Rotational Dynamics Numericals Class 12

You will enjoy the numerical problems of rotational dynamics if you are able to compare the rotational motion with linear motion.

Comparison Rotational motion with linear motion    

1.	Linear/Translational Motion	Rotational Motion
2.	Mass, m	Moment of inertia, I = $\Sigma $mr2
3.	Linear displacement, s	Angular displacement, $\theta $
4.	Linear velocity, v = $\frac{ds}{dt}$	Angular velocity, $\omega $ = $\frac{d\theta }{dt}$ Also, v = $\omega $r
5.	Linear acceleration, a = $\frac{dv}{dt}$ Also,   a = $\frac{v\,-u}{t}$	Angular velocity,  $\alpha $= $\frac{d\omega }{dt}$ Also,  $\alpha $ = $\frac{{{\omega }_{2}}\,-\,{{\omega }_{1}}}{t}$
6.	Force, F = ma Also, F = $\frac{dP}{dt}$	Torque, $\tau $ = I$\alpha $ Also,$\tau $ = $\frac{dL}{dt}$
7.	v = u + a t	$\omega $ = ${{\omega }_{o}}$ + $\alpha $t
8.	v2 = u2 + 2as	$\omega $2 = ${{\omega }_{o}}^{2}$ + 2$\alpha $$\theta $
9.	s = ut + $\frac{1}{2}$at2	$\theta $ = ${{\omega }_{o}}$t + $\frac{1}{2}$$\alpha $t2
10.	Linear K.E. = $\frac{1}{2}$m v2	Rotational K.E. = $\frac{1}{2}$I${{\omega }^{2}}$
11.	Linear momentum, P = constant (In the absence of external force)	Angular momentum, L = constant (In the absence of external force) I1${{\omega }_{1}}$ = I2${{\omega }_{2}}$

Type – 1 (Conservation of angular momentum)

Angular momentum L = constant (In the absence of external force) I₁${{\omega }_{1}}$ = I₂${{\omega }_{2}}$

1. I₁${{\omega }_{1}}$ = I₂${{\omega }_{2}}$
2. $\omega $ = 2$\pi $f
3. Moment of inertia, I = $\Sigma $mr²

Q.1. A ballet dancer spins with 2.4 rev/s with her arms outstretched when the moment of inertia about the axis of rotation is I. With her arms folded, the moment of inertia about the same axis becomes 0.6I. Calculate the new rate of spin.                                     [Ans : 4 rev/sec]

Solution:

Here,

Csae-1:

Frequency, f₁ = 2.4 rps 

Moment of inertia, I₁ = I

Csae-2:

Moment of inertia, I₂ = 0.6I

Frequency,        f₂ = ?

Form principle of conservation of angular momentum. 

I₁${{\omega }_{1}}$ = I₂${{\omega }_{2}}$

Or, I₁ × 2$\pi $f₁ = I₂ ×2$\pi $f₂

Or, f₂ = $\frac{{{I}_{1}}.{{f}_{1}}}{{{I}_{2}}}$= $\frac{I\times 2.4}{0.6I}$= 4 rps

$\therefore $ The new rate of her spin is 4rps.

Q.2. A ballet dancer spins about a vertical axis at 1 revolution per second with her arms stretched. when the moment of inertia about the axis of rotation is I. With her arms folded the moment of inertia about the axis decreases by 40 %. Calculate the new rate of revolution.                                     [Ans: 1.667 rev/sec]

Solution: 

Here,

Csae-1:

Frequency, f₁ = 1 rps 

Moment of inertia, I₁ = I (say) 

Csae-2:

Frequency,        f₂ = ?

Moment of inertia, I₂ = I₁ – 40% of I₁

Or, I₂  = I₁ – $\frac{40}{100}$ × I₁  = 0.6 I

Form principle of conservation of angular momentum. 

I₁${{\omega }_{1}}$ = I₂${{\omega }_{2}}$

Or, I₁ × 2$\pi $f₁ = I₂ ×2$\pi $f₂

$\therefore $ f₂ =$\frac{{{I}_{1}}.{{f}_{1}}}{{{I}_{2}}}$= $\frac{I\times 1}{0.6I}$= 1.667 rps

Q.3. A disc of moment of inertia 5×10^–4 kg m² is rotation freely about its axis through its centre at 40 rpm. Calculate the new revolution per minute if some wax of mass 0.02 kg is dropped gently on the disc 0.08m from the axis. [Ans: 32 rpm] 

Solution:

Here,

Case-1:

Moment of inertia, I₁ = 5×10^–4 kg m²

Frequency, f₁ = 40 rpm

Case-2:

Frequency,        f₂ = ?

