NCERT Solutions for Class 11th: Ch 7 System Of Particles And Rotational Motion Physics Science

Page No: 178

**Excercises**

7.1. Give the location of the centre of mass of a (i) sphere, (ii)

cylinder, (iii) ring, and (iv) cube, each of uniform mass density.

Does the centre of mass of a body necessarily lie inside the body?

**Answer**

In all the four cases, as the mass density is uniform, centreof mass is located at their respective geometrical centres.

No, it is not necessary that the centre of mass of a body should lie on the body. For example, in case of a circular ring, centre of mass is at the centre of the ring, where there is no mass.

7.2. In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 Å (1 Å = 10^{–10}

m). Find the approximate location of the CM of the molecule, given that

a chlorine atom is about 35.5 times as massive as a hydrogen atom and

nearly all the mass of an atom is concentrated in its nucleus.

**Answer**

Mass of H atom = *m*

Mass of Cl atom = 35.5*m*

Let the centre of mass of the system lie at a distance *x*

from the Cl atom.

Distance of the centre of mass from the H atom = (1.27 – *x*)

Let us assume that the centre of mass of the given molecule lies

at the origin. Therefore, we can have:

[ *m*(1.27 – *x*) + 35.5*mx* ] **/** (*m* + 35.5*m*) = 0*m*(1.27 – *x*) + 35.5*mx* = 0

1.27 – *x* = -35.5*x*

∴ *x* = -1.27 / (35.5 – 1) = -0.037 Å

Here, the negative sign indicates that the centre of mass lies at

the left of the molecule. Hence, the centre of mass of the HCl

molecule lies 0.037Å from the Cl atom.

7.3. A child sits stationary at one end of a long trolley moving

uniformly with a speed *V *on a smooth horizontal floor. If the

child gets up and runs about on the trolley in any manner, what is

the speed of the CM of the (trolley + child) system?

**Answer**

The child is running arbitrarily on a trolley moving with velocity*v*. However, the running of the child will produce no effect on

the velocity of the centre of mass of the trolley. This is because

the force due to the boy’s motion is purely internal. Internal

forces produce no effect on the motion of the bodies on which they

act. Since no external force is involved in the boy–trolley

system, the boy’s motion will produce no change in the velocity

of the centre of mass of the trolley.

7.4. Show that the area of the triangle contained between the vectors **a** and **b** is one half of the magnitude of **a** × **b**.

**Answer**

Consider two vectors OK = vector |a| and OM = vector |b|, inclined at an angle *θ*, as shown in the following figure.

**a**. (

**b**×

**c**) is equal in magnitude to the volume of the parallelepiped formed on the three vectors,

**a**,

**b**and

**c**.

**Answer**

A parallelepiped with origin O and sides a, b, and c is shown in the following figure.

7.6. Find the components along the x, y, z axes of the angular momentum **l** of a particle, whose position vector is **r** with components x, y, z and momentum is **p** with components px, py and pz. Show that if the particle moves only in the x-yplane the angular momentum has only a z-component.

**Answer**

7.7. Two particles, each of mass *m* and speed *v*, travel in opposite directions along parallel lines separated by a distance d. Show that the vector angular momentum of the two particle system is the same whatever be the point about which the angular momentum is taken.

**Answer**

Let at a certain instant two particles be at points P and Q, as shown in the following figure.

momentum of the system about point P:

*L*_{p} = *mv* × 0 + *mv* × *d* = *mvd* …**(i)**

Angular

momentum of the system about point Q:*L*_{Q} = *mv* × *d* + *mv* × 0 = *mvd* ….**(ii)**

Consider

a point R, which is at a distance *y*

from point Q, i.e.,

QR =*y*

∴ PR

= *d – y*

Angular

momentum of the system about point R:*L*_{R} = *mv* × (*d* – *y*) + *mv* ×* y**mvd* – *mvy* + *mvy*

= *mvd* ….**(iii)**

Comparing equations**(i)**, **(ii)**, and **(iii)**, we get:*L*_{P} = *L*_{Q} = *L*_{R} …**(iv)**

We

infer from equation **(iv)**

that the angular momentum of a system does not depend on the point

about which it is taken.

7.8. A non-uniform bar of weight *W* is suspended at rest by two strings of negligible weight as shown in Fig.7.39. The angles made by the strings with the vertical are 36.9° and 53.1° respectively. The bar is 2 m long. Calculate the distance d of the centre of gravity of the bar from its left end.

**Answer**

The free body diagram of the bar is shown in the following figure.

Length

of the bar, *l*

= 2 m*T*_{1}and *T*_{2}

are the tensions produced in the left and right strings respectively.

At

translational equilibrium, we have:*T*_{1}*Sin* 36.9^{0} = *T*_{2}*Sin* 53.1^{0}*T*_{1} / *T*_{2} = 4 / 3

⇒ *T*_{1} = (4/3) *T*_{2}

For

rotational equilibrium, on taking the torque about the centre of

gravity, we have:*T*_{1} (*Cos* 36.9) × *d* = *T*_{2}*Cos* 53.1 (2 – d)*T*_{1} × 0.800 *d* = *T*_{2} × 0.600 (2 – *d*)

(4/3) × *T*_{2} × 0.800d = *T*_{2} (0.600 × 2 – 0.600*d*)

1.067*d* + 0.6*d* = 1.2

∴ *d* = 1.2 / 1.67

= 0.72 m

Hence,

the C.G. (centre

of gravity) of the given bar lies

0.72 m from its left end.

