direction of the velocity is tangential to the arc traced by the oscillating bob. Thus, if it is cut at the mean position then it will follow a projectile path due to the presence of the horizontal velocity.
Question 13.
A man of mass 70 kg stands on a weighing scale in a lift which is moving
A. upwards with a uniform speed of 10 m s-1,
B. downwards with a uniform acceleration of 5 m s-2,
C. upwards with a uniform acceleration of 5 m s-2.
What would be the readings on the scale in each case?
D. What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?
Answer:
Let the force experienced by the weighing scale needle be F,
Given,
Mass of the man = 70 kg
Acceleration due to gravity, g = 10 m/s2
A. If the lift if moving upwards with a uniform velocity,
The acceleration of the lift ‘a’ is zero,
Using the Newton’s second law of motion, The net force is thus gravitational force acting downward,
F – m× g=0
⇒ F= m×g =70 kg × 10 m/s2
⇒ F=700 N
The apparent weight m’ shown on the weighing machine is that of mass which is calculated as,
⇒ m’= 70 kg
which is same as the original weight of the man.
B. If the lift is moving downwards
Given,
Acceleration of the lift, ‘a’=5 m/s2
The motion is in the same direction as the acceleration due to gravity,
Using the Newton’s second law of motion, the net force thus is,
F + mg = ma
⇒ F = m× (a-g) = 70 kg× (5– 10) m/s2
⇒ F = 70 kg× (-5 m/s2) = -350 N
Thus, the apparent weight of the man is,
C. If the lift is moving upwards,
Given,
Acceleration of the lift, ‘a’= 5 m/s2
The motion is in the opposite direction as the acceleration due to gravity,
Using the Newton’s second law of motion, the net force thus is,
F – mg = ma
⇒ F = m (a+g) = 70 kg × (5+10) m/s2
⇒ F = 70 kg × 15 m/s2 =1050 N
The apparent weight of the am on the weighing machine thus is,
D. When the lift is moving downward freely under gravity,
The acceleration of the lift, ‘a’ = g= 10 m/s2
Using the Newton’s second law of motion, the net force thus is,
F + mg = mg
⇒ F = 0 N
Thus, the body is under the state of free fall with a reading of zero in the weighing machine.
Question 14.
Figure 5.16 shows the position-time graph of a particle of mass 4 kg. What is the (a) force on the particle for t < 0, t > 4 s, 0 < t < 4 s? (b) impulse at t = 0 and t = 4 s ? (Consider one-dimensional motion only).
Answer:
For the particle with the trajectory as shown in the graph, the force on the particle,
A. For t<0
We can observe from the graph that the line displaying the position of the particle is coinciding with the x axis, i.e. There is no displacement and hence no net force on the body before t=0.
For t >4 s
We can observe that the position-time graph attains a constant value after t=4, this means that the displacement is not changing, it has a constant value of 3 m and thus the particle is at rest. Thus the net force on the body is zero.
For 0<t<4 s
We can observe that the position-time graph has a constant slope at the given time. The velocity is thus constant and so the acceleration is zero. The net force acting on the particle is thus zero.
B. From the second Newton’s law of motion, Impulse, ‘I’ is defined as the change in the momentum of an object.
I = mv-mu = m(v-u)
Where,
‘m’ is the mass of the particle
‘v’ is the final velocity of the particle
‘u’ is the initial velocity of the particle.
At t=0
Given,
Mass of the particle, m = 4 kg
Initial velocity of the particle, u= 0
Final velocity of the particle can be obtained by calculating the slope of the position-time graph at t =4 s
⇒ 
Therefore,
⇒ I = 3 kg m/s
At t=4 s
Given,
Mass of the particle, m = 4 kg
Initial velocity of the particle, 
Final velocity of the particle can be obtained by calculating the slope of the position-time graph after t =4, i.e. v = 0
Therefore,
⇒ I = -3 kg m/s
Question 15.
Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a light string. a horizontal force F = 600 N is applied to (i) A, (ii) B along the direction of string. What is the tension in the string in each case?
Answer:
The Newton’s second law of motion is defined as,
F = m× a
⇒
Where
‘a’ is the acceleration of the system
‘F’ is the force applied
‘m’ is the mass of the system.
