Updated By SaraNextGen

On March 11, 2024, 11:35 AM

Question 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,

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

We have the relation:

Where,

α = Angular acceleration

τ = Torque

I = Moment of inertia

For the hollow cylinder,

For the solid sphere,

As an equal torque is applied to both the bodies, 

Now, using the relation:

Where,

ω₀ = Initial angular velocity

t = Time of rotation

ω = Final angular velocity

For equal ω₀ and t, we have:

ω ∝ α … (ii)

From equations (i) and (ii), we can write:

ω_II > ω_I

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

Question 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:

∴Kinetic energy 

∴Angular momentum, L = Iω

= 6.25 × 100

= 62.5 Js

Question 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 reduces his moment of inertia to 2/5 times the initial value? Assume that the turntable rotates without friction.

(b) Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?

Answer:

(a) 100 rev/min

Initial angular velocity, ω₁= 40 rev/min

Final angular velocity = ω₂

The moment of inertia of the boy with stretched hands = I₁

The moment of inertia of the boy with folded hands = I₂

The two moments of inertia are related as:

Since no external force acts on the boy, the angular momentum L is a constant.

Hence, for the two situations, we can write:

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

Final kinetic rotation, E_F 

Initial kinetic rotation, E_I

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

Question 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 = mr²

= 3 × (0.4)² = 0.48 kg m²

Torque,

= 30 × 0.4 = 12 Nm

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

Linear acceleration = rα = 0.4 × 25 = 10 m s^–2

Question 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 required 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 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 × 10³

= 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:

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 = πR²σ

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

Radius of the smaller disc = 

Mass of the smaller disc, M^’ = 

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′= 

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′   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:

For the given system, we can write:

(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.

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?

Answer:

Answer: (a) Yes (b) Yes (c) On the smaller inclination

(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 = mgh

At the bottom of the plane, the sphere has both translational and rotational kinetic energies.

Hence, total energy = 

Using the law of conservation of energy, we can write:

For a solid sphere, the moment of inertia about its centre, 

Hence, equation (i) becomes:

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), (c)Consider two inclined planes with inclinations θ₁ and θ₂, related as:

θ₁ < θ₂

The acceleration produced in the sphere when it rolls down the plane inclined at θ₁ is:

g sin θ₁

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

R₁ is the normal reaction to the sphere.

Similarly, the acceleration produced in the sphere when it rolls down the plane inclined at θ₂ is:

g sin θ₂

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

R₂ is the normal reaction to the sphere.

θ₂ > θ₁; sin θ₂ > sin θ₁ … (i)

∴ a₂ > 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

From equations (ii) and (iii), we get:

t₂ < t₁

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?

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 KE + Rotational KE

Moment of inertia of the hoop about its centre, I = mr²

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 = mv² = 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 m² 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 m²

Velocity of the oxygen molecule, v = 500 m/s

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

Mass of each oxygen atom = 

Hence, moment of inertia I, is calculated as:

It is given that:

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.

(a) How far will the cylinder go up the plane?

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

Answer:

A solid cylinder rolling up an inclination is shown in the following figure.

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

Angle of inclination, θ = 30°

Height reached by the cylinder = h

(a) Energy of the cylinder at point A:

Energy of the cylinder at point B = mgh

Using the law of conservation of energy, we can write:

Moment of inertia of the solid cylinder, 

In ΔABC:

Hence, the cylinder will travel 3.82 m up the inclined plane.

(b) For radius of gyration K, the velocity of the cylinder at the instance when it rolls back to the bottom is given by the relation:

The time taken to return to the bottom is:

Therefore, the total time taken by the cylinder to return to the bottom is (2 × 0.764) 1.53 s.

Question 7.22:

As 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/s²)

(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:

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

In ΔADH:

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

N_c + N_B = mg = 392 … (iii)

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

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

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