7.9. A

car weighs 1800 kg. The distance between its front and back axles is

1.8 m. Its centre of gravity is 1.05 m behind the front axle.

Determine the force exerted by the level ground on each front wheel

and each back wheel.

**Answer**

Mass

of the car, *m*

= 1800 kg

Distance

between the front and back axles, *d*

= 1.8 m

Distance

between the C.G. (centre

of gravity) and the back axle = 1.05

m

The

various forces acting on the car are shown

in the following figure.

*R*

_{f}

and

*R*

_{b}are the forces exerted by the level

ground on the front and back wheels respectively.

At

translational equilibrium:*R*_{f} + *R*_{b} = mg

= 1800 × 9.8

= 17640 N ….**(i)**

For

rotational equilibrium, on taking the

torque about the C.G., we have:*R*_{f}(1.05) = *R*_{b}(1.8 – 1.05)*R*_{b} / *R*_{f} = 7 / 5*R*_{b} = 1.4 *R*_{f} ……**(ii)**

Solving

equations (*i*)

and (*ii*),

we get:

1.4*R*_{f} + *R*_{f} = 17640*R*_{f} = 7350 N

∴ *R*_{b}

= 17640 – 7350 = 10290 N

Therefore,

the force exerted on each front wheel = 7350 / 2 = 3675 N,

and

The

force exerted on each back wheel = 10290 / 2 = 5145 N

7.10. (a) Find the moment of

inertia of a sphere about a tangent to the sphere, given the moment

of inertia of the sphere about any of its diameters to be 2*MR*^{2}/5,

where *M *is

the mass of the sphere and *R*is the

radius of the sphere.

(b) Given the moment of

inertia of a disc of mass

*M*

and radius

*R*about

any of its diameters to be

*MR*

^{2}/4,

find its moment of inertia about an axis normal to the disc and

passing through a point on its edge.

**Answer**

(a)** **The

moment of inertia (M.I.) of a sphere about its diameter = 2*MR*^{2}/5

to the theorem of parallel axes, 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

centre of mass and the product of its mass and the square of the

distance between the two parallel axes.

The

M.I. about a tangent of the sphere = 2*MR*^{2}/5 + *MR*^{2 } = 7*MR*^{2} / 5

(b)** **The

moment of inertia of a disc about its diameter = *MR*^{2} / 4

According

to the theorem of perpendicular axis, 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.

The

M.I. of the disc about its centre = *MR*^{2} / 4 + *M*R^{2} / 4 = *MR*^{2} / 2

The

situation is shown in the given figure.

the theorem of parallel axes:

The

moment of inertia about an axis normal to the disc and passing

through a point on its edge

= *MR*^{2} / 2 + *MR*^{2} = 3*MR*^{2} / 2

Page No: 179

7.11. Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time?

**Answer**

Let *m* and *r* be the respective masses of the hollow cylinder and the solid sphere.

The moment of inertia of the hollow cylinder about its standard axis,* I*_{1} = *mr*^{2}

The moment of inertia of the solid sphere about an axis passing through its centre, *I*_{2} = (2/5)mr^{2}

We have the relation:

τ = *I**α*

Where,

α = Angular acceleration

τ = Torque*I* = Moment of inertia

For the hollow cylinder, τ_{1} = I_{1}α_{1}

For the solid sphere, τ_{n} = I_{n}α_{n}

As an equal torque is applied to both the bodies, τ_{1 }= τ_{2}

∴ α_{2 } / α_{1} = *I*_{1} / *I*_{2} = *mr*^{2} / (2/5)*mr*^{2}

α_{2} > α_{1} …**(i)**

Now, using the relation:

ω = ω_{0} + α*t*

Where,*ω*_{0} = Initial angular velocity*t* = Time of rotation*ω* = Final angular velocity

For equal *ω*_{0} and *t*, we have:*ω *∝ α … **(ii)**

From equations **(i)** and **(ii)**, we can write:*ω*_{2}*> ω*_{1}

Hence, the angular velocity of the solid sphere will be greater than that of the hollow cylinder.

7.12. A

solid cylinder of mass 20 kg rotates about its axis with angular

speed 100 rad

s^{–1}.

The radius of the cylinder is 0.25 m. What is the kinetic energy

associated with the rotation of the cylinder? What is the magnitude

of angular momentum of the cylinder about its axis?

**Answer**

Mass

of the cylinder, *m*

= 20 kg

Angular

speed, *ω*

= 100 rad s^{–1}

Radius

of the cylinder, *r*

= 0.25 m

The

moment of inertia of the solid cylinder:*I* = *mr*^{2} / 2

= (1/2) × 20 × (0.25)^{2}

= 0.625 kg m^{2}

∴ Kinetic

energy = (1/2)* I ω ^{2}*

= (1/2) × 6.25 × (100)

^{2}= 3125 J

∴Angular

momentum,

*L*

=

*Iω*

= 6.25 × 100

= 62.5 Js

7.13. (a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby blueuces his moment of inertia to 2/5 times the initial value? Assume that the turntable rotates without friction.