Given,
Horizontal force applied on the body, F = 600 N
Mass of the body A, mA = 10 kg
Mass of the body B, mB = 20 kg
The mass of the system, M = mA+mB=10+20 kg= 30 kg
Thus, the acceleration ‘a’ of the system is given as,
⇒ a = 20 m/s2
(i) When the force is applied to the body A,
Using the newton’s second law of motion, The equation of motion is given as,
F – T= mA a
⇒ T= F – mA a
⇒ T= 600 N – 10 kg × 20 m/s2
⇒ T= 600 N – 200 N = 400 N
Tension when the force is applied on body A is 400 N.
(ii) When the force is applied to the body B
Using the newton’s second law of motion, the equation of motion is given as,
F – T= mB a
⇒ T= F – mB a
⇒ T= 600 N – 20 kg × 20 m/s2
⇒ T= 600 N – 400 N = 200 N
Tension when the force is applied on body B is 200 N.
Question 16.
Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.
Answer:
Given:
m= 8 kg
M= 12 kg
g(acceleration due to gravity)=10 m/ s2
The system of masses and pulley has one smaller mass,’m’ and one larger mass, ‘M’.
‘T’ is the tension in the strings,
‘a’ is the acceleration of the masses.
The larger mass, ‘M’ moves downward with supposed acceleration ‘a’ and thus mass, ‘m’ moves upward.
According to the Newton’s second law of motion, we get equation of motion for each masses,
For mass m:
T – mg = ma …..(i)
For mass M:
Mg - T = Ma ……(ii)
We add both of the equation(i) and (ii), to get:
(M – m)g = (M+m)a
⇒ 
……(iii)
⇒ 
= 2 m/s2
The acceleration of the masses is 2 m/s2
To find tension, We substitute the value of a from eq(iii) in eq(ii), we get:
⇒ 
⇒ 
⇒ T=96 N
The tension in the string when the masses are released is 96 N.
Question 17.
A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei, the products must move in opposite directions.
Answer:
We suppose,
Mass of the parent nucleus as M and the mass of two daughter nuclei as m1 and m2 .
The final velocities of the respective two daughter nuclei after disintegration as v1,v2 respectively.
Since, the parent nucleus before disintegration was at rest,
Initial momentum of the system i.e. parent nucleus = 0
Final momentum of the system i.e. daughter nuclei = m1v1+m2v2
According to the law of conservation of linear momentum:
Total initial momentum = Total final momentum
Therefore,
0 = m1v1+m2v2
⇒ v1 = - (m1/m2)× v2
The negative sign indicates that both the velocities of the daughter nuclei are in opposite directions.
Question 18.
Two billiard balls each of mass 0.05 kg moving in opposite directions with speed 6 m s-1 collide and rebound with the same speed. What is the impulse imparted to each ball due to the other?
Answer:
The momentum of the ball can be calculated as
p = m× v
Where,
‘m’ is the mass of the ball
‘v’ is the velocity of the ball.
Given,
Mass of each billiard ball, m = 0.05 kg
Initial velocity of each ball, u = 6 m/s
The initial momentum, pi, of each ball is
Pi = m× u= 0.05 kg × 6 m/s
⇒ pi = 0.3 kg m/s
The balls rebound after the collision thus, they change directions but the magnitude of the velocity remains the same,
Now,
Final velocity of each ball, v = -6 m/s
The final momentum, pf of each ball is,
pf= m× v = 0.05 kg × -6 m/s
⇒ pf = -0.3 kg m/s
Impulse imparted to each ball, ‘I’ = change in the momentum of the system
⇒ I= pf - pi = (-0.3 – 0.3) kg m/s
⇒ I= -0.6 kg m/s
The negative sign is because the impulse imparted to each of the balls are in opposite directions.
the impulse imparted to each ball due to the other is 0.6 N in opposite directions.
Question 19.
A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 m s-1, what is the recoil speed of the gun?
Answer:
The initial velocity of both the gun and the shell system is zero
The initial momentum of the system = 0
Given,
Mass of the gun, M = 100 kg
Mass of the shell, m =0.02 kg
Final speed of the muzzle of the shell, v = 80 m/s
We suppose the recoil speed of the gun to be V.
The final momentum of the system = mv – MV
The negative sign is because the direction of motion of gun and shell is opposition to each other.
According to the law of conservation of linear momentum:
Total initial momentum = Total final momentum
Therefore,
0 = mv – MV
⇒ 
⇒
= 0.016 m/s
The recoil speed of the gun is 0.016 m/s
Question 20.
A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball? (Mass of the ball is 0.15 kg.)
