Circular Motion

Question 1:

Question 2:

Yes, it is possible to accelerate the motorcycle without putting higher petrol input rate into the engine by driving the motorcycle on a circular track.

Question 3:

The person rotating with the drum will observe that centrifugal force and coriolis force act on the water particles and the person washing the cloth will observe that water particles are thrown outward (away from the drum) and no pseudo force is acting on the particles.

Question 4:

The coin gets the required centripetal force from the frictional force between the coin and the record.

Question 5:

The bird tilts its body and tail in such a way that the air around offers a dragging force in left direction, perpendicular to its initial direction of motion. This dragging force provides the necessary centripetal force to take the left turn.

Question 6:

No, it is not necessary to express all the angles in radian while using the equation ω = ω₀ + at.
If ω (angular velocity) and α (angular acceleration) are in rad/s and rad/s², respectively, we will get the angle in radian.

Question 7:

While shaking, our hand moves on a curved path with some angular velocity and water on our hand feels centrifugal force in the outward direction. Therefore, water get detached from our hand and leaves it.

Question 8:

The outer wall will exert a non-zero normal contact force on the block. As the block moves in a uniform circular motion, centrifugal force in radially outward direction acts on it and it comes in contact with the outer wall of the tube.

Question 9:

(a) Gravitational attraction of the Sun on the Earth is equal to the centripetal force. This statement is more appropriate.

Gravitational attraction of the Sun on the Earth provides the necessary centripetal force required for the circular motion of the Earth around the Sun.

Question 10:

As the wall is at a distance r, the driver should either take a circular turn of radius r or apply brakes to avoid hitting the wall.

Question 1:

When the same mass is set into oscillation, the tension in the string increases because of the additional centripetal force of the mass oscillating in a curved path.

Question 2:

(d) its velocity and acceleration both change.

In a circular motion, the direction of particle changes. Therefore, velocity, being a vector quantity, also changes.
As the velocity changes, acceleration also changes.

Question 3:

(d) 1

Time period (T) is same for both the cars.
We know that:
$$\begin{gathered} \omega \propto \frac{1}{T} \\ \mathrm{So}, \frac{{\omega }_1}{{\omega }_2}=1 \end{gathered}$$

Question 4:

(c) N_A < N_B

From the figure in the question, it is clear that $r_B>r_A$.
Here, normal reaction is inversely proportional to the centrifugal force acting on the car, while taking turn on the curve track. Also, centrifugal force is inversely proportional to the radius of the circular track.
Therefore, we have: 
N_A < N_B

Question 5:

(d)  zero

The centrifugal force is a pseudo force and can only be observed from the frame of reference, which is non-inertial w.r.t. the particle.

Question 6:

(b) $m{\mathrm{\omega }}_0^2\mathrm{a}$

The centrifugal force on the particle depends on the angular speed (ω₀) of the frame and not on the angular speed (ω) of the particle. Thus, the value of centrifugal force on the particle is $m{\mathrm{\omega }}_0^2\mathrm{a}$.

Question 7:

(a) 10 cm/s, 10 cm/s²

It is given that the turntable is rotating with uniform angular velocity. Let the velocity be $\omega$.
We have:
$$\begin{gathered} v=r\omega \\ \Rightarrow v\propto r (\mathrm{for} \mathrm{constant} \omega ) \\ \frac{v}{v\text{'}}=\frac{r}{r\text{'}} \\ \Rightarrow v\text{'}=\frac{v}{2}=10 \mathrm{cm}/\mathrm{s} \end{gathered}$$
Similarly, we have:
$a\text{'}=\frac{a}{2}=10 \mathrm{cm}/{\mathrm{s}}^2$

Question 8:

(c) mg is not greater than $\frac{m{v}^2}{r}$

At the top of the path, the direction of mg is vertically downward and for centrifugal force $\left(\frac{m{v}^2}{r}\right)$, the direction is vertically upward. If the vertically downward force is not greater, water will not fall.

Question 9:

(c) along a tangent

The stone will move in a circle and the direction of velocity at any instant is always along the tangent at that point. Therefore, the stone will move along the tangent to the circle at a point where the string breaks.