Mass of wax, m = 0.02 kg

Perpendicular distance from axis of rotation, r = 0.08 m

When some wax is dropped gently on the disc then

New moment of inertia, I₂ = I₁ + mr²

Or, I₂  = 5×10^–4  + 0.02 × (0.08)²  = 6.28×10^–4  kg m²

Form principle of conservation of angular momentum. 

I₁${{\omega }_{1}}$ = I₂${{\omega }_{2}}$

Or, I₁ × 2$\pi $f₁ = I₂ ×2$\pi $f₂

$\therefore $ f₂ = $\frac{{{I}_{1}}.{{f}_{1}}}{{{I}_{2}}}$=$\frac{5\times {{10}^{4}}\times 40}{6.28\times {{10}^{4}}}$= 31.85 rpm

Type-2:

1. Torque, $\tau $ = I$\alpha $
2. $\omega $ = ${{\omega }_{o}}$ + $\alpha $t
3. ${{\omega }^{2}}$= ${{\omega }_{o}}^{2}$ + 2$\alpha $$\theta $
4. K.E. = $\frac{1}{2}$I${{\omega }^{2}}$
5. Total K.E. of a rolling body is given by
   1. K.E_total = K.E_tans+ K.E._rot
   2. K.E_total = $\frac{1}{2}$mv² + $\frac{1}{2}$I${{\omega }^{2}}$

Q.4. A constant torque of 500 Nm turns a wheel, which has a moment of inertia 20 kgm² about its center. Find the angular velocity gained in 2 second and the kinetic energy gained.      [Ans : 50 rad/sec, 25000 J]

Solution:

Here,

Torque, $\tau $ = 500 Nm

Moment of inertia, I = 20 kg m²

Time, t = 2 sec

(i) K.E. gained = ?

We have,

Torque, $\tau $ = I$\alpha $

$\alpha $ = $\frac{\tau }{I}$ = $\frac{500}{20}$ = 25 rad sec^–2

Now 

$\omega $ = $\omega $_o + $\alpha $t  

For gain, we can take initial angular velocity as zero.

$\omega $ = $\alpha $t = 25 × 2 = 50 rad sec^–1

Again K.E. gained = $\frac{1}{2}$I $\omega $²

$\therefore $ K.E. gained = $\frac{1}{2}$×20× 50² = 25000 J 

Q.5. A constant torque of 200Nm turns a wheel about its centre. The moment of inertia about this axis is 100 kgm². Find the kinetic energy gained after 20 revolutions when it starts from rest.                           [Ans : 22.42 rad/sec, 25132.82 J]

Solution:

Here,

Torque, $\tau $ = 200 Nm

Moment of inertia, I = 100 kgm²

K.E. gained = ? after n = 20 revolutions

Torque, $\tau $= I$\alpha $

Or, $\alpha $ = $\frac{\tau }{I}$= $\frac{200}{100}$ = 2 rad sec^–2

Now, $\omega $² = $\omega $_o² + 2$\alpha $$\theta $

$\omega $² =  2$\alpha $ ×n ×2$\pi $         [since, $\theta $ = n × 2$\pi $ ]        

 $\omega $  =  $\sqrt{2\times 2\times 20\times 2\pi }$ = 22.42 rad sec^–1

Then, K.E. gained = $\frac{1}{2}$I $\omega $² –  ^{$\frac{1}{2}$}I $\omega $_o²

$\therefore $ K.E. gained = $\frac{1}{2}$×100× 22.42² = 25132.82 J 

Q. 6. A constant torque of 500Nm turns a wheel about its centre. The moment of inertia about this axis is 100 kgm². Find the angular velocity gained in 4 seconds and kinetic energy gained after 20 revolutions.  [Ans : 20 rad sec^–1, 62831.85 J]

Solution:

Here,

Torque, $\tau $ = 500 Nm

Moment of inertia, I = 100 kgm²

(i) Angular velocity gained (in t = 4 sec), $\omega $ = ?

(ii) K.E. gained = ? after n = 20 revolutions

Torque, $\tau $ = I$\alpha $

Or, $\alpha $ = $\frac{\tau }{I}$= $\frac{500}{100}$ = 5 rad sec^–2

Now, $\omega $ = $\omega $_o + $\alpha $t

For gain, we can take initial angular velocity as zero.