**Answer**

(a) 100 rev/min

Initial angular velocity, ω_{1}=

40 rev/min

Final angular velocity = ω_{2}

The moment of inertia of the boy with stretched hands = *I*_{1}

The moment of inertia of the boy with folded hands = *I*_{2}

The two moments of inertia are related as:*I*_{2} = (2/5) *I*_{1}

Since no external force acts on the boy, the angular momentum *L*

is a constant.

Hence, for the two situations, we can write:*I*_{2}ω_{2} = *I*_{1} ω_{1}

ω_{2} = (*I*_{1}/*I*_{2}) ω_{1}

= [ *I*_{1} / (2/5)*I*_{1}] × 40 = (5/2) × 40 = 100 rev/min

(b)** **Final K.E. = 2.5 Initial K.E.

Final kinetic rotation, *E*_{F} = (1/2) *I*_{2}ω_{2}^{2}

Initial kinetic rotation, *E*_{I} = (1/2) *I*_{1}ω_{1}^{2}

E_{F} / E_{I} = (1/2) *I*_{2}ω_{2}^{2 }/ (1/2) *I*_{1}ω_{1}^{2}

= (2/5) *I*_{1} (100)^{2} / *I*_{1} (40)^{2}

= 2.5

∴ *E*_{F} = 2.5 *E*_{1}

The increase in the rotational kinetic energy is attributed to the

internal energy of the boy.

7.14. A

rope of negligible mass is wound round a hollow cylinder of mass 3 kg

and radius 40 cm. What is the angular acceleration of the cylinder if

the rope is pulled with a force of 30 N? What is the linear

acceleration of the rope? Assume that there is no slipping.

**Answer**

Mass

of the hollow cylinder, *m*

= 3 kg

Radius

of the hollow cylinder, *r*

= 40 cm = 0.4 m

Applied

force, *F* =

30 N

The

moment of inertia of the hollow cylinder

about its geometric axis:

*I*

= *mr*^{2}

=

3 × (0.4)^{2}

= 0.48 kg m^{2}

Torque, τ = *F* × *r* = 30 × 0.4 = 12 Nm

For

angular acceleration α,

torque is also given by the relation:

τ = Iα

α = τ / *I* = 12 / 0.48 = 25 rad s^{-2}

Linear

acceleration = τα

= 0.4 × 25 = 10 m s^{–2}^{}

7.15. To maintain a rotor at a uniform angular speed of 200 rad s^{–1},

an engine needs to transmit a torque of 180 Nm. What is the power

requiblue by the engine?

(Note: uniform angular velocity in the absence of friction implies

zero torque. In practice, applied torque is needed to counter

frictional torque). Assume that the engine is 100 % efficient.

**Answer**

Angular speed of the rotor, ω =

200 rad/s

Torque requiblue, τ = 180

Nm

The power of the rotor (*P*) is related to torque and angular

speed by the relation:*P* = τω

= 180 × 200 = 36 × 10^{3}

= 36 kW

Hence, the power requiblue by the engine is 36 kW.

7.16. From a uniform disk of radius *R*, a circular hole of radius*R*/2 is cut out. The centre of the hole is at *R*/2 from

the centre of the original disc. Locate the centre of gravity of the

resulting flat body.

**Answer***R*/6; from the original centre of the body and opposite to the centre of the cut portion.

Mass per unit area of the original disc = σ

Radius of the original disc = *R*

Mass of the original disc, *M* = π*R*^{2}σ

The disc with the cut portion is shown in the following figure:

*R*/2

Mass of the smaller disc, *M*^{’} = π (*R*/2)^{2}σ = π *R*^{2}σ / 4 = *M* / 4

Let O and O′ be the respective centres of the original disc and the

disc cut off from the original. As per the definition of the centre of

mass, the centre of mass of the original disc is supposed to be

concentrated at O, while that of the smaller disc is supposed to be

concentrated at O′.

It is given that:

OO′= *R*/2

After the smaller disc has been cut from the original, the remaining

portion is consideblue to be a system of two masses. The two masses are:*M* (concentrated at O), and –*M*′ (=M/4) concentrated at O′

(The negative sign indicates that this portion has been removed from the original disc.)

Let *x* be the distance through which the centre of mass of the remaining portion shifts from point O.

The relation between the centres of masses of two masses is given as:*x* = (*m*_{1}*r*_{1} + *m*_{2}*r*_{2}) / (*m*_{1 }+ *m*_{2})

For the given system, we can write:*x* = [ *M* × 0 – *M*‘ × (*R*/2) ] / ( *M* + (-*M*‘) ) = –*R* / 6

(The negative sign indicates that the centre of mass gets shifted toward the left of point O.)

7.17. A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick?

**Answer**

Let *W* and *W′* be the respective weights of the metre stick and the coin.

of the metre stick is concentrated at its mid-point, i.e., at the 50

cm mark.

Mass

of the meter stick = *m*^{’}

Mass

of each coin, *m*

= 5 g

When the

coins are placed 12 cm away from the end P, the centre of mass gets

shifted by 5 cm from point R toward the end P. The centre of mass is

located at a distance of 45 cm from point P.