Question 10:

(a) 1 cm

Let the force of friction between the coin and the rotating turntable be F.
For the coin to just slip, we have:
$F=m{\omega }^{2}r$
Here, $m{\omega }^{2}r$is the centrifugal force acting on the coin.
For constant F and m, we have:
$r\propto \frac{1}{{\omega }^2}$
Therefore,
$$\begin{gathered} \frac{r\text{'}}{r}={\left({\displaystyle \frac{\omega }{\omega \text{'}}}\right)}^2 \\ \Rightarrow r\text{'}=1 \mathrm{cm} \end{gathered}$$

Question 11:

(a) increases


The normal force on the motorcycle, $N=mg\cos \theta -\frac{m{v}^2}{R}$
As the motorcycle is ascending on the overbridge, θ decreases (from $\frac{\pi }{2}$ to 0).
So, normal force increases with decrease in θ.

Question 12:

(c) F_C is maximum of the three forces.

At the middle of bridge, normal force can be given as:
$N_A=mg$
$$\begin{gathered} N_B=\frac{m{v}^2}{r}-mg \\ {N}_C=\frac{m{v}^2}{r}+mg \end{gathered}$$

So, F_C is maximum.

Question 13:

(a) F₁ > F₂

When the trains are moving, effective angular velocity of both the trains are different (as shown in the figure).



Effective angular velocity of train B is more than that of train A.

Normal force with which both the trains push the tracks is given as:

$N=mg-mR\omega {\text{'}}^2$

From the above equation, we can conclude that F₁ > F₂.

Question 14:

(d) increase at some places and remain the same at some other places

If the Earth stops rotating on its axis, there will be an increase in the value of acceleration due to gravity at the equator. At the same time, there will be no change in the value of g at the poles.

Question 15:

(a) T₁ > T₂



Let the angular velocity of the rod be $\omega$.
Distance of the centre of mass of portion of the rod on the right side of L/4 from the pivoted end:
$r_1=\frac{L}{4}+\frac{1}{2}\left(\frac{3L}{4}\right)=\frac{5L}{8}$
Mass of the rod on the right side of L/4 from the pivoted end:
$m_1=\frac{3}{4}M$
At point L/4, we have:
$$\begin{gathered} T_1=m_1{\omega }^2{r}_1 \\ =\frac{3}{4}M{\omega }^2\frac{5}{8}L=\frac{15}{32}M{\omega }^{2}L \end{gathered}$$

Distance of the centre of mass of rod on the right side of 3L/4 from the pivoted end:
$r_1=\frac{1}{2}\left(\frac{L}{4}\right)+\frac{3L}{4}=\frac{7L}{8}$
Mass of the rod on the right side of L/4 from the pivoted end:
$m_1=\frac{1}{4}M$
At point 3L/4, we have:
$$\begin{gathered} T_2=m_2{\omega }^2{r}_2 \\ =\frac{1}{4}M{\omega }^2\frac{7}{8}L=\frac{7}{32}M{\omega }^{2}L \end{gathered}$$
∴ T₁ > T₂

Question 16:

(d) zero

When the car is in air, the acceleration of bob and car is same. Hence, the tension in the string will be zero.

Question 1:

(c) at the extreme positions

Tension is the string, $T=\frac{m{v}^2}{r}-mg\cos \theta$
When v = 0, $\left|T\right|=mg\cos \theta$
That is, at the extreme positions, the tension is the string is mgcosθ.

Question 2:

(a) speed
(d) magnitude of acceleration

When an object follows a curved path, its direction changes continuously. So, the scalar quantities like speed and magnitude of acceleration may remain constant during the motion.

Question 3:

(d) The instantaneous acceleration of the Earth points towards the Sun.

The speed is constant; therefore, there is no tangential acceleration and the direction of radial acceleration is towards the Sun. So, the instantaneous acceleration of the Earth points towards the Sun.