Answer:
Given,
Mass of the ball = 0.15 kg
Initial speed of the ball = final speed of the ball= 54 km/h = 15 m/s
From the figure,
The ball is deflected at an angle 45°. After deflection by the batsman,
∠XOA = ∠AOY = 22.5° = θ
The horizontal components of the initial and final velocities are 
The vertical components are of the initial and final velocities are 
The vertical components do not survive as they are in opposite directions,
The horizontal component of the velocity contributes in the momentum of the system,
The impulse ‘I’ imparted to the ball= change in the momentum of the ball
⇒ 
⇒ 
= 2× 0.15 kg × 15 m/s × cos 22.5°
⇒ I = 4.16 kg m/s
The impulse imparted to the ball is 4.16 kg m/s
Question 21.
A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev./min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?
Answer:
The angular velocity of the stone in circular motion is given as,
ω = 
where,
‘v’ is the linear velocity
‘r’ is the radius of the circle.
‘n’ is the number of revolutions per second
The centripetal force for the stone is provided by the tension T of the string,
The centripetal force ‘Fc’ can be given as
Fc
ω2r = m (2πn)2 r
And Fc = Tension in the string
Where, m
Given,
Mass of the stone, m= 0.25 kg
Radius of the circle, r= 1.5 m
Number of the revolution per second, n = 
⇒ n= 
Thus,
T= Fc = 
⇒ T = 6.57 N
The tension in the string is 6.57 N
Given,
The maximum tension that the string can withstand is, T’ =200 N
T’= 
⇒ v’= 
Where, v’ is the maximum velocity of the stone
⇒ v’ =
The maximum speed of the stone is 34.64 m/s
Question 22.
If, in Exercise 5.21, the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks:
A. the stone moves radially outwards,
B. the stone flies off tangentially from the instant the string breaks,
C. the stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle?
Answer:
If the speed of the stone is increased beyond the maximum permissible value and the string breaks suddenly, the stone will follow the path of the velocity at the instant that it breaks.
According to the first law of motion, the velocity of the stone at every point is tangential to path followed by it at that instant.
Thus, the stone will fly off tangentially the instant the string breaks.
Ans: B.
Question 23.
Explain why
A. a horse cannot pull a cart and run in empty space,
B. passengers are thrown forward from their seats when a speeding bus stops suddenly,
C. it is easier to pull a lawn mower than to push it,
D. a cricketer moves his hands backwards while holding a catch.
Answer:
A. The Newton’s third law of motion explains that to move forward and equal and opposite reaction force is acted on the ground. If the horse pushes the ground with some force, the ground in turn exerts equal and opposite force on the feet of the horse. This reaction force is the reason for the motion of the horse. An empty space does not have a surface for friction and thus no reaction force.
Therefore, a horse cannot pull a cart and run in empty space.
B. When a speeding bus suddenly stops, the feet of the passengers are still in the contact with bus, which suddenly comes to rest. Whereas, according to Newton’s first law of motion, the upper body of the passengers which is not in contact with the bus, continues to be in motion.
As a result, the passenger’s upper body is thrown forward while the lower body comes to halt with the bus.
C. When we pull a lawn mower, a force at some angle is applied on it. The vertical component is in the upward direction. This upward component reduces the effective weight of the mower, thus making it easier to pull.
Whereas, when we push a lawn mower, a force acting at some angle is applied on it, the vertical components is in the downward direction and act thus, increases the effective weight of the land mower.
Due to the less effective weight in the first case, the land mower is easier to pull than push.
D. According to the second law of motion, we have the equation of motion as,
F = ma = 
Where
‘F’ is the force experienced by the cricketer as he catches the ball.
‘m’ is the mass of the ball
‘Δt’ is the short time of the impact with the hand of cricketer.
We can thus see from the equation that impact force is inversely proportional to the impact time, Thus, if the impact is for a shorter period of time then the force will be large.
It also shows that the force experienced by the cricketer decreases with the increase in the impact time.
Therefore, the cricketer moves his hand backward while taking a catch to increase the impact time, and hence decrease the impact force on his hand and prevent it from getting hurt.
Question 24.
Figure 5.17 shows the position-time graph of a body of mass 0.04 kg. Suggest a suitable physical context for this motion. What is the time between two consecutive impulses received by the body? What is the magnitude of each impulse?
