# Rotational Motion

## NCERT Solution

### Part 2

Question 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 center. Which of the two will acquire a greater angular speed after a given time?

**Answer:** Let us assume that mass = m and radius = r

Moment of inertia of hollow cylinder about its standard axis `l_1=mr^2`

Moment of inertia of solid sphere about an axis passing through its center `l_2=2/5mr^2`

We know that `τ=lα`

Where, τ is torque, l is moment of inertia and α is angular momentum.

For hollow cylinder `τ_1=l_1α_1`

For solid sphere `τ_2=l_2α_2`

As per question, τ_{1} = τ_{2}

Or, `l_1α_2=l_2α_2`

Or, `(α_2)/(α_1)=(l_1)/(l_2)`

Or, `(α_2)/(α_1)=(mr^2)(2/5mr^2)=5/2`

Or, `α_2>α_1`

So, angular momentum of solid sphere is more than that of hollow cylinder.

Question 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 rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis?

**Answer:** Given: mass of cylinder m = 20 kg, angular speed ω = 100 rad s^{-1}, radius of cylinder r = 0.25 m

Moment of Inertia of solid cylinder `l=(r^2)/2`

`=1/2xx20xx0.25^2=0.625` kg m^{2}

Kinetic energy `=1/2lω^2`

`=1/2xx6.25xx100^2=3125` J

Angular momentum `L=lω

`= 6.25xx100-6.25` J s

Question 13:

(a) A child stands at the center of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev per 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.

**Answer:** Initial angular velocity ω_{1} = 40 rev/min

Final angular velocity = ω_{2}, moment of inertia with stretched hands = l_{1} and moment of inertia with folded hands = l_{2}

As per question: `l_2=2/5l_1`

There is no external force acting on the body, so angular momentum L is constant. So,

`l_2ω_2=l_1ω_1`

Or, `ω_2=(l_1)/(l_2)ω_1`

`=(l_1)/(2/5l_1)xx40`

`=5/2xx40=100` rev/min

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

**Answer:** Final kinetic rotation, `E_F=1/2l_2ω_2^2`

Initial kinetic rotation, `E_I=1/2l_1ω_1^2`

Or, `(E_F)/(E_I)=(l_2ω_2^2)/(l_1ω_1^2)`

`=(2/5xx100^2)/(40^2)`

`=(2xx20xx100)/(40xx40)=2.5`

Or, `E_F=2.5E_I`

Question 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 30N? What is the linear acceleration of the rope? Assume that there is no slipping.

**Answer:** Given, mass of hollow cylinder m = 3 kg, radius r = 40 cm = 0.4 m, applied force 30 N

Moment of inertia of hollow cylinder about its geometric axis: `l=mr^2`

`=3xx0.4^2=0.48` kg m^{2}

Torque τ = F × r

= 30 × 0.4 = 12 Nm

Torque is also given by following equaqtion:

`τ = lα`

Or, `α=(τ)/l`

`=(12)/(0.48)= 25` rad s^{-2}

Linear acceleration = τ × α = 0.4 × 25 = 10 m s^{-2}

Question 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:** Given, angular speed ω = 200 rad/s and required torque τ = 180 N m

Power can be calculated as follows:

`P=τω`

= 180 × 200 = 36000 = 36 kW

Question 16: From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The center of the hole is at R/2 from the center of the original disc. Locate the center of gravity of the resulting body.

**Answer:** The following figure shows the original disc and the disc that is cut out from it. Since radius of smaller disc is half the radius of original disc, mass of smaller disc will be in duplicate ratio.

So, if M_{2} is the mass of smaller disc, it will be = `M/4`

Now, the relation between centers of masses can be calculated as follows:

`X=(m_1r_1+m_2r_2)/(m_1+m_2)`

`=(M×0-M_2R/2)/(M-M_2)`

Negative sign for M_{2} is taken because this portion has been removed.

`=(-M/4×R/2)/((3M)/4)`

`=-(MR)/8×4/(3M)=-R/6`

Negative sign means that the center of mass shifts towards left of the origin O.

Question 17: A meter stick is balanced on a knife edge at its center. 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 meter stick?

**Answer:** Following figure shows the meter stick. Let us assume that mass of meter stick = m. As per question, it is balanced at 50 cm mark.

When coins are placed 12 cm from point P, the balance point becomes 45 cm from point P. The net torque will be conserved for rotational equilibrium about point R (original balance point).

So, 10 × g (45 – 12) – mg (50 – 45) = 0

Or, 10 × 33 g – 5mg = 0

Or, 330g – 5 mg = 0

Or, g(330 – 5m) = 0

Or, 330 – 5m = 0

Or, 5m = 330

Or, m = 66 g

Question 18: A solid sphere rolls down two different inclined planes of the same heights but at different angles of inclination.

