System Of Particles And Rotational Motion

Centre of mass: It is defined as a point t which whole the mass of the body is concentrated.

For Sphere, Cylinder, Ring and Cube each of uniform mass density they all have their centre of masses at their geometric centres.

Fig. A ring with centre of mass (CM) at O

There is no need to be Centre of mass always lies inside the body. It may also lies outside the body. For example, In Ring Centre of mass lies outside the body (in space).

Let, Mass of the Hydrogen = m

Mass of the Chlorine = 35.5m

and given that, Separation between atoms = 1.27 Å

= 1.27×10^-10 m

Let centre of mass for the molecule is a ‘x’ Å distance from the Cl atom then distance from Hydrogen atom be (1.27 – x) Å.

(As shown in the above diagram)

For one dimension, centre of mass CM can calculate by,

m

Where, m₁ = mass of the first body,

m₂ = mass of the second body,

x₁ = Distance from centre of mass to the first body,

x₂ = Distance from centre of mass to the second body,

Suppose Centre of mass of the molecule lies at origin (Zero) then,

(∵ It is in 1 dimension)

⇒

⇒ 1.27m – mx + 35.5mx = 0

⇒ 1.27 – x = -35.5x

∴ Å = -0.037 Å = -0.037×10^-10 m

But, negative sign just indicates that CM lies left to chlorine (as assumed in figure) and it lies at a distance of 0.037×10^-10 m.

Unlike External forces, internal forces produce no effect on the motion of bodies which they act. Since, child is running on the trolley, the force generate by him is internal to the (trolley + child) system.

So, the speed of the CM of the (trolley + child) has no effect even though the child is running.

Consider two vectors (OK) and (OM) are making angle θ which each other as shown in following figure,

Now,

In ΔOMN we can write the equation,

⇒

We know that magnitude of cross product of vectors a and b is,

= (∵)

= ×OK×MN (∵ = 1)

= 2×Area of ΔOMK

∴ Area of ΔOMK =

Hence Proved.

Consider a parallelepiped with origin O and sides a, b, and C as shown in the below figure,

If is unit vector which is perpendicular to both vectors b and c then direction of a and be same.

Therefore, cross product

= bc (∵ θ = 90⁰)

Since a is in same direction as n,

∴ a.( ) = abc

= abc (∵ angle between bc and a is 0⁰)

Hence, a.( ) = Volume of the parallelepiped

From the given data,

Linear momentum vector,

Position vector,

We have, Angular momentum,

=

=

∴

By comparing respective components, we get,

Since, the particle moves in x-y plane, then the vectors of both position and linear momentum vectors be Zero.

Thus,

Hence, the particle moves in x-y plane the angular momentum acts towards z-direction.

Let at a point of time particles P and Q are in some position (exactly collinear, perpendicular to paths) as shown in figure below,

Angular momentum, I = mvr

Where, m = mass of the particle,

v = velocity of the particle,

r = distance from rotating point.

Thus,

Angular momentum about P, I_P = mv×0 + mv×d = mvd …..(1)

Angular momentum about Q, I_Q = mv×d + mv×0 = mvd …..(2)

If the rotating point is R as shown in figure above,

I_R = [mv× (d-y)] + mv×y

= mvd …………(3)

From equations 1, 2 and 3 we can conclude that angular momentum of system doesn’t depend on the point at which it is taken.

Given,

Length of the bar, l = 2 m

Let T₁ and T₂ be the tensions produced by the strings.

The free body diagram can be drawn as,

For transitional equilibrium we have,

⇒

∴

For rotational equilibrium, on taking the torque about the centre of gravity. We have,

⇒ T₁×0.8×d = T₂×0.6×(2-d)

⇒

⇒ 1.067d + 0.6d = 1.2

∴ m = 0.72 m

Hence, Centre of gravity lies at a distance of 0.72 m from the left side of the bar.