Answer:
Given:
Mass of body = 0.04 kg
From the graph (position- time) we observe that, a body is moving linearly to a displacement of 2 cm, then it comes back to 0 position, each cycle consumes 4 seconds. This graph might represent a ball bouncing between two walls located 2 cm apart. Ball starts at time 0 from position x = 0 with some velocity, after 2 seconds it reaches the second wall at a distance of 2 cm, then bounces back to the zero position in next 2 seconds.
We can observe that after every two seconds the ball is changing direction, so we can conclude that it strikes the walls every two seconds, so impulse is imparted to the ball every 2 seconds.
We know that,
Impulse = Change of momentum
→ I = final momentum – initial momentum
→ I = mv- mu …(1)
Where,
M = mass of body
v = velocity of body after collision with wall
u = velocity of body before collision with wall
Velocity of ball can be found as the slope of the curve,
i.e. 
…(2)
Where, v = velocity
x0 = initial displacement in metres
x1 = final displacement in metres
t0 = time at beginning of event
t1 = time at final displacement
By putting the values in equation (2),
→ v = 10-2 ms-1
Initial velocity, u is also the same but with a negative sign,
u = -10-2 ms-1
By putting the values in equation (1) ,
I = |(0.04 kg× 10-2ms-1)- (0.04 kg× (-10-2ms-1))
I = 0.08× 10-2Kg ms-1
The magnitude of impulse is 0.08× 10-2Kg ms-1.
Question 25.
Figure 5.18 shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with 1 m s-2. What is the net force on the man? If the coefficient of static friction between the man’s shoes and the belt is 0.2, up to what acceleration of the belt can the man continue to be stationary relative to the belt?
(Mass of the man = 65 kg.)
Fig 5.18
Answer:
Given:
Mass of person = 65 kg
Co-efficient of static friction, � = 0.2
Acceleration of the belt, a = 1 ms-2
The net force F, acting on the man is given by Newton’s second law of motion as,
Fnet = ma …(1)
Where,
m = mass of body
a = acceleration of body
By putting the values in equation (1) , we get
Fnet = 65 Kg× 1 ms-2
Fnet = 65 N
The man on the conveyer will remain stationary with respect to conveyer until the force exerted on him is less than the frictional force between him and conveyer,
We can write,
Fnet = fs
m× amax = �× m× g
→ amax = �× g …(2)
Where,
� = coefficient of static friction
g = acceleration due to gravity
By putting the values in equation (2) we get,
→ amax = 0.2× 10ms-2
→ amax = 2 ms-2
As long as the acceleration of conveyer doesn’t exceed 2ms-2, the man will stand stationary.
Question 26.
A stone of mass m tied to the end of a string revolves in a vertical circle of radius R.
The net forces at the lowest and highest points of the circle directed vertically downwards are: [Choose the correct alternative]
T1 and v1 denote the tension and speed at the lowest point. T2 and v2 denote corresponding values at the highest point.
Answer:
Given:
Mass of body = m Kg
Radius of vertical circle = R m
From the free body diagram of body at lowest point, we can write,
According Newton’s 2nd law of motion,
Fnet = mg-T
Fnet = mv12/R (Centrifugal force)
mg-T = mv12/R …(1)
Where,
v1 = Velocity at lowest point.
m = mass of body
R = radius of path
T = tension in string
From the free body diagram of body at highest point , we can write,
Fnet = T + mg
Fnet = mv22/ R
Where,
v2 = velocity of body at highest point,
T + mg = mv22/ R …(2)
From equation (1) and (2) we conclude that option (1) is correct.
Question 27.
A helicopter of mass 1000 kg rises with a vertical acceleration of 15 m s-2. The crew and the passengers weigh 300 kg. Give the magnitude and direction of the
A. force on the floor by the crew and passengers,
B. action of the rotor of the helicopter on the surrounding air,
C. force on the helicopter due to the surrounding air.
Answer:
Given:
Mass of helicopter, M = 1000 Kg
Mass of the crew, mp = 300 Kg
Total mass of helicopter, mh = 1300 Kg
Vertical acceleration, a = 15ms-2
(a) Using Newton’s second law of motion, we can write the normal reaction, R as,
R = mpg + mpa
R = mp(g + a) …(1)
By putting values in equation 1, we get,
R = 300 Kg (10 ms-2 + 15 ms-2)
R = 7500 N
This reaction force from the floor works in upward direction. We can conclude the force on
by floor on crew is 7500 N. (From Newton’s third law).