(a) Will it reach the bottom with the same speed in each case?

**Answer:** Let us assume that mass of sphere = m, height of plane = h and velocity of sphere at the bottom of plane = v

Total energy of sphere at the top of the plane = potential energy = mgh

Once the sphere reaches the bottom of the plane, it has both translational and rotational kinetic energies.

Total energy at bottom = `1/2mv^2+1/2lω^2`

As per the law of conservation of energy:

`1/2mv^2+1/2lω^2=mgh`

In case of a solid sphere, moment of inertia about its center = `l=2/5mr^2`

Substituting this value in previous equation, we get:

`1/2mv^2+1/2(2/5mr^2)ω^2=mgh`

`1/2v^2+1/5r^2ω^2=gh`

We know that `v=rω

So, above equation can be written as follows:

`1/2v^2+1/5v^2=gh`

Or, `7/(10)v^2=gh`

Or, `v^2=(10)/7gh`

Or, `v=sqrt((10)/7gh)`

This equation shows that the velocity at the bottom depends only on height and acceleration due to gravity. So, velocity at the bottom remains the same in case of both the inclines.

(b) Will it take longer to roll down one plane than the other?

(c) If so, which one and why?

**Answer:** Let us take angles of two inclined planes as θ_{1} and θ_{2} where θ_{1} < θ_{2}

When the sphere rolls down the plane inclined at θ_{1}, acceleration in sphere = g sin θ_{1}

Similarly, acceleration produced when the sphere rolls down the second plane = g sin θ_{2}

Following figure shows various forces acting on the sphere.

Since θ_{2} > θ_{1}

So, sin θ_{2} > sin θ_{1}

Or, a_{2} > a_{2}

Initial velocity u = 0 and final velocity v = constant (as shown in answer to part (a) of this question)

Now, time can be calculated as follows:

`v=u+at`

Or, `t=(v-u)/a`

This shows that time varies inversely as acceleration.

So, t_{2} < t_{1}

This means that it will take longer for the sphere to roll down the incline with smaller angle.

Question 19: A hoop of radius 2 m weight 100 kg. It rolls along a horizontal floor so that its center of mass has a speed of 20 cm/s. How much work has to be done to stop it?

**Answer:** Given, radius r = 2m, mass m = 100 kg, velocity v = 20 cm/s = 0.2 m/s

Total energy of hoop = Translational KE + Rotational KE

`=1/2mv^2+1/2mr^2ω^2`

Since `v=rω`

So, above equation can be written as follows:

`E=1/2mv^2+1/2mv^2=mv^2`

`=100xx0.2^2=4` J

Required work to stop the hoop = 4 J

Question 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 center 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 oxygen molecule m = 5.30 × 10^{-26} kg

Moment of inertial l = 1.94 × 10^{-46} kg m^{2}

Velocity v = 500 m/s

Separation between two atoms of oxygen = 2r and mass of each oxygen atom = `m/2`

Moment of inertia can be calculated as follows:

`l=m/2r^2+m/2r^2=mr^2`

Or, `r^2=l/m`

Or, `r=sqrt(l/m)`

`=sqrt((1.94xx10^(-46))/(5.36xx10^(-26)))`

`=sqrt(0.36xx10^(-20))`

`r=0.6xx10^(-20)` m

It is given that:

Rotational Energy = 2/3 of Translational Energy

Or, `1/2lω^2=2/3xx1/2xxmv^2`

Or, `mr^2ω^2=2/3mv^2`

Or, `ω=sqrt(2/3)v/r`

`=sqrt(2/3)xx(500)/(0.6xx10^(-10))`

= 6.80 × 10^{12} rad/s

Question 21: A solid cylinder rolls up an inclined plane of angle of inclination 30°. At the bottom of the inclined plane the center of mass of the cylinder has a speed o 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:** Given, initial velocity of cylinder v = 5 m/s, angle of inclination θ = 30° and height = h

As per theory of conservation of energy:

`1/2mv^2+1/2lω^2=mgh`

Or, `1/2mv^2+1/2(1/2mr^2)ω^2=mgh`

Or, `3/4mv^2=mgh`

Or, `3/4v^2=gh`

Or, `h=(3v^2)/(4g)`

`=(3xx5)/(4xx9.8)=1.913` m

If s is the distance up the inclined plane then

Sin θ = h/s

Or, s = `h/(si\n\θ)`

`=(1.913)/(si\n\30°)=3.826` m

Time taken to return to the bottom

`t=sqrt((2s(1+(K^2)/(r^2)))/(g si\n\θ))`

`=(2xx3.826xx(1+1/2))/(9.8 si\n\30°)=1.53` s