Given,

Weight of the car, W = (1800×g) N = 1800×9.8 = 17640 N

Distance between front and back wheels, d = 1.8 m

Distance between centre of gravity from front axle = 1.05 m

The free body diagram of car can be drawn as,

From the figure,

For transitional equilibrium,

R_f + R_b = W

R_f + R_b = 17640 N ……..(1)

For rotational equilibrium, on taking torque about centre of gravity as,

R_b = 1.4R_f ……..(2)

By solving equations 1 and 2 we get,

1.4R_f + R_f = 17640 N

∴

R_b = 17640 – 7350 = 10290 N

Hence, The force exerted on each of the front wheel =

The force exerted on each of the rear wheel =

(A)The moment of inertia (M.I) of sphere about its diameter = kg m⁴

According to the parallel axis theorem, the moment of inertia of a body about an axis is equal to the sum of the moment of inertia of the about a parallel axis passing through its center of mass and product of square of distance between axis of its mass.

The M.I about tangent of the sphere


kg m⁴

(B)The moment of inertia of a disc about its diameter = kg m⁴

According to the perpendicular axis theorem, the moment of inertia of a planar body (lamina) about an axis perpendicular to its plane is equal to the sum of its moments of inertia about its planar axis in which the body lies.

The M.I of the disc about its center:-

The situation can be shown as,

Applying the parallel axis theorem,

The moment of inertia about an axis normal to the disc and passing through a point on its edg

Given that Torque acting on the both the bodies are same.

We know Torque on a body,

τ = Iα

Where

I is moment if inertia of the body

α is angular acceleration of the body

The body with greater angular acceleration will acquire greater angular speed in a given time.

Both hollow cylinder and solid sphere have the same mass and radius (let them be m and R respectively).

Moment of inertia of hollow cylinder, I_cyl = mR²

Moment of inertia of solid sphere, I_sph = 2mR²/5

Let, α_cyl and α_sph be the angular acceleration of hollow cylinder and solid sphere respectively on action of torque.

Since the torque acting on the bodies are same,

I_cyl α_cyl = I_Sph α_Sph

But I_sph = 2I_cyl/5

∴ α_cyl = 2α_sph/5

That is Solid sphere has greater angular acceleration than hollow cylinder when same torque acts on both bodies. Therefore, Solid sphere will attain greater angular speed than hollow cylinder in a given time.

We need to find Kinetic energy of the rotating cylinder

We know that, Rotational kinetic energy,

KE = 1/2I ω²

Where

I is moment of inertia

ω is angular velocity

Given

Mass of cylinder, M = 20 kg

Radius of cylinder, R = .25 m = 1/4 m

Angular speed of cylinder, ω = 100 rad s^-1

Moment of inertia of a solid cylinder, I = 1/2 MR²

I.e. I = 1/2 × 20 × (1/4)²

= 5/8 kg m²

∴ KE = 1/2 I ω²

= 5/16 (100)²

= 3125 J

= 3.125 kJ

We need to find angular momentum of rotating cylinder about its axis.

Angular momentum, L = I ω

We know that,

Moment of inertia of cylinder, I = 5/8 kg m²

Angular speed of cylinder, ω =100 rad s^-1

∴ Angular momentum, L = 5/8 × 100

= 62.5 kg m s^-1

= 62.5 J s

A

Given

Angular speed of child, ω = 40 rev min^-1

Angular momentum of child = I ω

Where I is moment of inertia of child

ω is angular velocity of child

When the child folds his hand his moment of inertia decreases, but angular momentum is conserved. Therefore, angular speed increases. We need to find the new angular speed (ω’)

Let new moment of inertia be I’

And new angular velocity be ω’

Given, I’ = 2/5×I

Since Angular momentum is conserved,

I ω = I’ ω’

ω’ = I ω / I’

= 5/2 × ω

= 5/2 × 40

= 100 rev min^-1

= 5/3 rev s^-1

=1.66 rev s^-1

B

Kinetic energy of rotation,

KE_rot = 1/2 I ω²

Let moment of inertia of child before folding hands be I

Given, angular velocity before folding hand, ω = 100 rev min^-1

∴ KE = 1/2 Iω²

= 1/2 I (40)²

= 800I

Let moment of inertia after folding hands be I’, Then,

I’ = 2/5 I

Angular speed after folding hands, ω’ = 10π /3

∴ KE’ = 1/2 I’ω’²

= 1/2 (2I/5) × (100)²

= 2000I

KE’/KE = 2000I/ 800I

= 5/2

i.e. KE’ is greater than KE. When the child folded the hands, Kinetic energy increased

We need to find angular acceleration of cylinder when force was applied tangential to hollow cylinder (by pulling rope wound in it).