The

net torque will be conserved for rotational

equilibrium about point R.

10 × *g*(45 – 12) – *m’g*(50 – 45) = 0

∴ *m’* = 66 *g*

Hence, the

mass of the metre stick is 66 *g*.

7.18. A solid sphere rolls down two different inclined planes of the same heights but different angles of inclination. (a) Will it reach the bottom with the same speed in each case? (b) Will it take longer to roll down one plane than the other? (c) If so, which one and why?

**Answer**

(a) Mass

of the sphere = *m*

Height

of the plane =* h*

Velocity

of the sphere at the bottom of the plane =

v

At

the top of the plane, the total energy of the sphere = Potential

energy = *m*g*h*

At

the bottom of the plane, the sphere has

both translational and rotational kinetic energies.

Hence,

total energy = (1/2)*mv*^{2} + (1/2) *I* ω^{2}

Using

the law of conservation of energy, we can write:

(1/2)*mv*^{2} + (1/2) *I* ω^{2} = *mgh *

For

a solid sphere, the moment of inertia about its centre, *I* = (2/5)mr^{2}

Hence,

equation (*i*)

becomes:

(1/2)*mr*^{2} + (1/2) [(2/5)*mr*^{2}]ω^{2} = *mgh*

(1/2)*v*^{2} + (1/5)*r*^{2}ω^{2} = *gh*

But we have the relation, *v* = rω

∴ (1/2)*v*^{2} + (1/5)*v*^{2} = *gh**v*^{2}(7/10) = *gh**v* = √*(10/7)gh*

Hence,

the velocity of the sphere at the bottom depends only on height (*h*)

and acceleration due to gravity (*g*). Both these values are constants.

Therefore, the velocity at the bottom remains the same from whichever

inclined plane the sphere is rolled.

(b) Consider

two inclined planes with inclinations *θ*_{1}and *θ*_{2},

related as:

θ_{1}

< *θ*_{2}

The

acceleration produced in the sphere when it rolls down the plane

inclined at *θ*_{1}

is:

g

sin *θ*_{1}

The various forces acting on the sphere are shown in the following

figure.

*R*

_{1}

is the normal reaction to the sphere.

Similarly,

the acceleration produced in the sphere when it rolls down the plane

inclined at *θ*_{2}

is:

g

sin *θ*_{2}

The

various forces acting on the sphere are

shown in the following figure.

*R*_{2}

is the normal reaction to the sphere.

θ_{2}

> *θ*_{1};

sin *θ*_{2}

> sin *θ*_{1}

… (*i*)

∴*a*_{2}

> *a*_{1
… }(*ii*)

Initial

velocity, *u *=

0

Final

velocity, *v*

= Constant

Using

the first equation of motion, we can obtain

the time of roll as:*v*

= *u* + *at*

∴ *t* ∝ (1/α)

For inclination θ_{1} : *t*_{1}∝ (1/α_{1})

For inclination θ_{2} : *t*_{2}∝ (1/α_{2})

From

above equations,

we get:*t*_{2}< *t*_{1}

Hence, the sphere will take a longer time to reach the bottom of the

inclined plane having the smaller inclination.

7.19. A

hoop of radius 2 m weighs 100 kg. It rolls along a horizontal floor

so that its centre of mass has a speed of 20 cm/s. How much work has

to be done to stop it?

**Answer**

Radius

of the hoop, *r*

= 2 m

Mass

of the hoop, *m*

= 100 kg

Velocity

of the hoop, *v*

= 20 cm/s = 0.2 m/s

Total

energy of the hoop = Translational K.E. + Rotational K.E.*E*_{T} = (1/2)*mv*^{2} + (1/2) *I* ω^{2}

Moment

of inertia of the hoop about its centre, *I*=

*mr*

^{2}

*E*

_{T}= (1/2)

*mv*

^{2}+ (1/2) (

*mr*

^{2})ω

^{2}

But we have the relation, v = rω

∴

*E*

_{T}= (1/2)

*mv*

^{2}+ (1/2)

*mr*

^{2}ω

^{2}

= (1/2)

*mv*

^{2}+ (1/2)

*mv*

^{2 }=

*mv*

^{2}

The

work requiblue to be done for stopping the

hoop is equal to the total energy of the hoop.

∴ Requiblue

work to be done,

*W*

=

*mv*

^{2}= 100 × (0.2)

^{2}= 4 J.

7.20. The oxygen molecule has a mass of 5.30 × 10^{–26} kg and a moment of inertia of 1.94×10^{–46} kg m^{2} about

an axis through its centre perpendicular to the lines joining the two

atoms. Suppose the mean speed of such a molecule in a gas is 500 m/s and

that its kinetic energy of rotation is two thirds of its kinetic energy

of translation. Find the average angular velocity of the molecule.