Question 4:

(b) speed remains constant
(d) tangential acceleration remains constant

If the speed is constant, the position vector of the particle sweeps out equal area in equal time in circular motion.
Also, for constant speed, tangential acceleration is zero, i.e., constant.

Question 5:

(c) The magnitude of acceleration is constant.

As the pitch and radius of the path is constant, it shows that the magnitude of tangential and radial acceleration is also constant.
Hence, the magnitude of total acceleration is constant.

Question 6:

(b) The magnitude of the frictional force on the car is greater than $\frac{m{v}^2}{r}$.
(c) The friction coefficient between the ground and the car is not less than a/g.

If the magnitude of the frictional force on the car is not greater than $\frac{m{v}^2}{r}$, it will not move forward, as its speed (v) is increasing at a rate a.

Question 7:

(b) If the car turns at a speed less than 40 km/hr, it will slip down.
(d) If the car turns at the correct speed of 40 km/hr, the force by the road on the car is greater than mg as well as greater than $\frac{m{v}^2}{r}$.

The friction is zero and the road is banked for a speed v = 40 km/hr. If the car turns at a speed less than 40 km/hr, it will slip down.

Question 1:

(b) There are other forces on the particle.
(d) The resultant of the other forces varies in magnitude as well as in direction.

We cannot move a particle in a circle by just applying a constant force. So, there are other forces on the particle.
As a constant force cannot move a particle in a circle, the resultant of other forces varies in magnitude as well as in direction (because $\overrightarrow{F}$ is constant).

Question 2:

Distance between the Earth and the Moon:
$r=3.85\times {10}^5 \mathrm{km}=3.85\times {10}^8 \mathrm{m}$

Time taken by the Moon to revolve around the Earth:

$$\begin{gathered} T=27.3 \mathrm{days} \\ =24\times 3600\times 27.3 \mathrm{s}=2.36\times {10}^6 \mathrm{s} \\ \mathrm{Velocity} \mathrm{of} \mathrm{the} \mathrm{Moon}: \\ v=\frac{2\pi r}{T} \\ =\frac{2\times 3.14\times 3.85\times {10}^8}{2.36\times {10}^6}=1025.42 \mathrm{m}/\mathrm{s} \\ \mathrm{Acceleration} \mathrm{of} \mathrm{the} \mathrm{Moon}: \\ a=\frac{v^2}{r}=\frac{(1025.42{)}^2}{2.36\times {10}^6}=0.00273 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$
$\Rightarrow a=2.73\times {10}^{-3} \mathrm{m}/{\mathrm{s}}^2$

Question 3:

Diameter of the Earth = 12800 km
So, radius of the Earth, R = 6400 km = 6.4 × 10⁶ m

Time period of revolution of the Earth about its axis:

$$\begin{gathered} T=24 \mathrm{hr}=24\times 3600 \mathrm{s} \\ v=\frac{2\mathrm{\pi }r}{T}=\frac{2\times 3.14\times 64\times {10}^6}{24\times 3600} \end{gathered}$$
^{$\Rightarrow v=465.185 \mathrm{m}/\mathrm{s}$}
$$\begin{gathered} \mathrm{Acceleration} \mathrm{of} \mathrm{the} \mathrm{particle}: \\ \mathrm{a}=\frac{v^{\mathit{2}}}{R}=\frac{{\left(465.185\right)}^2}{64\times {10}^5}=0.038 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Question 4:

Speed is given as a function of time. Therefore, we have:
v = 2t
Radius of the circle = r = 1 cm
At time t = 2 s, we get:
(a) Radial acceleration
$a=\frac{v^2}{r}=\frac{2{}^2}{1}=4 \mathrm{cm}/{\mathrm{s}}^2$
(b) Tangential acceleration
$$\begin{gathered} a=\frac{dv}{dt} \\ =\frac{d}{dt}\left(2t\right) =2 \mathrm{cm}/{\mathrm{s}}^2 \end{gathered}$$
(c) Magnitude of acceleration
$$\begin{gathered} a=\sqrt{4^2+2^2} \\ =\sqrt{20} \mathrm{cm}/{\mathrm{s}}^2 \end{gathered}$$