Torque produced by action of force on a body,

τ = F r

Where

F is force applied on body

r is perpendicular distance of point of application of force with axis of rotation.

Given

F = 30 N

r = 40 Cm = .4 m

∴ τ = 30 × .4

= 12 Nm

On action of this torque (or tangential force), body gains an angular acceleration, say, α. In terms of α,

Torque, τ = I α

Where

I is moment of inertia of body

α is angular acceleration

Moment of inertia of hollow cylinder,

I = Mr²

Where

M is mass of cylinder

r is radius of cylinder

Given

M = 3 kg

r = 40 Cm = .4 m

∴ I = 3 × (.4)²

= .48 kg m²

τ = I α = 12 Nm

I = .48 kg m²

∴ Angular acceleration, α = τ /I

= 12/.48

= 25 rad s^-2

Linear acceleration on point P,

a = r α

Where

α is angular acceleration

r is perpendicular distance of point P from axis of rotation

We need to find linear acceleration of rope, which is at distance of .4 m from axis of rotation,

Thus, r = .4 m

α = 25 s^-2

∴ Linear acceleration, a = r α

= .4 × 25

= 10m s^-2

The power needed,

P = τ × ω

Where

τ is torque needed to counter friction

ω is angular speed to be maintained

Given,

Torque needed, τ = 180 N m

Angular speed, ω = 200 rad s^-1

∴ power required to maintain rotation of 200 rad s^-1 by countering frictional torque of 180 N m,

P = 180 × 200

= 36 kW

From a big uniform disc of radius R with centre O, a smaller circular hole of radius R/2 with its centre O₁ (where OO₁ = R/2) is cut out.

Let the centre of gravity of remaining flat body be C and OC= d.

If Ω is mass per unit area, then mass of the whole disk of Radius R is,

M₁ = πR²Ω

and mass cut out part

M₂ = π(R/2)² Ω = πR²Ω/4 = M₁/4

Let centre of mass of original disc be origin(O)

It can be considered as centre of mass of system consisting of smaller carved out disc and remaining portion.

Thus, we have,

M₁ × (0) = (M₁ -M₂) × d + M₂ × (OO₁)

0 = (M₁ – M₂) × d + M₂ × (- R/2)

(Since, M₂ = M₁/4)

d = (M₂R/2)/(M₁-M₂)

= (M₁/4) × (R/2) / (M₁ - (M₁/4))

= R/6

∴ Cis at a distance R/6 from the centre of disc on diametrically opposite side of the centre of hole.

Let m be the mass of the scale. The centre of mass of scale is at 50 cm. thus scale is balanced at this point.

When 2 coins of mass 5 g are placed at 12 cm, the equilibrium has shifted to 45 cm mark.

Centre of mass of coins are at 12 cm mark and centre of mass of meter stick is at 50 cm mark.

For equilibrium about 45 cm mark,

W_coin × d_coin = W_{meter stick} × d_{meter stick}

Where

W_coin is weight of coin which is 10×g

W_{meter scale} is weight of meter stick which is mg

d_coin is distance of centre of mass of coin from equilibrium

d_{meter scale} is distance of centre of mass of meter scale from equilibrium

Thus,

10 g (45 - 12) = m g (50 - 45)

10 g × 33 = mg × 5

m = 10 × 33 /5

m = 66 grams

I.e. The mass of meter stick is 66 grams

Both spheres are identical.

∴ Total energy,

TE = KE + PE

Where

KE is kinetic energy

PE is potential energy

Since bodies are starting from rest, there initial KE is zero,

TE_i = PE_i = mgh

Where

m is mass of the bodies

g is acceleration due to gravity

h is height of inclined plane

On reaching bottom, the potential energy drops to zero (PE_f = 0)

Therefore, total energy of bodies at bottom are KE of bodies.

