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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.

Answer

Distance between H and Cl atoms = 1.27 Å

Mass of H atom = Mass of Cl atom = 35.5

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

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

Let us assume that the centre of mass of the given molecule lies at the origin. Therefore, we can have:

[

1.27 -

∴

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.

The child is running arbitrarily on a trolley moving with velocity

Answer

Consider two vectors OK = vector |a| and OM = vector |b|, inclined at an angle

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

Question 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

Question 7.7. Two particles, each of mass

Answer

Angular momentum of the system about point P:

Angular momentum of the system about point Q:

Consider a point R, which is at a distance

QR =

∴ PR =

Angular momentum of the system about point R:

=

Comparing equations

We infer from equation

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

Length of the bar,

At translational equilibrium, we have:

⇒

For rotational equilibrium, on taking the torque about the centre of gravity, we have:

(4/3) ×

1.067

∴

= 0.72 m

Hence, the C.G. (centre of gravity) of the given bar lies 0.72 m from its left end.

Mass of the car,

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.

= 1800 × 9.8

= 17640 N ....

For rotational equilibrium, on taking the torque about the C.G., we have:

Solving equations (

1.4

∴

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

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

(b) Given the moment of inertia of a disc of mass

(a) The moment of inertia (M.I.) of a sphere about its diameter = 2

According 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(b)

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 =

The situation is shown in the given figure.

Applying the theorem of parallel axes:

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

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

The moment of inertia of the hollow cylinder about its standard axis,

The moment of inertia of the solid sphere about an axis passing through its centre,

We have the relation:

τ =

Where,

α = Angular acceleration

τ = Torque

For the hollow cylinder, τ

For the solid sphere, τ

As an equal torque is applied to both the bodies, τ

∴ α

α

Now, using the relation:

ω = ω

Where,

For equal

From equations

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

Answer

Mass of the cylinder, *m* = 20 kg

Angular speed, Radius of the cylinder,

The moment of inertia of the solid cylinder:

= (1/2) × 20 × (0.25)

= 0.625 kg m

∴ Kinetic energy = (1/2)

= (1/2) × 6.25 × (100)

∴Angular momentum,

= 6.25 × 100

= 62.5 Js

Final angular velocity = ω

The moment of inertia of the boy with stretched hands =

The moment of inertia of the boy with folded hands =

The two moments of inertia are related as:

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

Hence, for the two situations, we can write:

ω

= [

Final kinetic rotation,

Initial kinetic rotation,

E

= (2/5)

= 2.5

∴

The increase in the rotational kinetic energy is attributed to the internal energy of the boy.

Answer

Mass of the hollow cylinder,

Radius of the hollow cylinder,

Applied force,

The moment of inertia of the hollow cylinder about its geometric axis:

= 3 × (0.4)

Torque, τ =

For angular acceleration α, torque is also given by the relation:

τ = Iα

α = τ /

Linear acceleration = τα = 0.4 × 25 = 10 m s

(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 required, 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 required by the engine is 36 kW.

**Question 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**

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:

Radius of the smaller disc = *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 considered 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.)

**Question 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.

The mass 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*.

**Question 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?**

**(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.

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.

**Question 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?**

Radius of the hoop,

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 required to be done for stopping the hoop is equal to the total energy of the hoop.

∴ Required work to be done, *W* = *mv*^{2 }= 100 × (0.2)^{2 }= 4 J.

**Question 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.**

Mass of an oxygen molecule,

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.

**Question 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,

Angle of inclination, θ = 30°

Let the cylinder go up the plane upto a height

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

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

The given situation can be shown as:

BA = CA = 1.6 m

DE = 0. 5 m

BF = 1.2 m

Mass of the weight,

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 …

D is the mid-point of AB.

Hence, we can write:

AD = (1/2) × BA = 0.8 m ...

Using equations

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

= (0.8

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

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

-N

-N

(N

N

Adding equations

N

N

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

-N

-245 × 0.5 + 40 X 9.8 × 0.125 + T × 0.76 = 0

∴ T = 96.7 N.

(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

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.

**Question 7.24. A bullet of mass 10 g and speed 500 m/s is fired 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}

^{}**Question 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}.**

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.

**Question 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^{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.

Moment 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:

Question 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:

*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}=* m*g*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.

**Question 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?**

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.

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

A torque is required 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 required for making the disc roll.

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 reduction 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.

**Question 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:

The 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°.

**Question 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**

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 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

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.

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