Question 5:

Given:
Mass = m = 150 kg
Speed = v = 36 km/hr = 10 m/s
Radius of turn = r = 30 m
Let the horizontal force needed to make the turn be F. We have:
$$\begin{gathered} F=\frac{m{v}^2}{r}=\frac{150\times (10{)}^2}{30} \\ =\frac{150\times 100}{30}=500 \mathrm{N} \end{gathered}$$

Question 6:

Given:
Speed of the scooter = v = 36 km/hr = 10 m/s
Radius of turn = r = 30 m
Let the angle of banking be $\theta$. We have:
$\tan \theta =\frac{v^2}{rg}$
$$\begin{gathered} \Rightarrow \tan \theta =\frac{100}{30\times 10}=\frac{1}{3} \\ \Rightarrow \theta ={\tan }^{-1}\left(\frac{1}{3}\right) \end{gathered}$$

Question 7:

Given:
Speed of the vehicle = v = 18 km/h = 5 m/s
Radius of the park = r = 10 m
Let the angle of banking be $\theta$.
Thus, we have:
$\tan \theta =\frac{v^2}{rg}$
$$\begin{gathered} \Rightarrow \theta ={\tan }^{-1}\left(\frac{25}{100}\right) \\ \Rightarrow \theta ={\tan }^{-1}\left(\frac{1}{4}\right) \end{gathered}$$
     

Question 8:

If the road is horizontal (no banking), we have:
$$\begin{gathered} \frac{m{v}^2}{R}=f_s \\ N=mg \end{gathered}$$
Here, f_s is the force of friction and N is the normal reaction.
If μ is the friction coefficient, we have:
$$\begin{gathered} \mathrm{Friction} \mathrm{force} = f_s=\mu N \\ \mathrm{So}, \frac{m{v}^2}{R}=\mu mg \end{gathered}$$

Here,
Velocity = v = 5 m/s
Radius = R = 10 m
$$\begin{gathered} \therefore \frac{25}{10}=\mu g \\ \Rightarrow \mu =\left(\frac{25}{100}\right)=0.25 \end{gathered}$$

Question 9:

Given:
Angle of banking = θ = 30°
Radius = r = 50 m
Assume that the vehicle travels on this road at speed v so that the friction is not used.
We get:
$$\begin{gathered} \tan \theta =\frac{v^2}{rg} \\ \Rightarrow \tan 30°=\frac{v^2}{rg} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{v^2}{rg} \\ \Rightarrow v^2=\frac{rg}{\sqrt{3}}=\frac{50\times 10}{\sqrt{3}} \\ \Rightarrow v=\sqrt{\frac{500}{\sqrt{3}}}=17 \mathrm{m}/\mathrm{s} \end{gathered}$$

Question 10:

$$\begin{gathered} \mathrm{Given}: \\ \mathrm{Radius} \mathrm{of} \mathrm{the} \mathrm{orbit} \mathrm{of} \mathrm{the} \mathrm{ground} \mathrm{state} = r=5.3\times {10}^{-11} \mathrm{m} \\ \mathrm{Mass} \mathrm{of} \mathrm{the} \mathrm{electron} = m=9.1\times {10}^{-31} \mathrm{kg} \\ \mathrm{Charge} \mathrm{of} \mathrm{electron} = q=1.6\times {10}^{-19} \mathrm{c} \\ \mathrm{We} \mathrm{know}: \\ \mathrm{Centripetal} force=\mathrm{Coulomb} attraction \\ \mathrm{Therefore}, \mathrm{we} \mathrm{have}: \\ \Rightarrow \frac{m{v}^2}{r}=\frac{1}{4{\mathrm{\pi \epsilon }}_{\circ }}\frac{q^2}{r^2} \\ \Rightarrow v^2=\frac{1}{4{\mathrm{\pi \epsilon }}_{\circ }}\frac{q^2}{rm} \\ =\frac{9\times {10}^9\times {\left(1.6\times {10}^{-19}\right)}^2}{5.3\times {10}^{-11}\times 9.1\times {10}^{-31}} \\ =\frac{23.04}{48.23}\times {10}^{13} \\ =0.477\times {10}^{13}=4.7\times {10}^{12} \\ \Rightarrow v=\sqrt{4.7}\times {10}^{12}=2.2\times {10}^6 \mathrm{m}/\mathrm{s} \end{gathered}$$