(b) Using Newton’s third law of motion,
R’ = mhg + mha
→ R’ = mh(g + a)
→ R’ = 1300 Kg (10 + 15) ms-2
→ R’ = 32500 N
Since surrounding air provides a Upward normal reaction of 32500N, using Newton’s 3rd law of motion, we conclude that rotor also applies a force on 32500 N on surrounding air in downward direction.
(c) The force on the helicopter due to surrounding air is 32500 N and upwards.
Question 28.
A stream of water flowing horizontally with a speed of 15 m s-1 gushes out of a tube of cross-sectional area 10-2 m2, and hits a vertical wall nearby. What is the force exerted on the wall by the impact of water, assuming it does not rebound?
Answer:
Given:
Speed of water stream, v = 15 ms-1
Cross-sectional area of the tube, A = 10-2 m2
Density of water, ρ = 1000 Kgm-3
Volume of water leaving the tube per second = A× v
V = A× v …(1)
V = 10-2 m2× 15 ms-1
V = 0.15 m3s-1
Mass of water flowing out of the pipe = density of fluid× volume
M = ρ× V …(2)
M = 1000 Kgm-3× 0.15 m3s-1
M = 150 Kgs-1
Since the water doesn’t rebound after striking the Wall, hence according to Newton’s 2nd law , we can write,
F = Rate of Change of momentum
F = change in momentum / time
F = |mv-mu|/ t …(3)
Where,
M = mass of water
v = final velocity = 0
u = initial velocity
t = time = 1sec
the equation (3) becomes,
F = mu/t
By putting values in the above equation,
F = 150 Kgs-1× 15 ms-1
F = 2250 N.
The water exerts a force of 2250 N on the wall.
Question 29.
Ten one-rupee coins are put on top of each other on a table. Each coin has a mass m. Give the magnitude and direction of
A. the force on the 7th coin (counted from the bottom) due to all the coins on its top,
B. the force on the 7th coin by the eighth coin,
C. the reaction of the 6th coin on the 7th coin.
Answer:
Given:
Mass of each coin = m Kg
Number of coins = 10
(a) Force on the 7the coin from bottom is exerted by the weight of the 3 coins above it.
F = 3× (m× g)
F = 3mg
The force is vertically downward.
(b) Force on the seventh coin due to the eighth coin is the sum of force by all the coins above it.
Since there are 2 more coins above coin 8, so the force on coin 7 due to 8 is given by,
F = mg + 2mg
F = 3mg
The force acts vertically downwards.
(c) The 6th coin has 4 coins above it hence it experiences the force,
F = 4 × mg
From Newton’s 3rd law of motion we conclude that coin 6, exerts an equal force in upward direction.
R = 4× mg
This reaction is in upward direction.
Question 30.
An aircraft executes a horizontal loop at a speed of 720 km/h with its wings banked at 15°. What is the radius of the loop?
Answer:
Given:
Speed of aircraft, v = 720 Km/h = 200ms-1
Banking of aircraft, Θ = 15°
Let radius of loop = R m
We know that, radius of loop has the following relation,
R = v2/ (g tan Θ) …(1)
→ R = 14925.37 m
→ R = 14.92 Km
Question 31.
A train runs along an unbanked circular track of radius 30 m at a speed of 54 km/h.
The mass of the train is 106 kg. What provides the centripetal force required for this purpose — The engine or the rails? What is the angle of banking required to prevent wearing out of the rail?
Answer:
Given:
Radius of track, R = 30 m
Speed of train, v = 54 Km/h = 15ms
Mass of the train, m = 106 Kg
The centripetal force is provided by the lateral thrust of the rail on the wheel. As per Newton’s third law of motion, the wheel exerts an equal and opposite force on the rail. This reaction force is responsible for the wear and tear of the rail.
The angle of banking Θ, is related to the radius (R) and speed (v) by the relation:
R = v2/ (g tan Θ) …(1)
→ tanΘ = 0.75
→ Θ = 36.87°
The angle of banking is about 37°.
Question 32.
A block of mass 25 kg is raised by a 50 kg man in two different ways as shown in Fig. 5.19. What is the action on the floor by the man in the two cases? If the floor yields to a normal force of 700 N, which mode should the man adopt to lift the block without the floor yielding?