TE_f = KE_f = 1/2 mv_f²+ 1/2 I ω²

Where

m is mass of body

v_f is velocity attained by body

ω is angular velocity

I is moment of inertia

By Law of conservation of energy, we have,

TE_i = TE_f

I.e. mgh = 1/2 mv_f²+ 1/2 I ω²

= 1/2 mv_f² + 1/2 (2mR²/5) ω²

V = Rω

∴ mgh = 1/2 mv_f²+ mv_f²/5

v_f = (10gh/7)^1/2

The final velocity is independent of angle of inclination. Therefore

both bodies will have same final speed on reaching ground.

B & C

When a body is moving on an inclined plane of inclination θ, acceleration acting on body is g (acceleration due to gravity vertically downward)

The acceleration along inclined plane is g sin(θ) component of acceleration.

Let height from which body fall be H.

Then length of inclined plane of inclination θ,

L = H/sin(θ)

We know the equation,

s = u t + 1/2 at²

Where

u is initial velocity

s is distance covered

a is acceleration acting on body

t is time

For case of motion along plane,

u = 0 (since body starts from rest)

s = L = H/sin(θ)

a = g sin(θ)

t is time

Substituting the values,

H/sin(θ) = 0× t + 1/2 g sin(θ) t²

∴ time for reaching ground,

t = (2H/g)^1/2 / sin(θ)

From above equation, we can infer that time taken by a body to roll down an inclined plane is inversely proportional to sin of angle of inclination (θ).

Sine increases with increase of angle, thus time for fall decreases for greater inclination (t ∝ 1/sin(θ))

Thus greater the angle of inclination, lower the time needed for reaching ground.

What we want to find is work done. We know that the work done to stop a moving body is the total kinetic energy of the body.

∴ Work Done = KE = KE_rot + KE_tran

Where KE_rot is rotational kinetic energy and

KE_tran is translational kinetic energy

KE_rot = 1/2 Iω²

KE_tran = 1/2 Mv²

Where M is mass of hoop

I is moment of inertia of hoop

v is velocity of centre of mass

ω is angular velocity

Now the given values are,

Mass of hoop, M = 100 Kg

Radius of hoop, R = 2 m

Speed of centre of mass of hoop, v = 20 Cm s^-1

= .2 m s^-1

The axis of rotation of hoop passes through point B, we know that velocity of point O (centre of loop) is v. If ω is angular velocity of hoop then,

v = R ω (R is Distance between axis and centre(O))

Where ω is angular velocity.

i.e. ω = v/R = 0.2/2 = .1 s^-1

Moment of inertia of hoop, I = MR²

Where, M is mass of hoop

R is Radius of hoop

∴ I = 100(2)²

I = 400 Kg m²

Total Kinetic Energy of hoop,

KE = KE_rot + KE_tran

KE_rot = 1/2 400 × (.1)²

= 2 J

KE_tran = 1/2 100 × (.2)²

= 2 J

∴ Total Kinetic energy,

KE = 2 + 2 = 4 J

Work done to stop the moving hoop is equal to its KE which is 4 J

We need to find average angular velocity, ω of oxygen molecule.

We know that,

V = r ω

Given

Mass of oxygen molecule, M = 5.30 × 10^-26 kg

Moment of inertia of oxygen molecule, I = 1.94 × 10^-46 kg m²

Mean speed of molecule, V = 500 m s^-1

To find ω we need value of r (distance between axis of rotation and oxygen atom in the molecule)

The equation for moment of inertia of an oxygen molecule (system consisting of 2 equal masses (m) separated at a distance of r from axis),

I = mr² + mr²

= 2mr²

But m is mass of an oxygen atom,

∴ m = M/2

Thus, I = 2(M/2) r²

= Mr²

i.e. r = (I/M)²

= [ (1.94 × 10^-46)/5.30 × 10²] 1/2

= .606 × 10^-10

KE_rot = 2KE_tran/3

1/2 I ω² = 2/3 × (m v²/2)

1/2 m r² ω = 2/3 × mv²/2

ω = (2/3)^1/2× v/r

we know,

v = 500 ms^-1

r = .606 × 10^-10 m

Thus, ω = (2/3)^1/2 × 500/ (.606 × 10^-10)

= 6.68 × 10¹² rad s^-1

Average angular velocity of molecule is 6.68 ×10¹² rad s^-1

At bottom of inclined plane centre of mass has speed 5 m s^-1 let this be v.

