1. [2081 GIE ‘B’] A disc of moment of inertia [latex]5\times 10^{-4}\ kgm^2[/latex] is rotating freely about the axis through its centre at 60 rpm. Calculate the new revolution per minute if some wax of mass 0.01 kg dropped gently on the disc 0.06 m from the axis.  [3]
   Solution:
   Moment of inertia (I₁) = 5 x 10^-4 kgm²
   Initial angular frequency (f₁) = 60 rpm = 1 revolutions/sec
   Final angular frequency (f₂) =?
   Mass of wax (m) = 0.01 kg
   Radius (r) = 0.06 m
   Moment of inertia of wax (I₂) = mr²
   = 0.01 x 0.06² = [latex]3.6\times 10^{-5}[/latex] kgm²
   Now, from principle of conservation of angular momentum,
   [latex]I_1\omega_1 = (I_1 + I_2)\omega_2[/latex]
   [latex]Or,\ (5\times 10^{-4})\times 2\pi f_1 = (5\times 10^{-4}+3.6\times 10^{-5})\times 2\pi f_2[/latex]
   Or, [latex]f_2 = \frac{3.141\times 10^{-3}}{(3.367\times 10^{-3})}[/latex]
   = 0.932 rev/sec
   = 0.932 x 60 = 55.9 rpm.

2. [2081 ‘B/C’] A flywheel of moment of inertia 0.32 kgm² is rotated steadily at 120 rad/sec by a 50 watt electric motor. Calculate
   i. the kinetic energy of the flywheel.
   ii. the frictional couple opposing the rotation.   [2] 
   Solution:
   Moment of inertia ([latex]I[/latex]) = 0.32 kgm²
   Angular velocity ([latex]\omega[/latex]) = 120 rad/sec
   Power (P) = 50 watt
   Frictional couple ([latex]\tau[/latex]) =?
   K.E. =?
   We know that,
   K.E. = [latex]\frac{1}{2}I\omega^2[/latex]
   = [latex]\frac{1}{2}\times 0.32\times (120)^2[/latex]
   = 2304 J
   Power (P) = [latex]\tau . \omega[/latex]
   Or, 50 = [latex]\tau . 120[/latex]
   Or, [latex]\tau = 0.42 Nm[/latex]
3. [2081 ‘D’] A wheel starts from rest and accelerates with constant angular acceleration to an angular velocity of 8 revolutions per second in 5 seconds. Calculate:
   i. angular acceleration and
   ii. angle which the wheel has rotated at the end of 3 sec. [2]
   Solution:
   Initial angular momentum ([latex]\omega_o[/latex]) = 0 rad/s
   Final frequency (f) = 8 rev/s
   Final angular frequency ([latex]\omega[/latex]) = [latex]2\pi f[/latex] = [latex]2\pi \times 8 = 16\pi[/latex] rad/s
   Time (t) = 5 s
   Angular acceleration ([latex]\alpha[/latex]) =?
   We know,
   [latex]\alpha = \frac{\omega – \omega_o}{t}[/latex]
   = [latex]\frac{16\pi – 0}{5}[/latex] = 10 rad/s²
   Angle ([latex]\theta[/latex]) =? at the end of 3 sec
   At t = 3s
   [latex]\theta = \omega_o t + \frac{1}{2}\alpha t^2[/latex]
   = [latex]\frac{1}{2}\times 10.05 \times 9[/latex]
   = 45.238 rad
4. [2080 GIE ‘B’] A playground merry-go-round of radius 2.0 m has a moment of inertia 250 kgm² and is rotating at 10 rev/min. A 25 kg child jumps onto the edge of the merry-go-round. What is the new angular speed of the merry-go-round?              [3]
   Solution:
   Radius (r) = 2 m
   Initial Moment of inertia ([latex]I[/latex]) = 250 kgm²
   Initial Frequency ([latex]f_o[/latex]) = 10 rev/min = [latex]\frac{10}{60}rev/s[/latex]
   = [latex]\frac{1}{6} rev/s[/latex]
   So, [latex]\omega_o = 2\pi f = 2\pi \times \frac{1}{6}[/latex]
   = [latex]\frac{\pi}{3}[/latex] rad/s
   Mass of child (m) = 25 kg
   First, Moment of inertia of the child ([latex]I[/latex]) = mr²
   = [latex]25\times 2^2 = 100 kgm^2[/latex]
   New angular speed ([latex]\omega[/latex]) =?