**Answer**

Mass

of an oxygen molecule, *m*

= 5.30 × 10^{–26}

kg

Moment

of inertia, *I*

= 1.94 × 10^{–46}

kg m^{2}

Velocity

of the oxygen molecule, *v*

= 500 m/s

The

separation between the two atoms of the oxygen molecule = 2*r*

Mass

of each oxygen atom = *m*/2

Hence,

moment of inertia *I*,

is calculated as:

(*m*/2)r^{2} + (*m*/2)*r*^{2} = *mr*^{2}*r* = ( *I* / *m*)^{1/2}

(1.94 × 10^{-46} / 5.36 × 10^{-26} )^{1/2} = 0.60 × 10^{-10} m

It

is given that:

KE_{rot} = (2/3)KE_{trans}

(1/2) I ω^{2} = (2/3) × (1/2) × *mv*^{2}*mr*^{2}ω^{2} = (2/3)*mv*^{2}

ω = (2/3)^{1/2} (*v*/*r*)

= (2/3)^{1/2} (500 / 0.6 × 10^{-10}) = 6.80 × 10^{12} rad/s.

7.21. A solid cylinder rolls up an inclined plane of angle of inclination 30°. At the bottom of the inclined plane the centre of mass of the cylinder has a speed of 5 m/s.

(b) How long will it take to return to the bottom?

**Answer**

Initial velocity of the solid cylinder, *v* = 5 m/s

Angle of inclination, θ = 30°

Let the cylinder go up the plane upto a height *h*.

From, 1/2 *mv*2 + 1/2 *I* ω2 = *mgh*

1/2mv2 + 1/2 (1/2 *mr*2) ω2 = *mgh*

3/4 *mv*2 = *mgh**h* = 3*v*2/4*g* = 3 × 52/4 × 9.8 = 1.913 m

If s is the distance up the inclined plane, then as

sin θ = *h*/s

s = h/sin θ = 1.913/*sin* 30° = 3.826 m

Time taken to return to the bottom

**Additional Excercises**

shown in Fig.7.40, the two sides of a step ladder BA and CA are 1.6 m

long and hinged at A. A rope DE, 0.5 m is tied half way up. A weight 40

kg is suspended from a point F, 1.2 m from B along the ladder BA.

Assuming the floor to be frictionless and neglecting the weight of the

ladder, find the tension in the rope and forces exerted by the floor on

the ladder. (Take

*g*= 9.8 m/s

^{2})

(Hint: Consider the equilibrium of each side of the ladder separately.)

**Answer**

The given situation can be shown as:

*N*

_{B}= Force exerted on the ladder by the floor point B

*N*

_{C}= Force exerted on the ladder by the floor point C

*T*= Tension in the rope

BA = CA = 1.6 m

DE = 0. 5 m

BF = 1.2 m

Mass of the weight,

*m*= 40 kg

Draw a perpendicular from A on the floor BC. This intersects DE at mid-point H.

ΔABI and ΔAIC are similar

∴BI = IC

Hence, I is the mid-point of BC.

DE || BC

BC = 2 × DE = 1 m

AF = BA – BF = 0.4 m …

**(i)**

D is the mid-point of AB.

Hence, we can write:

AD = (1/2) × BA = 0.8 m …

**(ii)**

Using equations

**(i)**and

**(ii)**, we get:

FE = 0.4 m

Hence, F is the mid-point of AD.

FG||DH and F is the mid-point of AD. Hence, G will also be the mid-point of AH.

ΔAFG and ΔADH are similar

∴ FG / DH = AF / AD

FG / DH = 0.4 / 0.8 = 1 / 2

FG = (1/2) DH

= (1/2) × 0.25 = 0.125 m

In ΔADH:

AH = (AD^{2} – DH^{2})^{1/2}

= (0.8^{2} – 0.25^{2})^{1/2} = 0.76 m

For translational equilibrium of the ladder, the upward force should be equal to the downward force.*N*_{c} + *N*_{B} = *m*g = 392 … **(iii)**

For rotational equilibrium of the ladder, the net moment about A is:

-N_{B} × BI + mg × FG + N_{C} × CI + T × AG – T × AG = 0

-N_{B} × 0.5 + 40 × 9.8 × 0.125 + N_{C} × 0.5 = 0

(N_{C} – N_{B}) × 0.5 = 49

N_{C} – N_{B} = 98 …..**(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 × 0.5 + 40 X 9.8 × 0.125 + T × 0.76 = 0

∴ T = 96.7 N.

Page No: 180

7.23. A

man stands on a rotating platform, with his arms stretched

horizontally holding a 5 kg weight in each hand. The angular speed of

the platform is 30 revolutions per minute. The man then brings his

arms close to his body with the distance of each weight from the axis

changing from 90cm to 20cm. The moment of inertia of the man together

with the platform may be taken to be constant and equal to 7.6 kg m^{2}.

(a) What

is his new angular speed? (Neglect friction.)

(b) Is

kinetic energy conserved in the process? If not, from where does the

change come about?