$\therefore $ $\omega $ = $\alpha $t = 5 × 4 = 20 rad sec^–1

And 

K.E. gained = $\frac{1}{2}$I$\omega $² –  $\frac{1}{2}$I$\omega $_o² ……..(i)

Angular velocity gained in n = 20 revolutions

        $\omega $² = $\omega $_o² + 2$\alpha $$\theta $

          $\omega $² =  2$\alpha $ × n ×2$\pi $         [since, $\theta $ = n×2$\pi $ ]        

          $\omega $  =  $\sqrt{2\times 5\times 20\times 2\pi }$ = $\sqrt{400\pi }$ rad sec^–1

From equation (i),

$\therefore $ K.E. gained = $\frac{1}{2}$×100 × ($\sqrt{400\pi }$)² = 62831.85 J

Q.7. A constant torque of 200Nm turns a wheel about its centre. The moment of inertia about this axis is 100 kgm². Find the angular velocity gained in 4 seconds and kinetic energy gained after 10 revolutions. [Ans: 8 rad/sec, 12566.4 J]

Solution:

Here,

Torque, $\tau $ = 200 Nm

Moment of inertia, I = 100 kgm²

(i) Angular velocity gained (in t = 4 sec) = ?

(ii) K.E. gained = ? after n = 10 revolutions

Torque, $\tau $ = I$\alpha $

Or, $\alpha $ = $\frac{\tau }{I}$= $\frac{200}{100}$ = 2 rad sec^–2

Now, $\omega $ = $\omega $_o + $\alpha $t

For gain, we can take initial angular velocity ($\omega $_o) as zero.

$\therefore $ $\omega $ = $\alpha $t = 2 × 4 = 8 rad sec^–1

And 

K.E. gained = $\frac{1}{2}$I$\omega $² –  $\frac{1}{2}$I$\omega $_o² ……..(i)

Angular velocity gained in n = 20 revolutions

$\omega $² =$\omega $_o² + 2$\alpha $$\theta $

Or,  $\omega $² =  0 + 2$\alpha $×n ×2$\pi $         [since, $\theta $ = n × 2$\pi $ ]        

 Or,  $\omega $  =  $\sqrt{2\times 2\times 10\times 2\pi }$ =$\sqrt{80\pi }$  rad sec^–1

From equation (i),

$\therefore $ K.E. gained = $\frac{1}{2}$×100 × ($\sqrt{80\pi }$)² = 12566.4 J

Q.8. A constant torque of 200Nm turns a wheel about its centre. The moment of inertia about this axis is 100 kgm². Find the kinetic energy gained after 20 revolutions.  [Ans: 22.42 rad sec^–1, 25132.82 J]

Solution:

Here,

Torque, $\tau $ = 200 Nm

Moment of inertia, I = 100 kgm²

K.E. gained = ? after n = 20 revolutions

Torque, $\tau $ = I $\alpha $

Or, $\alpha $ = $\frac{\tau }{I}$= $\frac{200}{100}$ = 2 rad sec^–2

We have,

  $\omega $² = $\omega $_o² + 2$\alpha $$\theta $

  $\omega $² =  2$\alpha $×n ×2$\pi $         [since, $\theta $ = n × 2$\pi $ ]        

   $\omega $  =  $\sqrt{2\times 2\times 20\times 2\pi }$ = 22.42 rad sec^–1

Now,

K.E. gained = $\frac{1}{2}$I$\omega $² –  $\frac{1}{2}$I$\omega $_o² = $\frac{1}{2}$I$\omega $² –  0

$\therefore $ K.E. gained = $\frac{1}{2}$×100× 22.42² = 25132.82 J

Q. 9. A disc of radius 1 m and mass 5 kg is rolling along a horizontal plane. Its moment of inertia about its centre 2.5 kg m². If its velocity along the plane is 2 ms^–1, find its angular velocity and the total energy. [Ans:2 rad/sec, 15 J]

Solution:

Here,

Radius, r = 1 m

Mass, m = 5 kg

Moment of inertia, I = 2.5 kgm²

Linear velocity, v = 2 ms^–1

(i) Angular velocity, $\omega $ = ?

(ii) Total energy = ?