We need to find how far the solid cylinder moves up on the plane.

By law of conservation of energy, energy is conserved.

Initial energy of body is the completely Kinetic (at bottom), On moving up the plane its kinetic energy gradually decreases. At top most point, Kinetic energy is zero and energy is completely gravitational potential energy.

The axis of rotation of cylinder passes through point B, we know that velocity of point O (axis of cylinder) is v. If ω is angular velocity of cylinder then,

v = R ω

(R is Distance between axis of rotation and axis of cylinder)

Initial KE,

KE_i = KE_r + KE_t

Where

KE_r is rotational kinetic energy

KE_t is translational kinetic energy

KE_t = 1/2 mv²

KE_r = 1/2 I ω²

= 1/2 (mR²/2) ω² = 1/4 mv² (since v = R ω)

∴ Total kinetic energy, KE_i = 1/4 mv² + 1/2 mv²

=3mv²/4

Let h be the maximum height attained.

Then potential energy at that point,

PE_f = mgh

By law of conservation of energy,

KE_i = PE_f

3mv²/4 = mgh

h = 3v²/4g

We know,

V=5 ms^-1

g=9.8 ms^-2

Thus,

h = 3(5)²/4× 9.8

= 1.91 m

Thus, maximum height the body would reach is 1.91 m from ground.

By law of conservation of energy,

3mv²/4 = mgh

v = (4gh/3)^1/2

Let h be maximum height attained and θ be angle of inclination.

Then, Length travelled along plane to reach maxim um height,

L = h/sin(θ)

When v is the velocity, the times taken to reach maximum height,

T = L/v

= h/sin(θ) ÷ (4gh/3)^1/2

= (3h/4g)^1/2 ÷ sin(θ)

= (3 × 1.91/ 4 × 9.8)^1/2 / sin (30)

= 0.765 s

the time taken for the body to move up and reach the bottom twice this time that is,

t = 2 × 0.765= 1.53 s

The diagram is shown below:

Given:

AB = AC = 1.6 m

DE = 0.5 m

BF = 1.2 m

Mass of the weight, m = 40 Kg

N_B = Normal reaction at point B

N_C = Normal reaction at point C

T = tension in the rope

Let us drop a perpendicular from A to imaginary line BC. This should bisect the line DE (ABC is Isosceles triangle). Let the point of intersection be H.

Let the point of intersection of perpendicular and BC be I.

ΔABI is similar to Δ AIC

So, BI = CI

Hence, I is the mid-point of BC.

⇒ DE||BC

⇒ BC = 2 × DE = 1m

And, AF = AB-BF = 0.4 m …(i)

Also, D is the mid-point of AB

So, we can write,

AD = 0.5 × BA

⇒ AD = 0.8 m …(ii)

From (i) and (ii) we conclude that,

FE = 0.4 m

Hence, F is the mid-point of AD.

Now, FG||DH and F is the mid-point of AH.

∴ G is the midpoint of AH.

⇒ ΔAFG and ΔADH are similar,

So

FG/DH = AF/AD

⇒ FG/DH = 0.4 m/0.8 m

⇒ FG/DH = 0.5

⇒ FG = 0.125 m

Now, In ΔADH

AH = (AD²- DH²)^1/2

⇒ AH = ((0.8m)²-(0.25m)²)^1/2

⇒ AH = 0.76 m

For static equilibrium of ladder, the upward force should be equal to the downward force.

⇒ N_c + N_B = mg = 392 N …(iii)

For rotational equilibrium,

The net moment of all forces about A should be Zero,

⇒ -N_B × BI + mg × FG + N_C × CI + T × AG = 0

⇒ -N_B × 0.5m + 40kg × 9.8ms^-2 × 0.125m + N_C × 0.5m = 0

(N_C - N_B) × 0.5 = 49

N_C - N_B = 98 N …(iv)

Adding equations (iii) and (iv), we get,

N_C = 245 N

N_B = 147 N

For rotational equilibrium of the side AB, consider the moment about A,

⇒ -N_B × BI + mg × FG + T × AG = 0

⇒ -245 N × 0.5 m + 40Kg × 9.8 ms^-2 × 0.125m + T × 0.76m = 0

⇒ T = 96.7 N

Given:

(a) Moment of Inertia of the man-platform system = 7.6 Kgm²

Moment of inertia when the man stretches his hands to a distance of 90 cm.