Answer:
Given:
Mass of block, m = 25 Kg
Mass of man, M = 50 Kg
Yielding normal force of floor, Fm = 700 N
Acceleration due to gravity, g = 10 ms-2
In both the cases we will focus on finding the normal force on the floor.
Normal force will change according to the apparent weight of man.
(a) When the block is lifted directly,
The man applies a force in upward direction, which increases his apparent weight as the weight of the block is added to weight of man. This apparent weight is responsible for the normal force on floor.
F = Weight of man + weight of block
F = Mg + mg
F = (50 Kg + 25 Kg) × 10 ms-2
F = 750 N
(b) When the block is lifted with pulley,
The man applies a force in downward direction, which decreases his apparent weight. This apparent weight is responsible for the normal force on floor.
F = Weight of man - weight of block
F = Mg- mg
F = (50 Kg- 25 Kg) × 10 ms-2
F = 250 N
The man should use the pulley to easily lift the block. If block is lifted directly, the force will be greater than the yield force of floor hence the floor can break.
Question 33.
A monkey of mass 40 kg climbs on a rope (Fig. 5.20) which can stand a maximum tension of 600 N. In which of the following cases will the rope break: the monkey
A. climbs up with an acceleration of 6 m s-2
B. climbs down with an acceleration of 4 m s-2
C. climbs up with a uniform speed of 5 m s-1
D. falls down the rope nearly freely under gravity?
(Ignore the mass of the rope).
Answer:
Given:
Mass of monkey = 40Kg
Maximum tension that rope can withstand, Tm = 600 N
We need to find the case in which rope will break, i.e. the force should be more than 600 N.
(A) acceleration (Upwards) = 6ms-2
The monkey is moving in upward direction,
From Newton’s 2nd law we can write,
F = ma = T- mg
→ ma = T-mg
→ T = m(a + g)
→ T = 40 Kg( 6 ms-2 + 10 ms-2)
→ T = 640 N
Since T> Tm, the rope is destined for a failure, it will break.
(B) Acceleration (Downwards) = 4 ms-2
The monkey is moving in upward direction,
From Newton’s 2nd law we can write,
F = ma = mg-T
→ ma = mg-T
→ T = m (g - a)
→ T = 40 Kg (10 ms-2 - 4 ms-2)
→ T = 240 N
Since T< Tm, the rope will not break.
(B) Acceleration, a = 0ms-2
Climbing velocity, v = 5ms-1
Using Newton’s 2rd law we can write,
F = ma = T-mg
→ ma = T-mg
→ T = mg
→ T = 40Kg × 10 ms-2
→ T = 400 N
Since T<Tm, the rope will not break.
(C) If the monkey is free falling, i.e. a = g;
Using Newton’s 2nd law of motion, we can write,
F = mg-T = mg
T = mg-mg
T = 0 N
Since, T< Tm, the rope will not break.
Question 34.
Two bodies A and B of masses 5 kg and 10 kg in contact with each other rest on a table against a rigid wall (Fig. 5.21). The coefficient of friction between the bodies and the table is 0.15. A force of 200 N is applied horizontally to A. What are (a) the reaction of the partition (b) the action reaction forces between A and B? What happens when the wall is removed? Does the answer to (b) change, when the bodies are in motion? Ignore the difference between μs and μk.
Answer:
Given:
Mass of body A, MA = 5 Kg
Mass of body B, MB = 10Kg
Coefficient of friction between bodies and surface, � = 0.15
Force applied on block A, F = 200N
(A) The reaction force at partition, R
The force of 200 N is acting rightwards, a frictional force will act in leftward direction against the impending motion.
Using Newton’s second law,
F’ = F- fs
F’ = 200- ( �× (M �a + MB) g)
F’ = 200N- (0.15 × (10 + 5) Kg× 10ms-2)
F’ = 177.5 N
Net force acting in rightward direction at the partition is 177. 5 N.
We can conclude using 3rd law of motion that the normal force at the partition is 177.5 N in leftward direction.
(B) Force of friction on body A,
fA = �MAg
→ fA = 0.15× 5Kg × 10ms-2
→ fA = 7.5 N
This force acts leftwards.
Net force exerted by A on B ,
FNet = F- fA
FNet = 200N - 7.5N
FNet = 192.5 N
This force acts rightwards.
According to Newton’s 3rd law we can conclude that force by B on A is also 192.5 N in leftward direction.
If the Wall is removed, the bodies will start in the direction of applied force.