Question 11:

Let m be the mass of the stone.
Let v be the velocity of the stone at the highest point.
R is the radius of the circle.
Thus, in a vertical circle and at the highest point, we have:
$$\begin{gathered} \frac{m{v}^2}{R}=mg \\ \Rightarrow v^2=Rg \\ \Rightarrow v=\sqrt{Rg} \end{gathered}$$

Question 12:

Diameter of the fan = 120 cm
∴ Radius of the fan = r = 60 cm = 0.6 m
Mass of the particle = M = 1 g = 0.001 kg
Frequency of revolutions = n = 1500 rev/min = 25 rev/s
Angular velocity = ω = 2$\pi$n = 2$\pi$ × 25 = 157.14 rev/s
Force of the blade on the particle:
F = Mrw²
   = (0.001) × 0.6 × (157.14)²
   =14.8 N
The moving fan exerts this force on the particle.
The particle also exerts a force of 14.8 N on the blade along its surface.

Question 13:

 $\mathrm{Frequency} \mathrm{of} \mathrm{disc}= n=33\frac{1}{3}\mathrm{rev}/\mathrm{m}=\frac{100}{3\times 60}\mathrm{rev}/\mathrm{s}$
$\mathrm{Angular} \mathrm{velocity} = \mathrm{\omega }=2\mathrm{\pi }n=2\mathrm{\pi }\times \frac{100}{180}=\frac{10\mathrm{\pi }}{9}\mathrm{rad}/\mathrm{s}$
$$\begin{gathered} r=10 \mathrm{cm}=0.1 \mathrm{m} \\ g=10 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$       

It is given that the mosquito is sitting on the L.P. record disc. Therefore, we have:
Friction force ≥ Centrifugal force on the mosquito
⇒ μmg ≥ mrω²
⇒ μ ≥ rω²/g
$$\begin{gathered} \Rightarrow \mu \ge \mathrm{0.1\times }{\left({\displaystyle \frac{\mathrm{10\pi }}{9}}\right)}^2\frac{1}{10} \\ \Rightarrow \mu \ge \frac{{\mathrm{\pi }}^2}{81} \end{gathered}$$

Question 14:

Speed of the car = v = 36 km/hr = 10 m/s
Acceleration due to gravity = g = 10 m/s²

 

Let T be the tension in the string when the pendulum makes an angle θ with the vertical.
From the figure, we get:
$$\begin{gathered} T\sin \theta =\frac{m{v}^2}{r} ...\left(\mathrm{i}\right) \\ T\cos \theta =mg ...\left(\mathrm{ii}\right) \\ \Rightarrow \frac{\sin \theta }{\cos \theta }=\frac{m{v}^2}{rmg} \\ \Rightarrow \tan \theta =\frac{v^2}{rg} \\ \Rightarrow \theta ={\tan }^{-1}\left(\frac{v^2}{rg}\right) \\ ={\tan }^{-1}\left[\frac{100}{(10\times 10)}\right] \\ ={\tan }^{-1}\left(1\right) \\ \Rightarrow \theta =45° \end{gathered}$$

Question 15:

 
Given:
Mass of the bob = m = 100 gm = 0.1 kg
Length of the string = r = 1 m
Speed of bob at the lowest point in its path = 1.4 m/s
Let T be the tension in the string.
From the free body diagram, we get:
$$\begin{gathered} T=mg+\frac{m{v}^2}{r} \\ =\left(\frac{1}{10}\right)\times 9.8+\frac{(1.4{)}^2}{10} \\ =0.98+0.196 \\ =1.176\approx 1.2 \mathrm{N} \end{gathered}$$

Question 16:

Given:
Mass of the bob = m = 0.1 kg
Length of the circle = R = 1 m
Velocity of the bob = v = 1.4 m/s
Let T be the tension in the string when it makes an angle of 0.20 radian with the vertical.