We know, I = 2 × mr²

Where,

m = mass

r = distance of mass from axis of rotation

Using the above formula, we get,

⇒ I = 2 × 5Kg × (0.9m)²

⇒ I = 8.1 Kgm²

Now, the Initial moment of inertia of the system,

I_i = 7.6 + 8.1 = 15.7 Kgm²

Angular speed, ω_i = 30 rev/min

Angular momentum, L_i = I_iω_i

L_i = 15.7Kgm² × 30rev/min

⇒ L_i = 471 Kgm²/sec …(i)

Moment of inertia when the man folds his hands to a distance of 20 cm,

I = 2 × mr²

Where,

m = mass

r = distance of mass from axis of rotation

⇒ I = 2 × 5Kg × (0.2m)²

⇒ I = 0.4 Kgm²

Final moment of inertia, I_f = 7.6 + 0.4 = 8 Kgm²

Final angular speed = ω_f

Final angular momentum = L_f = I_fω_f

L_f = 0.79 ω_f …(ii)

From the conservation of angular momentum, We have

I_iω_i = I_fω_f

From equation (i) and (ii),

ω_f = 15.7 Kg m² × ((30rev/sec)/8 Kgm²)

ω_f = 58.88 rev/min

(b) Kinetic energy is not conserved in the given process. In fact, with the decrease in moment of inertia, kinetic energy increases. The additional kinetic energy comes from the work done by the man to fold his hands towards himself.

Given:

Mass of Bullet, m = 10g = 10^-3 kg

Speed of bullet, v = 500m/s

Width of the door, w = 1m

Distance from hinge where bullet hits, r = 0.5 m

Mass of door, M = 12 Kg

We know that angular moment is given by,

α = mvr

where,

α = angular momentum

m = mass of the body

v = velocity of the body

r = radius of the body

by putting the value, we get,

⇒ α = 10 × 10^-3 kg × 500 ms^-1 × 0.5 m

⇒ α = 2.5 Kgm²s^-1 …(i)

Moment of inertia is given by,

I = ML²/3 (For rectangle)

Where,

M = mass of door

L = width of door

I = 1/3 × 12 Kg × (1m)²

I = 4 Kgm²

We also have,

α = Iω

where is α is the angular momentum

l is the linear momentum

ω is the angular velocity

⇒ ω = α /I

By putting the values, we get,

⇒ ω = 2.5/ 4

⇒ ω = 0.625 rad s^-1

Note: It is a good practice to memorise the moment of inertia for common shapes like rectangle, circle, triangle, cylinder and sphere.

Given:

Moment of inertia of disc 1 = I₁

Angular speed of disc 1 = ω₁

∴ Angular momentum of disc 1 , L₁ = I₁ω₁

Angular speed of disc 2 = ω₂

∴ Angular momentum of disc 2, L₂ = I₂ω₂

⇒ Total initial angular momentum, L_i = I₁ω₁ + I₂ω₂

When the two discs are joined together, their moments of inertia get added up.

Moment of inertia of the system of two discs, I = I₁ + I₂

Let ω be the angular speed of the system.

Let total final angular momentum, L_f = (I₁ + I₂) × ω_f

Using the law of conservation of angular momentum, we have:

L_i = L_f

⇒ I₁ω₁ + I₂ω₂ = (I₁ + I₂) × ω_f

⇒ ω_f = (I₁ω₁ + I₂ω₂) / (I₁ + I₂) …(i)

(b) Kinetic energy of disc I, E₁ = 1/2 I₁ω₁²

Kinetic energy of disc II, E₂ = 1/2 I₂ω₂²

⇒ Total initial kinetic energy, E_i = 1/2 I₁ω₁² + 1/2 I₂ω₂² …(ii)

When the discs are joined, their moments of inertia get added up.