For motion of combined body,
From (A), we see that Net force on the bodies, F = 177.5 N
According to Newton’s second law, we can write,
F = (MA + MB)× a
a = F/ (MA + MB)
a = 177.5/ (5 + 10)
a = 11.83 ms-2
Net force on body A,
F = MA× a
F = 5Kg× 11.83 ms-2
F = 59.15 N
Net force on body, B by Body A,
F = 192.5- 59.15
F = 133.35 N
This force acts in the direction of motion of body. An equal amount of force is applied by body B on A, in opposite direction.
Question 35.
A block of mass 15 kg is placed on a long trolley. The coefficient of static friction between the block and the trolley is 0.18. The trolley accelerates from rest with 0.5 m s-2 for 20 s and then moves with uniform velocity. Discuss the motion of the block as viewed by (a) a stationary observer on the ground, (b) an observer moving with the trolley.
Answer:
Given:
Mass of block, M = 15 Kg
Coefficient of static friction, � = 0.18
Acceleration of the trolley, a = 0.5 ms-2
Time for which the trolley accelerates, t = 20 s
To check whether the block will slide or not, let us compare the force due to trolley and
trolley and Block.
Using the Newton’s second law of motion,
Force, F on the block due to accelerating trolley is given by,
F = M× a
F = 15 Kg× 0.5 ms-2
F = 7.5 N
Frictional force, fs
fs = �× M× g
fs = 0.18× 15Kg × 10 ms-2
fs = 27 N
F<fs
The force of friction between the block and the trolley is greater than the applied force by trolley. Hence for an observer on ground the block will seem to be moving at the same velocity as that of trolley.
b) For an observer moving with the trolley, the trolley and the box will seem to be at rest.
Question 36.
The rear side of a truck is open and a box of 40 kg mass is placed 5 m away from the open end as shown in Fig. 5.22. The coefficient of friction between the box and the surface below it is 0.15. On a straight road, the truck starts from rest and accelerates with 2 m s-2. At what distance from the starting point does the box fall off the truck? (Ignore the size of the box).
Answer:
Given:
Mass of the block, M = 40 Kg
Co-efficient of friction, � = 0.15
Initial velocity, u = 0ms-1
Acceleration, a = 2ms-2
Distance of Box from the end of truck, s = 5m
Using the Newton’s 3rd law of motion, the force on the Box, F can be written as,
F = M× a
F = 40Kg × 2 ms-2
F = 80 N
Since there is friction between the truck and the body, we can write frictional force as,
fs = �× M× g
fs = 0.15× 40Kg × 10ms-2
fs = 60 N.
Since, force of friction is less than that of exerted by the truck, there exists a net force on Box.
FNet = F-fs
FNet = 80 N – 60 N
FNet = 20 N
This force acts in backward direction.
We can find the acceleration of the Box using Newton’s second Law,
FNet = M× a
a = 20 N/ 40 Kg
a = 0.5 ms-2
The box accelerates in backward direction at the rate of 0.5 ms -2.
Since, box is 5 m away from end, i.e. 5 m away from falling down, so we can use the second equation of motion to find the time of fall.
Second equation of motion is given as,
s = ut + 1/2 at2 …1
Where,
s = distance travelled
u = initial velocity
t = time
a = acceleration
Since box was at rest initially so u = 0ms-1
Equation 1 can be re-written as,
S = 1/2 at2
→ t = (2s/a)1/2
→ t = 201/2 s
Now, the distance travelled by truck till the time of fall can also be calculated by equation 1
s = ut + 1/2 at2
s = 0 + 0.5 × 2× (201/2)2
s = 20 m
Question 37.
A disc revolves with a speed of 
rev/min, and has a radius of 15 cm. Two coins are placed at 4 cm and 14 cm away from the centre of the record. If the co-efficient of friction between the coins and the record is 0.15, which of the coins will revolve with the record?
Answer:
Given:
Speed of revolution of disc, v = 33.3 rev/min = 0.55 rev/sec
Radius of Disc = 15 cm = 0.15 m
Distance of coin 1 from centre, r = 4 cm = 0.04 m
Distance of coin 2 from centre, R = 14 cm = 0.14 m
Coefficient of friction, � = 0.15
To decide which coin will rotate, we shall calculate the net force on each coin,
For coin 1, at 4 cm distance
Radius of path = 0.04 m
Angular frequency, ω = 2πv
ω = 2× 3.14× 0.55 rev/sec
→ ω = 3.49 s-1
Frictional force, f = �× m× g
→ f = 0.15× m Kg× 10 ms-2
→ f = 1.5 m N
Centrifugal force on the coin:
F = m× r× ω2
F = m× 0.04m × 3.492
F = 0.49m N
Since force of friction is larger than the centrifugal force, the coin will not slide and will revolve along the record.