From the free body diagram, we get:
$$\begin{gathered} T-mg\cos \theta =\frac{m{v}^2}{R} \\ T=\frac{m{v}^2}{R}+mg \cos \theta \\ \mathrm{For} \mathrm{small} \mathrm{\theta }, \mathrm{it} \mathrm{is} \mathrm{given} \mathrm{that}: \\ \cos \theta =1-\frac{{\theta }^2}{2} \\ \therefore T=\frac{0.1\times (1.4{)}^2}{1}+(0.1)\times 9.8\left(1-\frac{{\theta }^2}{2}\right) \\ =0.196+0.98\times \left(1-\frac{{\left(0.2\right)}^2}{2}\right) \\ =0.196+0.9604 \\ =1.156 \mathrm{N}\approx 1.16 \mathrm{N} \end{gathered}$$

Question 17:

Let T be the tension in the string at the extreme position.
Velocity of the pendulum is zero at the extreme position.
So, there is no centripetal force on the bob.
∴ T = mgcosθ₀


    

Question 18:

(a) Balance reading = Normal force on the balance by the Earth.
At equator, the normal force (N) on the spring balance: 
N = mg − mω²r

True weight = mg
Therefore, we have:
$$\begin{gathered} \mathrm{Fraction} \mathrm{less} \mathrm{than} \mathrm{the} \mathrm{true} \mathrm{weight}=\frac{mg-(mg-m{\mathrm{\omega }}^{2}r)}{mg} \\ =\frac{{\mathrm{\omega }}^{2}r}{g}={\left(\frac{2\mathrm{\pi }}{24\times 3600}\right)}^2\left(\frac{6.4\times {10}^6}{10}\right) \end{gathered}$$
$=3.5\times {10}^{-3}$

(b) When the balance reading is half, we have:
$\mathrm{True} weight = \frac{mg-m{\mathrm{\omega }}^{2}r}{mg}=\frac{1}{2}$



$$\begin{gathered} \Rightarrow {\mathrm{\omega }}^{2}r=\frac{g}{2} \\ \Rightarrow \mathrm{\omega }=\sqrt{\frac{g}{2r}} \\ =\sqrt{\frac{10}{2\times 6400\times {10}^3}} \mathrm{rad}/\mathrm{s} \end{gathered}$$

$$\begin{gathered} \therefore \mathrm{Duration} \mathrm{of} \mathrm{the} \mathrm{day}=2\pi \times \sqrt{\frac{2\times 6400\times {10}^3}{9.8}}\mathrm{s} \\ =2\mathrm{\pi }\sqrt{\frac{6.4\times {10}^7}{49}}\mathrm{s} \\ =\frac{2\mathrm{\pi }\times 8000}{7\times 3600} \mathrm{h}=2 \mathrm{h} \end{gathered}$$

Question 19:

Given:
Speed of vehicles = v = 36 km/hr = 10 m/s
Radius = r = 20 m
Coefficient of static friction = μ = 0.4
Let the road be banked with an angle $\theta$. We have:
$$\begin{gathered} \theta ={\tan }^{-1}\frac{v^2}{rg} \\ ={\tan }^{-1}\frac{100}{20\times 10} \\ ={\tan }^{-1}\left(\frac{1}{2}\right) \\ \Rightarrow \mathrm{tan\theta }=0.5 \end{gathered}$$



When the car travels at the maximum speed, it slips upward and μN₁ acts downward.
Therefore we have:
$$\begin{gathered} N_1-mg\cos \theta -\frac{m{v}_1^2}{r}\sin \theta =0 ...\left(\mathrm{i}\right) \\ \mu N_1+mg\sin \theta \mathit{-}\frac{m{v}_{\mathit{1}}^{\mathit{2}}}{r}\cos \theta =0 ...\left(\mathrm{ii}\right) \end{gathered}$$