**Answer**

(a)** **Moment

of inertia of the man-platform system = 7.6 kg m^{2}

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

90 cm:

2

×*m* *r*^{2}

=

2 ×

5 ×

(0.9)^{2}

=

8.1 kg m^{2}

Initial

moment of inertia of the system, *I*_{i} = 7.6 + 8.1 = 15.7 kg m^{2}

Angular

speed, ω_{i} = 300 rev/min

Angular

momentum, *L*_{i} = *I*_{i}ω_{i} = 15.7 × 30 ….(i)

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

cm:

2

×*mr*^{2}

=

2 ×

5 (0.2)^{2}

= 0.4 kg m^{2}

Final

moment of inertia, *I*_{f} = 7.6 + 0.4 = 8 kg m^{2}

Final

angular speed = ω_{f}

Final

angular momentum, *L*_{f} = *I*_{f}ω_{f} = 0.79 ω_{f} …… (ii)

From the conservation of angular momentum, we have:*I*_{i}ω_{i} = *I*_{f}ω_{f}

∴ ω_{f} = 15.7 × 30 / 8 = 58.88 rev/min

(b)** **Kinetic

energy is not conserved in the given process. In fact, with the

decrease in the moment of inertia, kinetic energy increases. The

additional kinetic energy comes from the work done by the man to fold

his hands toward himself.

7.24. A

bullet of mass 10 g and speed 500 m/s is fiblue into a door and gets

embedded exactly at the centre of the door. The door is 1.0 m wide

and weighs 12 kg. It is hinged at one end and rotates about a

vertical axis practically without friction. Find the angular speed of

the door just after the bullet embeds into it.

(Hint:

The moment of inertia of the door about the vertical axis at one end

is *ML*^{2}/3.)

**Answer**

Mass

of the bullet, *m*

= 10 g = 10 ×

10^{–3}

kg

Velocity

of the bullet, *v*

= 500 m/s

Thickness

of the door, *L*

= 1 m

Radius

of the door,

r

= *m* / 2

Mass

of the door, *M*

= 12 kg

Angular

momentum imparted by the bullet on the door:

α =

mvr

= (10 × 10^{-3} ) × (500) × (1/2) = 2.5 kg m^{2} s^{-1} …(i)

Moment

of inertia of the door:

I = ML^{2} / 3

= (1/3) × 12 × 1^{2} = 4 kgm^{2}

But α = Iω

∴ ω = α / I

= 2.5 / 4

= 0.625 rad s^{-1}^{}

7.25. Two discs of moments of inertia *I*_{1} and *I*_{2}

about their respective axes (normal to the disc and passing through

the centre), and rotating with angular speeds ω_{1}

and ω_{2} are

brought into contact face to face with their axes of rotation

coincident. (a) What is the angular speed of the two-disc system? (b)

Show that the kinetic energy of the combined system is less than the

sum of the initial kinetic energies of the two discs. How do you

account for this loss in energy? Take ω_{1}

≠ ω_{2}.

**Answer**

(a)** **Moment of inertia of disc *I* = *I*_{1}

Angular speed of disc *I* = ω_{1}

Moment of inertia of disc *I* = *I*_{2}

Angular speed of disc *I* = ω_{2}

Angular momentum of disc I, *L*_{1} = *I*_{1}ω_{1}

Angular momentum of disc II, *L*_{2} = *I*_{2}ω_{2}

Total initial angular momentum *L*_{i} = *I*_{1}ω_{1} + *I*_{2}ω_{2}

When the two discs are joined together, their moments of inertia

get added up.

Moment of inertia of the system of two discs, *I* = *I*_{1} + *I*_{2}

Let ω be the angular speed of the system.

Total final angular momentum, *L*_{T} = (*I*_{1} + *I*_{2}) ω

Using the law of conservation of angular momentum, we have:*L*_{i} = *L*_{T}*I*_{1}ω_{1} + *I*_{2}ω_{2} = (*I*_{1} + *I*_{2})ω

∴ ω = (*I*_{1}ω_{1} + *I*_{2}ω_{2}) / (*I*_{1} + *I*_{2})

(b) Kinetic energy of disc I, *E*_{1} = (1/2) *I*_{1}ω_{1}^{2}

Kinetic energy of disc II, *E*_{1} = (1/2) *I*_{2}ω_{2}^{2}

Total initial kinetic energy, *E*_{i} = (1/2) ( *I*_{1}ω_{1}^{2} + *I*_{2}ω_{2}^{2})

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

Moment of inertia of the system, *I* = *I*_{1} + *I*_{2}

Angular speed of the system = ω

Final kinetic energy *E*_{f}: = (1/2) ( *I*_{1} + *I*_{2}) ω^{2}

= (1/2) ( *I*_{1} + *I*_{2}) [ (*I*_{1}ω_{1} + *I*_{2}ω_{2}) / (*I*_{1} + *I*_{2}) ]^{2}

= (1/2) (*I*_{1}ω_{1} + *I*_{2}ω_{2})^{2} / (*I*_{1} + *I*_{2})

∴ *E*_{i} – *E*_{f}

= (1/2) ( *I*_{1}ω_{1}^{2} + *I*_{2}ω_{2}^{2}) – (1/2) (*I*_{1}ω_{1} + *I*_{2}ω_{2})^{2} / (*I*_{1} + *I*_{2})

Solving the equation, we get

= *I*_{1}*I*_{2} (ω_{1} – ω_{2})^{2} / 2(*I*_{1} + *I*_{2})

All the quantities on RHS are positive

∴ *E*_{i} – *E*_{f} > 0*E*_{i} > *E*_{f}

The loss of KE can be attributed to the frictional force that

comes into play when the two discs come in contact with each other.