For coin 2, at 14 cm distance
Radius of path = 0.14 m
Angular frequency, ω = 2πv
ω = 2× 3.14× 0.55 rev/sec
→ ω = 3.49 s-1
Frictional force, f = �× m× g
→ f = 0.15× m Kg× 10 ms-2
→ f = 1.5 m N
Centrifugal force on the coin:
F = m× r× ω2
F = m× 0.14m × 3.492
F = 1.7m N
Since force of friction is smaller than the centrifugal force, the coin will slip on the surface of record.
Question 38.
You may have seen in a circus a motorcyclist driving in vertical loops inside a ‘deathwell’ (a hollow spherical chamber with holes, so the spectators can watch from outside). Explain clearly why the motorcyclist does not drop down when he is at the uppermost point, with no support from below. What is the minimum speed required at the uppermost position to perform a vertical loop if the radius of the chamber is 25 m?
Answer:
Given:
Radius of vertical loop, R = 25 m
A motor-cyclist should not fall from the top most position of the vertical loop provided that the weight of motorcycle and the normal reactions are balanced by the centrifugal force.
Using Newton’s second law of motion we can write,
F = mac
Where,
ac = Centripetal acceleration
F = Normal force + weight of motorcycle
F = FN + FW
FN + FW = mv2/R
FN + mg = mv2/R …(1)
If we consider for the minimum speed the normal force becomes equivalent to Zero.
FN = 0
We can rewrite equation (1) as,
mg = mvmin2 /R
i.e.
Vmin = (Rg)1/2
Vmin = (25 m× 10ms-2)1/2
Vmin = 15.8 ms-1
Question 39.
A 70 kg man stands in contact against the inner wall of a hollow cylindrical drum of radius 3 m rotating about its vertical axis with 200 rev/min. The coefficient of friction between the wall and his clothing is 0.15. What is the minimum rotational speed of the cylinder to enable the man to remain stuck to the wall (without falling) when the floor is suddenly removed?
Answer:
Given:
Mass of man, M = 70 Kg
Radius of the drum, r = 3m
Co-efficient of friction, � = 0.15
Frequency of rotation, v = 200 rev/min = 3.3 rev s-1
The necessary centrifugal force required for the rotation of man is provided by the Normal force (FN).
When the floor rotates, the man sticks to the wall of drum. So the friction between his cloth and wall prevents him from falling. i.e frictional force balances the weight.
Frictional force, f = � × FN
The man will not fall until,
Weight< f
mg<f
→ mg< �× FN
→ mg< �× mrω2
→ ω>( g/ �r )1/2
→ ωmin = (10/(0.15× 3) )1/2
→ ωmin = 4.71 rad/s
The minimum angular speed is 4.7 rad/sec.
Question 40.
A thin circular loop of radius R rotates about its vertical diameter with an angular frequency w. Show that a small bead on the wire loop remains at its lowermost point for
. What is the angle made by the radius vector joining the centre to the bead with the vertical downward direction for
? Neglect friction.
Answer:
Let the radius vector joining the bead with the centre makes an angle Θ, with the vertical downward direction.
OP = R = Radius of the circle
N = Normal reaction
The respective Vertical and Horizontal equations of forces can be written as:
Mg = NCosΘ …(1)
m/ω2 = Nsin Θ …(2)
In triangle OPQ, we have,
sinΘ = I/R
→ I = RSin Θ …(3)
Substituting equation (3) in equation (1) , we have,
m(R SinΘ )ω2 = NsinΘ
→ mRω2 = N …(4)
Substituting equation 4 in equation 1, we have,
mg = mRω2Cos Θ
CosΘ = g/ Rω2 …(5)
Since, CosΘ ≤ 1, the bead will remain at the lowermost point,
for g/ Rω2≤ 1,
i.e. ω≤(g/R)1/2
For ω = (2g/R)1/2 …(6)
On equation equations 5 and 6, we have
2g/R = g/RcosΘ,
CosΘ = 1/2
Θ = Cos-1(0.5)
→ Θ = 60°