On solving the above equations, we get:
$$\begin{gathered} v_1=\sqrt{rg\frac{\mu +\tan \theta }{1-\mu \tan \theta }} \\ =\sqrt{20\times 10\times \frac{0.9}{0.8}} \\ =15 \mathrm{m}/\mathrm{s}=54 \mathrm{km}/\mathrm{hr} \end{gathered}$$

      

Similarly, for the other case, it can be proved that:
$$\begin{gathered} v_2=\sqrt{rg\frac{\mathrm{tan\theta }-\mathrm{\mu }}{\sqrt{1-\mathrm{\mu } \mathrm{tan\theta }}}} \\ =\sqrt{20\times 10\times \frac{0.1}{1.2}} \\ =4.08 \mathrm{m}/\mathrm{s}=14.7 \mathrm{km}/\mathrm{hr} \end{gathered}$$

Thus, the possible speeds are between 14.7 km/hr and 54 km/hr so that the car neither slips down nor skids up.

Question 20:

R = Radius of the bridge
L = Total length of the over bridge

(a) At the highest point:
Let m be the mass of the motorcycle and v be the required velocity.


$$\begin{gathered} mg=\frac{m{v}^2}{R} \\ \Rightarrow v^2=Rg \\ \Rightarrow v=\sqrt{Rrg} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{b}\right) \mathrm{Given}: \\ v=\left(\frac{1}{\sqrt{2}}\right)\sqrt{Rg} \end{gathered}$$


Suppose it loses contact at B.
$$\begin{gathered} \mathrm{At} \mathrm{point} \mathrm{B}, \mathrm{we} \mathrm{get}: \\ mg\cos \theta =\frac{m{v}^2}{R} \\ \Rightarrow v^2=Rg\cos \theta \\ \mathrm{Putting} \mathrm{the} \mathrm{value} \mathrm{of} v, \mathrm{we} \mathrm{get}: \\ \sqrt{{\left(\frac{Rg}{2}\right)}^2}=Rg\cos \theta \\ \Rightarrow \frac{Rg}{2}=Rg\cos \theta \\ \Rightarrow \cos \theta =\frac{1}{2} \\ \Rightarrow \mathrm{\theta }=60°=\frac{\mathrm{\pi }}{3} \\ \because \theta =\frac{L}{R} \\ \therefore L=R\theta =\frac{\pi R}{3} \end{gathered}$$
So, it will lose contact at a distance $\frac{\pi R}{3}$ from the highest point.
(c) Let the uniform speed on the bridge be v. The chances of losing contact is maximum at the end bridge. We have:

$$\begin{gathered} \alpha =\frac{L}{2R} \\ \mathrm{So},\frac{m{v}^2}{R}=mg\cos \alpha \\ \Rightarrow v=\sqrt{g\mathrm{Rcos}\left(\frac{L}{2R}\right)} \end{gathered}$$

Question 21:

Let v be the speed of the car.
Since the motion is non-uniform, the acceleration has both radial (a_r) and tangential (a_t) components.
$$\begin{gathered} a_r=\frac{v^2}{R} \\ {a}_t=\frac{dv}{dt}=a \\ \mathrm{Resultant} \mathrm{magnitude}=\sqrt{{\left(\frac{v^2}{R}\right)}^2+a^2} \end{gathered}$$



From free body diagram, we have:
$$\begin{gathered} mN=m\sqrt{{\left(\frac{v^2}{R}\right)}^2+a^2} \\ \Rightarrow \mu mg=m\sqrt{{\left(\frac{v^2}{R}\right)}^2+a^2} \\ \Rightarrow {\mu }^2{g}^2=\frac{v^4}{R^2}+a^2 \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{v^4}{R^2}=({\mathrm{\mu }}^2{g}^2-a^2) \\ \Rightarrow v^4=({\mathrm{\mu }}^2{g}^2-a^2)R^2 \\ \Rightarrow v=\left[\right({\mathrm{\mu }}^2{\mathrm{g}}^2-{\mathrm{a}}^2)R^2{]}^{1/4} \end{gathered}$$

Question 22:

(a) Given:
Mass of the block = m
Friction coefficient between the ruler and the block = μ
Let the maximum angular speed be ω₁for which the block does not slip
Now, for the uniform circular motion in the horizontal plane, we have:
$$\begin{gathered} \mu mg=m{{\omega }_1}^{2}L \\ \therefore {\omega }_1=\sqrt{\frac{\mu g}{L}} \end{gathered}$$



(b) Let the block slip at an angular speed ω_2.
For the uniformly accelerated circular motion, we have:
$$\begin{gathered} \mu mg=\sqrt{{\left(m{\omega }_2^{2}L\right)}^2+{\left(mL{\alpha }^2\right)}^2} \\ \Rightarrow {\omega }_2^4+{\alpha }^2=\frac{{\mu }^2{g}^2}{L^2} \end{gathered}$$
$\Rightarrow {\omega }_2={\left[{\left(\frac{\mu g}{L}\right)}^2-{\alpha }^2\right]}^{1/4}$

Question 23:

Given:
Radius of the curves = r = 100 m
Mass of the cycle = m = 100 kg
Velocity = v = 18 km/hr = 5 m/s

$$\begin{gathered} \left(a\right) \mathrm{At} \mathrm{B}, \mathrm{we} \mathrm{have}: \\ mg-\frac{m{v}^2}{r}=N \\ \Rightarrow N=(100\times 10)-\left(100\times \frac{25}{100}\right) \\ =1000-25=975 \mathrm{N} \\ \mathrm{At} \mathrm{D}, \mathrm{we} \mathrm{have}: \\ N=mg+\frac{m{v}^2}{r} \\ =1000+25=1025 \mathrm{N} \end{gathered}$$

(b) At B and D, we have:
Tendency of the cycle to slide is zero.
So, at B and D, frictional force is zero.
At C, we have:
mgsinθ = f
$\Rightarrow 1000\times \left(\frac{1}{\sqrt{2}}\right)=707 \mathrm{N}$
$$\begin{gathered} \left(\mathrm{c}\right) \left(i\right) \mathrm{Before} \mathrm{C}, \\ mg\cos \theta -N=\frac{m{v}^2}{r} \\ \Rightarrow N=mg \cos \theta -\frac{m{v}^2}{r} \\ =707-25=682 \mathrm{N} \\ \left(ii\right) N-mg\cos \theta =\frac{m{v}^2}{r} \\ \Rightarrow \mathrm{N}=\frac{m{v}^2}{r}+mg\cos \theta \\ =25+707=732 \mathrm{N} \end{gathered}$$

(d) To find the minimum coefficient of friction, we have to consider a point where N is minimum or a point just before c .
$$\begin{gathered} \mathrm{Therefore}, \mathrm{we} \mathrm{have}: \\ \mu \mathrm{N}=mg\sin \theta \\ \Rightarrow \mathrm{\mu }\times 682=707 \\ \Rightarrow \mathrm{\mu }=1.037 \end{gathered}$$

Question 24:

Given:
$$\begin{gathered} \mathrm{Frequency} \mathrm{of} \mathrm{rod}=n=20 \mathrm{rev} \mathrm{per} \min \\ \Rightarrow n= \frac{20}{60}=\frac{1}{3}\mathrm{rev}/\mathrm{s} \end{gathered}$$
Therefore, we have:
angular velocity of rod,
$\mathrm{Angular} \mathrm{velocity} \mathrm{of} \mathrm{rod}=\omega =2\pi n=\frac{2\pi }{3}\mathrm{rad}/\mathrm{s}$
Mass of each kid = $m=15 \mathrm{kg}$
Radius = $r=\frac{3}{2}=1.5 \mathrm{m}$



$$\begin{gathered} \therefore \mathrm{Frictional} \mathrm{force}=F=mr{\mathrm{\omega }}^2 \\ \Rightarrow F=15\times (1.5)\times \frac{(2\mathrm{\pi }{)}^2}{9} \\ =5\times (0.5)\times 4{\mathrm{\pi }}^2=10{\mathrm{\pi }}^2 \mathrm{N} \end{gathered}$$
Thus, the force of frictional on one of the kids is 10$\pi$².