7.26. (a) Prove

the theorem of perpendicular axes.

(Hint:

Square of the distance of a point (*x,
y*)

in the

*x–y*

plane

from an axis through the origin perpendicular to the plane is

*x*

^{2}

*+*

y

y

^{2}).

(b) Prove

the theorem of parallel axes.

(Hint:

If the centre of mass is chosen to be the origin ∑ m

_{i}r

_{i}= 0).

**Answer**

(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 centre O and a point mass *m*,in the *x*–*y*

plane at (*x*,*y*) is

shown in the following figure.

of inertia about

*x*-axis,

*I*

_{x}

=

*mx*

^{2}

Moment

of inertia about *y*-axis,*I*_{y}

= *my*^{2}

Moment

of inertia about *z*-axis,*I*_{z}

= m(x^{2} + y^{2})^{1/2}

*I*_{x}

+ *I*_{y}

= *mx*^{2}

+ *my*^{2}

= *m*(*x*^{2}

+ *y*^{2})

= *m* [(*x*^{2}

+ *y*^{2})^{1/2}]^{1/2}*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 centre of mass and the

product of its mass and the square of the distance between the two

parallel axes.

Suppose a rigid body is made up of n particles, having masses *m*_{1}, *m*_{2}, *m*_{3}, … , *m*_{n}, at perpendicular distances *r*_{1}, *r*_{2}, *r*_{3}, … , *r*_{n} respectively from the centre of mass O of the rigid body.

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

7.27. Prove the result that the velocity *v *of translation of a rolling body (like a ring, disc, cylinder or sphere) at the bottom of an inclined plane of a height *h *is given by *v*^{2} = 2*gh* / [1 + (*k*^{2}/*R*^{2}) ]

Using dynamical consideration (i.e. by consideration of forces and torques). Note *k *is

the radius of gyration of the body about its symmetry axis, and R is

the radius of the body. The body starts from rest at the top of the

plane.

**Answer**

A

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

*m*

= Mass of the body

*R*= Radius of the body

*K*

= Radius of gyration of the body

*v*

= Translational velocity of the body

*h*

=Height

of the inclined plane

*g*=

Acceleration due to gravity

Total

energy at the top of the plane,

*E*

_{1}=

*g*

m

m

*h*

Total

energy at the bottom of the plane,

*E*

_{b}= KE

_{rot}+ KE

_{trans}

= (1/2)

*I*ω

^{2}+ (1/2)

*mv*

^{2}

But

*I*=

*mk*

^{2}and ω =

*v*/

*R*

∴

*E*

_{b}= (1/2)(

*mk*

^{2})(

*v*

^{2}/

*R*

^{2}) + (1/2)

*mv*

^{2}

= (1/2)

*mv*

^{2}(1 +

*k*

^{2}/

*R*

^{2})

From the

law of conservation of energy, we have:

*E*

_{T}=

*E*

_{b}

*mgh*= (1/2)

*mv*

^{2}(1 +

*k*

^{2}/

*R*

^{2})

∴

*v*= 2

*gh*/ (1 +

*k*

^{2}/

*R*

^{2})

Hence,

the given result is proved.

9.28. A disc rotating about its axis with angular speed ω_{o}is

placed lightly (without any translational push) on a perfectly

frictionless table. The radius of the disc is *R*. What are the

linear velocities of the points A, B and C on the disc shown in Fig.

7.41? Will the disc roll in the direction indicated?

**Answer**

*v*

_{A}=

*R*ω

_{o};

*v*

_{B}=

*R*ω

_{0}, v

_{c}= (R/2)ω

_{0}

The disc will not roll

Angular speed of the disc = ω_{o}

Radius of the disc = *R*

Using the relation for linear velocity, *v* = ω_{o}*R*__For point A:__*v*_{A} = *R*ω_{o};

in the direction tangential to the right__For point B:__*v*_{B} = *R*ω_{o};

in the direction tangential to the left__For point C:__

v_{c} = (*R*/2)ω_{o} in

the direction same as that of *v*_{A}

The directions of motion of points A, B, and C on the disc are

shown in the following figure

Since the disc is placed on a frictionless table, it will not roll. This is because the presence of friction is essential for the rolling of a body.

7.29. Explain why friction is necessary to make the disc in Fig. 7.41 roll in the direction indicated. (a) Give the direction of frictional force at B, and the sense of frictional torque, before perfect rolling begins.

(b) What is the force of friction after perfect rolling begins?

**Answer**

A torque is requiblue to roll the given disc. As per the definition of torque, the rotating force should be tangential to the disc. Since the frictional force at point B is along the tangential force at point A, a frictional force is requiblue for making the disc roll.

horizontal table simultaneously, with initial angular speed equal to

10 π rad s

^{-1}. Which of the two will start to roll

earlier? The co-efficient of kinetic friction is

*μ*

_{k}

= 0.2.