Moment of inertia of the system, I = I₁ + I₂

Angular speed of the system = ω_f

Final kinetic energy,

E_f = 1/2 (I₁ + I₂)ω_f²

Substituting the value of ω from equation (i)

⇒ E_f = 1/2 (I₁ + I₂) {(I₁ω₁ + I₂ω₂) / (I₁ + I₂)}² …(iii)

From (ii) and (iii)

E_i- E_f = [1/2 I₁ω₁² + 1/2 I₂ω₂²] - 1/2 (I₁ + I₂) {(I₁ω₁ + I₂ω₂) / (I₁ + I₂)}²

By solving the equation we get,

E_i- E_f = I₁I₂(ω₁-ω₂)²/ 2(I₁ + I₂)

Since all Quantities on RHS are positive so we conclude that,

E_i-E_f> 0

i.e. E_i> E_f

The loss can be attributed to the frictional forces that arise when discs come into contact with each other.

Given:

(a)The theorem of perpendicular axes states that the moment of inertia of a planar body (lamina) about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes concurrent with perpendicular axis and lying in the plane of the body.

A physical body with center O and a point mass m, in the x–y plane at (x, y) is shown in the following figure.

Moment of inertia about x-axis, I_x = mx²

Moment of inertia about y-axis, I_y = my²

Moment of inertia about z-axis, _z

I_x + I_y = mx² + my²

⇒ I_x + I_y = m(x² + y²)

We can write the following equation as,

⇒ I_x + I_y = m{(x² + y²)^1/2}²

⇒ I_x + I_y= I_z

Hence, the theorem is proved.

(b)The theorem of parallel axes states that the moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its center of mass and the product of it ass and the square of the distance between the two parallel axes.

Suppose a rigid body is made up of n particles, having massesm₁, m₂, m₃ …m_n , at perpendicular distances r₁, r_{2 �,} r₃, … r_n respectively from the center of mass O of the rigid body.

The moment of inertia about axis RS passing through the point O:

The perpendicular distance of mass mi, from the axis QP = a + r_i

Hence, the moment of inertia about axis QP:

⇒

⇒

⇒

Now, at the center of mass, the moment of inertia of all the particles about the axis passing through the center of mass is zero, that is,

∵ a≠0

⇒ ∑m_ir_i = 0

Also,

∑m_i = M

Where,

M = Total mass of the rigid body

I_QP = I_RS + Ma²

Hence, the theorem is proved.

Given:

A body rolling on an inclined plane of height h, is shown in the following figure:

Mass of the body = m

Radius of the body = R

Radius of gyration of the body = K

Translational velocity of the body = v

Height of the inclined plane = h

Acceleration due to gravity = g = 9.8 ms^-2

Total energy at the top of the plane, E₁ = mgh

Total energy at the bottom of the plane, E_T = KE_Rotational + KE_{Translational}

E_T = 1/2 Iω ² + 1/2 mv² …(i)

We know that ,

I = mK²

ω = v/R

Using in equation (i)

E_T = 1/2 mK²(v/R) ² + 1/2 mv²

E_T = 1/2 mv² (1 + K²/R²)

We know that total energy at the top will be equal to total energy at the bottom, by law of conservation of energy, so we can write,

E₁ = E_T

mgh = 1/2 mv² (1 + K²/R²)

⇒ v = 2gh/(1 + K²/R²)

Hence proved.

Velocity of a point on a rolling disc is given by,

"V= dω"

Where,

d = Distance from center of rotation

ω = Angular velocity of the body

Now, Velocity at points A and B are same and equal to Rω₀

(∵ distance between points A& B and centre of rotation = R)

⇒ v_A = v_B = Rw₀

Now,

Velocity at point C,

(∵ distance between point C and centre of rotation = )

No, Disc will rotate in the shown direction below.

For rotation of the disc torque is required. If friction is absent at B due to tangential force (Torque) at A disc will rotate in the same place but won’t move forward. So, friction is necessary to make the disc to move forward.

A. Friction opposes motion. So, frictional force is opposite to the direction of linear velocity at point B, in rightward direction. Sense of frictional torque, before perfect rolling begins is perpendicular to plane of disc in the outward direction.