We have, 

(i)      v = $\omega $r

Or, $\omega $ = $\frac{v}{r}$= $\frac{2}{1}$= 2 rad sec^–1 

(ii) Total K.E. of a rolling body is given by

       K.E_total = K.E_tans+ K.E._rot

Or,  K.E_total = $\frac{1}{2}$mv² + $\frac{1}{2}$I$\omega $²

 Or,  K.E_total = $\frac{1}{2}$ × 5 × 2² + $\frac{1}{2}$ × 2.5 × 2²

$\therefore $ K.E_total = 10 + 5 = 15 J

Type-3:

1. Angular acceleration, $\alpha $ = $\frac{{{\omega }_{2}}-{{\omega }_{1}}}{t}$
2. $\theta $ = $\omega $_ot + $\frac{1}{2}$$\alpha $t²
3. $\omega $ = $\omega $_o + $\alpha $t
4. Torque, $\tau $ = I $\alpha $

Q.10. An electric fan is turned off and its angular velocity decreases from 500 rev/min to 200 rev/min in 4 seconds. Find the angular acceleration and the number of revolutions made by the motor in 4 sec interval?         [Ans: -7.85 rad/s² , 23.34]

Solution:

Here,

Initial frequency, f₁ = 500 rpm = $\frac{500}{60}$rps = $\frac{25}{3}$ rps

Final frequency, f₂ = 200 rpm = $\frac{200}{60}$rps = $\frac{10}{3}$rps

Time interval, t = 4sec

(a) Angular acceleration, $\alpha $ = ?

(b) Number of revolutions (in 4 sec), n = ?

We know,

(a) Angular acceleration, $\alpha $ = $\frac{{{\omega }_{2}}-{{\omega }_{1}}}{t}$

Or, $\alpha $ = $\frac{2\pi {{f}_{2}}-2\pi {{f}_{1}}}{t}$

Or, $\alpha $ = $\frac{2\pi ({{f}_{2}}-{{f}_{1}})}{t}$

Or, $\alpha $ = $\frac{2\pi (10/3-25/3)}{4}$

$\therefore $ $\alpha $ = -7.85 rad/s²

Now, Angular displacement made in 4 sec

 $\theta $ = $\omega $₁t + $\frac{1}{2}$$\alpha $t² 

Or, $\theta $ = 2$\pi $ f₁ ×t + $\frac{1}{2}$$\alpha $t² 

Or, $\theta $ = 2$\pi $ $\frac{25}{3}$×4 + $\frac{1}{2}$(–7.85) ×4² = 146.64 rad

$\therefore $ Number of revolution (n) = $\frac{\theta }{2\pi }$= 23.34

Q.11. A wheel starts from rest and accelerates with constant angular acceleration to an angular velocity of 15 revolutions per second in 10 seconds. Calculate the angular acceleration and angle which the wheel has rotated at the end of 2 second. [Ans : 9.42 rad/sec², 18.84 radian]

Solution:

Here,

Initial frequency, f₁ = 0 rps

Final frequency, f₂ = 15 rps

Time interval, t = 10 sec

(a) Angular acceleration, $\alpha $ = ?

(b) Angle through the wheel has rotated (in 2 sec), $\theta $ = ?

We know,

(a) Angular acceleration, $\alpha $ = $\frac{{{\omega }_{2}}-{{\omega }_{1}}}{t}$

Or, $\alpha $ = $\frac{2\pi {{f}_{2}}-2\pi {{f}_{1}}}{t}$

Or, $\alpha $ = $\frac{(2\pi {{f}_{2}}-0)}{t}$

Or, $\alpha $ = $\frac{2\pi \times 15}{10}$

$\therefore $ Angular acceleration, $\alpha $ = 9.42 rad/s²

Now, Angular displacement made in 2 sec

 $\theta $ = $\omega $₁t + $\frac{1}{2}$$\alpha $t² 

Or, $\theta $ = 2$\pi $ f₁ ×t + $\frac{1}{2}$$\alpha $t² 

$\therefore $ $\theta $ = 0 + $\frac{1}{2}$(9.42) ×2² = 18.84 rad

Q.12. A computer disc drive is turned on starting from the rest and has constant angular acceleration, (a) how long did it take to make complete rotation and (b) what is its angular acceleration? Given that the disc took 0.750 sec for the drive to make its second revolution. [Ans: 1.81 sec, 3.83 rad sec^–2]

Solution:

Here,

Initial angular velocity, $\omega $₀ = 0 rad/sec 

Let time to complete first revolution is t₁

Then (a) t₁ = ? 

(b) Angular acceleration, α = ?