**Answer**

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

Initial angular speed, ω_{0}=10 π rad s^{–1}

Coefficient of kinetic friction, *μ*_{k} = 0.2

Initial velocity of both the objects, *u* = 0

Motion of the two objects is caused by frictional force. As per

Newton’s second law of motion, we have frictional force, *f*

= *ma*

μ_{k}*m*g= *ma*

Where,*a* = Acceleration produced in the objects*m *= Mass

∴ *a* = *μ*_{k}*g* … **(i)**

As per the first equation of motion, the final velocity of the

objects can be obtained as:*v* = *u* + *at*

= 0 + *μ*_{k}*gt*

= *μ*_{k}*gt* … **(ii)**

The torque applied by the frictional force will act in

perpendicularly outward direction and cause blueuction in the initial

angular speed.

Torque, τ= –*Iα*

α = Angular acceleration

μ_{k}*m*g*r* = –*Iα*

∴ *α = –*μ_{k}*mgr* / *I* …..**(iii)**

Using the first equation of rotational motion to obtain the final

angular speed:

ω = ω_{0} + α*t*

= ω_{0} + (*–*μ_{k}*m*g*r* / *I*)*t* ….**(iv)**

Rolling starts when linear velocity, *v* = *r*ω

∴ *v* = *r* (ω_{0}*– *μ_{k}*mgrt* / *I*) …**(v)**

Equating equations **(ii)** and **(v)**, we get:

μ_{k}*gt* = *r* (ω_{0}*– *μ_{k}*mgrt* / *I*)

= rω_{0} – μ_{k}*mgr*^{2}*t* / *I* ….**(vi)**

For the ring:*I* = *mr*^{2}

∴ μ_{k}*gt* = *r*ω_{0} – μ_{k}*m**gr*^{2}t / *mr*^{2}

= *r*ω_{0} – μ_{k}gt

2μ_{k}*gt* = *r*ω_{0}

∴ *t* = rω_{0} / 2μ_{k}*g*

= 0.1 × 10 × 3.14 / 2 × 0.2 × 9.8 = 0.80 s ….**(vii)**

For the disc: I = (1/2)*mr*^{2}

∴ μ_{k}*gt* = *r*ω_{0} – μ_{k}*m**gr*^{2}*t* / (1/2)*mr*^{2}

= *r*ω_{0} – 2μ_{k}*gt*

3μ_{k}*gt* = rω_{0}

∴ *t* = *r*ω_{0} / 3μ_{k}*g*

= 0.1 × 10 × 3.14 / 3 × 0.2 × 9.8 = 0.53 s …..**(viii)**

Since *t*_{d} > *t*_{r}, the disc

will start rolling before the ring.

7.31. A

cylinder of mass 10 kg and radius 15 cm is rolling perfectly on a

plane of inclination 30°.

The coefficient of static friction *µ*_{s}

= 0.25.

(a) How

much is the force of friction acting on the cylinder?

(b) What

is the work done against friction during rolling?

(c) If

the inclination *θ*_{}of

the plane is increased, at what value of *θ* does

the cylinder begin to skid, and not roll perfectly?

**Answer**

Mass

of the cylinder, *m*

= 10 kg

Radius

of the cylinder,* r*

= 15 cm = 0.15 m

Co-efficient

of kinetic friction, *µ*_{k}= 0.25

Angle

of inclination, *θ*

= 30°

Moment

of inertia of a solid cylinder about its geometric axis, I = (1/2)*mr*^{2}

The

various forces acting on the cylinder are

shown in the following figure:

acceleration of the cylinder is given as:

*a* = *mg* Sinθ / [m + (I/r^{2}) ]

= *mg *Sinθ / [m + {(1/2)mr^{2}/ r^{2}} ]

= (2/3) *g* Sin 30°

= (2/3) × 9.8 × 0.5 = 3.27 ms^{-2}^{}

(a)** **Using

Newton’s second law of motion, we can write net force as:*f*_{net}

= *ma*

*mg* Sin 30° – *f* = *ma**f* = *mg* Sin 30° – *ma*

= 10 × 9.8 × 0.5 – 10 × 3.27

49 – 32.7 = 16.3 N

(b) During

rolling, the instantaneous point of contact with the plane comes to

rest. Hence, the work done against frictional force is zero.

(c) For

rolling without skid, we have the relation:

μ = (1/3) tan θ

tan θ = 3μ = 3 × 0.25

∴ θ = tan^{-1} (0.75) = 36.87°.

7.32. Read each statement below carefully, and state, with reasons, if it is true or false;

(a) During rolling, the force of friction acts in the same direction as the direction of motion of the CM of the body.

(b) The instantaneous speed of the point of contact during rolling is zero.

(c) The instantaneous acceleration of the point of contact during rolling is zero.

(d) For perfect rolling motion, work done against friction is zero.

(e) A wheel moving down a perfectly frictionless* *inclined plane will undergo slipping (not rolling) motion.

**Answer**

(a)** **False

Frictional force acts opposite to the direction of motion of

the centre of mass of a body. In the case of rolling, the direction of

motion of the centre of mass is backward. Hence, frictional force acts

in the forward direction.

(b)** **True

Rolling can be consideblue 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

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

(d)** **True

This is because once the perfect rolling begins, force of friction becomes zero. Hence work done against friction is zero.

(e)** **True

This is because rolling occurs only on account of friction which is a tangential force capable of providing torque. When the inclined plane is perfectly smooth, the wheel will simply slip under the effect of its own weight.