B. Perfect rolling means the linear velocity at point B is zero. So, frictional force is also zero.

Given,

Radius of the disc and ring, r = 10 cm = 0.1 m

Angular velocity of the disc and ring, ω₀ = 10 π rad/s

Initial velocity of both the objects, u = 0 m/s

Coefficient of kinetic friction, μ_k = 0.2

Motion of the objects start due to frictional force,

f = ma

Where f = frictional force ,

⇒ f = μ_kmg

m = mass of the body

a = acceleration of the body

∴ μ_kmg = ma

⇒ a = μ_kg ------ > (1)

We have final velocity as per first equation of motion,

v = u + at

Where,

v = final velocity

u = initial velocity

a = acceleration

t = time

∴ v = μ_kg ------- > (2)

The torque generated by friction initially causes reduction in angular velocity. Thus,

T = -Iα

Where,

I = moment of inertia of the body

α = Angular acceleration

and, T = fr

Where,

f = frictional force

r = radius

∴ T = -μ_kmgr

⇒ ------- > (3)

As per the frist equation of rotational motion, final angular velocity,

ω = ω₀ + αt

Where,

ω₀ = intial angular velocity

α = angular acceleration

t = time

∴ ------ > (4)

Rolling starts when linear velocity, v = rω

∴ -------- > (5)

From the equations 2 & 5 we get,

⇒ ------- > (6)

We know that,

For the ring, I = mr²

∴

⇒ μ_kgt = rω₀ – μ_kgt

⇒ 2μ_kgt = rω₀

∴

⇒ t_ring

⇒ t_ring = 0.8s

For the disc, I = 0.5mr²

∴

⇒

⇒ 3μ_kgt = rω₀

∴

⇒ t_disc

Since the t_disc < t_ring.

Therefore, the disc will start rolling before the ring.

Given,

Mass of the cylinder, m = 10 kg

Radius of the cylinder, r = 15 cm = 0.15 m

Inclination angle, θ = 30⁰

Coefficient of static friction, μ_s = 0.25

The moment of inertia of the cylinder about its geometric axis is given by, I = 0.5mr²

The free diagram of the body is given by,

We have,

A. From the newton second law of motion, force net,

F_net = ma

Where,

m = mass of the body

a = acceleration of the body

∴ mg sinθ – f = ma

⇒ f = ma – mg sinθ

= (10×3.27)-(10×9.81×sin30)

= 16.3 N

B. During rolling the instantaneous point of contact have zero velocity. Thus, work done against frictional force is zero.

C. For rolling without skid, we have,

⇒ 3μ = tan θ

(a) False

Frictional force acts opposite to the direction of motion of the center of mass of a body. In the case of rolling, the direction of motion of the center of mass is backward. Hence, frictional force acts in the forward direction.

(b) True

Rolling can be considered as the rotation of a body about an axis passing through the point of contact of the body with the ground. Hence, its instantaneous speed is zero.

(c) False

When a body is rolling, its instantaneous acceleration is not equal to zero. It has some value.

(d) True

When perfect rolling begins, the frictional force acting at the lowermost point becomes zero. Hence, the work done against friction is also zero.

(e) True

The rolling of a body occurs when a frictional force acts between the body and the surface. This frictional force provides the torque necessary for rolling. In the absence of a frictional force, the body slips from the inclined plane under the effect of its own weight.

A. Take a system of I moving particles:

Mass of the i^th particle = m_i

Velocity of the i^th particle = v_i

∴ momentum of the i^th particle, p_i = m_iv_i

Velocity of the canter of the mass = V

Thus, velocity of the i^th particle with respect to canter of the mass, v_i’ = v_i – V …..(1)

Multiplying m_i throughout the equation (1) we get,

⇒ m_iv_i’ = m_iv_i – m_iV

⇒ p_i’ = p_i - m_iV

Where,

p_i’ = momentum of the i^th particle with respect to the canter of the mass.

Taking the summation of the momentums of the all the particle with respect to the canter the mass of the body,

…..(2)

Where,

r_i’ = position of the i^th particle with respect to the canter of mass and,

As per the definition of the canter of the mass, we have,

∴ from equation 2,

B. We have the relation for velocity of the particle as,

…..(3)

Taking the dot product with itself will give,

Here, For the canter of mass of the system of the particle,

Therefore,

C. Total angular momentum of the system of the particles,

The las two terms vanish for both contain the factor which is equal to zero from the definition of the center of the mass. Also,

So that,

D. From the previous solution,

(∵ ) Total toque,