And let time to complete second revolution is t₂, here t₂ = 0.750 sec (given)

We know,

$\theta $ = $\omega $_ot + $\frac{1}{2}$$\alpha $t²

For one complete revolution, $\theta $ = 2$\pi $

So from above equation

2$\pi $=  $\frac{1}{2}$$\alpha $t₁²………….. (i)

For two complete revolutions, θ = 4π

$\therefore $  4$\pi $ = $\frac{1}{2}$$\alpha $ (t₁ + t₂)² …………. (ii)

Dividing (ii) by (i)

Or, 2 = ${{\left( \frac{{{t}_{1}}+{{t}_{2}}}{{{t}_{1}}} \right)}^{2}}$

Or, 2 = ${{\left( \frac{{{t}_{1}}+0.75}{{{t}_{1}}} \right)}^{2}}$

Or, $\sqrt{2}$ = $\left( \frac{{{t}_{1}}+0.75}{{{t}_{1}}} \right)$

Or, $\sqrt{2}$ t₁ = t₁ + 0.75

Or, t₁ ($\sqrt{2}$–1) = 0.75

Or, t₁ = $\frac{0.75}{\sqrt{2}\text{ –1}}\text{ }$ 

$\therefore $ time to complete first revolution = 1.81 sec

Equation (i) becomes,

$\alpha $ = $\frac{4\pi }{{{t}_{1}}^{2}}$= 3.83 rad/s²

$\therefore $  The angular acceleration is 3.83 rad/s²

Q.13. Speed of a body spinning about an axis increases from rest to 100 rev. min^–1 in 5 sec, if a constant torque of 20 Nm is applied. The external torque is then removed and the body comes to rest in 100 sec. due to friction. Calculate the frictional torque. [Ans: 1 Nm]

Solution:

Here,

Initial frequency, f₁ = 0 rpm (starting from rest)

After t₁ = 5 sec, frequency, f₂ = 100 rpm = $\frac{100}{60}$rps

Torque, $\tau $ = 20 Nm

After time interval, t₂ = 100 sec

Final frequency, f₃ = 0 rpm = 0 rps 

Frictional torque, ${{\tau }_{F}}$ = ?

We know,

(a) Angular acceleration, $\alpha $ = $\frac{{{\omega }_{2}}-{{\omega }_{1}}}{t}$

Or, $\alpha $ = $\frac{2\pi {{f}_{2}}-2\pi {{f}_{1}}}{t}$

Or, $\alpha $ =$\frac{2\pi {{f}_{2}}-0}{t}$

Or, $\alpha $ = $\frac{2\pi }{5}$×$\frac{100}{60}$

Or, $\alpha $ = 2.094 rad/s²

Now,

Torque, $\tau $ = I$\alpha $

Or, I = $\frac{\tau }{\acute{a}}$= $\frac{20}{2.094}$ = 9.55 kgm² 

Again, when torque is removed the body comes to rest in 100 sec

Then angular retardation due to friction 

 $\alpha $_r = $\frac{{{\omega }_{3}}-{{\omega }_{2}}}{{{t}_{2}}}$

Or, $\alpha $_r = $\frac{2\pi {{f}_{3}}-2\pi {{f}_{2}}}{{{t}_{2}}}$

Or, $\alpha $_r =$\frac{0-2\pi {{f}_{2}}}{{{t}_{2}}}$

Or, $\alpha $_r = – $\frac{2\pi }{100}$ × $\frac{100}{60}$

Or, $\alpha $_r = – 0.105 rad sec^–2

$\therefore $ Frictional torque, ${{\tau }_{F}}$ = I$\alpha $_r = 9.55 × 0.105 = 1.002 Nm

Q.14. A string is wrapped around the rim of a wheel of moment of inertia 0.20 kg m² and radius 20 cm. The wheel is free to rotate about its axis as in figure. Initially, the wheel is at rest. The string is now pulled by a force of 20 N. Find the angular velocity of the wheel after 5.0 s. [100 rad sec^–1]

Solution:

Here,

Moment of inertia, I = 0.2 kgm²

Radius, r = 20 cm = 0.2 m

Initial angular velocity, ω_o = 0 rad/sec (rest)

Force, F  = 20 N

Angular velocity gained $\omega $ = ? (After t = 5 sec)

We have, 

Torque, $\tau $ = I$\alpha $

Or,        F × r   = I$\alpha $       [Since, $\tau $ = F × r]

Or, $\alpha $ = $\frac{F\times r}{I}$= $\frac{20\times 0.2}{0.2}$ = 20 rad sec^–2

Now, $\omega $ =$\omega $_o +  $\alpha $t

$\omega $ = 0 + $\alpha $t = 20 × 5 

$\therefore $ Angular velocity gained, $\omega $ = 100 rad sec^–1 

Also read: Rotational Dynamics Notes