   We know, from the principle of conservation of angular momentum,
   [latex]I_o\omega_o = (I_o + I)\omega[/latex]
   Or, [latex]250\times \frac{\pi}{3} = (250 + 100)\times \omega[/latex]
   Or, 261.799 = [latex]350\times \omega[/latex]
   Or, [latex]\omega = \frac{261.799}{350} = 0.748[/latex] rad/s Ans.
   Again, [latex]\omega = 2\pi f[/latex]
   Or, f = [latex]\frac{\omega}{2\pi} = \frac{0.748}{2\pi} = 0.119 rps[/latex]
   = [latex]0.119\times 60 rpm[/latex]
   = 7.14 rpm Ans.
5. [2080 ‘P’] The speed of a motor engine decreases from 900 rev/min. to 600 rev/min. in 10 seconds. Calculate:
   i. The angular acceleration
   ii. Number of revolutions made by the motor during this interval
   iii. How many additional seconds are required for motor to come to rest in the same rate. [3]
   Solution:
   Initial frequency ([latex]f_o[/latex]) = 900 rev/m = [latex]\frac{900}{60} = 15[/latex] rev/s
   So, [latex]\omega_o = 2\pi f_o = 2\pi\times 15 = 30\pi[/latex] rad/s
   Final frequency (f) = 600 rev/m = [latex]\frac{600}{60} = 10[/latex] rev/s
   So, [latex]\omega = 2\pi f = 2\pi \times 10 = 20\pi[/latex] rad/s
   Time (t) = 10 s
   i. Angular acceleration ([latex]\alpha[/latex]) =?
   So, [latex]\alpha = \frac{\omega – \omega_o}{t} = \frac{20\pi – 30\pi}{10} = – 3.14[/latex] rad/s²
   ii. No. of revolutions (n) =?
   First, [latex]\theta = \omega_o t + \frac{1}{2}\alpha t^2[/latex]
   = [latex]30\pi \times 10 + \frac{1}{2}\times (-3.14)\times 10^2[/latex]
   = 785.47 rad
   Now, [latex]\theta = 2\pi n[/latex]
   Or, n = [latex]\frac{\theta}{2\pi} = \frac{785.47}{2\times 3.14} = 125.07[/latex]
   iii. Additional time (t) =?
   Using, [latex]\omega_2 = \omega_1 + \alpha t[/latex]
   Or, 0 = 20[latex]\pi + (-3.14)\times t [/latex]                    [latex][∵ \omega_2 = 0[/latex] & [latex]\omega_1 = \omega][/latex]
   Or, t = 20.01 s
6. [2080 ‘R’] A physics teacher stands on a freely rotating platform. He holds a dumbbell in each hand of his outstretched arms while a student gives him a push until his angular velocity reaches 1.5 rad/s. When the freely spinning teacher pulls his hands in close to his body, his angular velocity increases to 5.0 rad/s. What is the ratio of his final kinetic energy to initial kinetic energy?         [2] 
   Solution:
   Initial angular velocity ([latex]\omega_o[/latex]) = 1.5 rad/s
   Final angular velocity ([latex]\omega[/latex]) = 5 rad/s
   Ratio of final K.E. to initial K.E. [latex](\frac{K.E.}{K.E._o})[/latex] =?
   From principle of conservation of angular momentum,
   [latex]I_o\omega_o = I\omega[/latex]
   Or, [latex]\frac{I}{I_o} = \frac{\omega_o}{\omega} = \frac{1.5}{5}[/latex]
   = 0.3
   Now, Final K.E. = [latex]\frac{1}{2}I\omega^2[/latex]
   Initial K.E. [latex](K.E._o) = \frac{1}{2}I_o\omega_o^2[/latex]
   So, [latex]\frac{K.E.}{K.E._o}[/latex] = [latex]\frac{I\omega^2}{I_o\omega_o^2}[/latex]
   = [latex]\frac{I}{I_o}\times \frac{\omega^2}{\omega_o^2}[/latex]
   = [latex]0.3\times \frac{5^2}{1.5^2} = 3.33[/latex]
7. [2079 GIE ‘A’/ 2075 GIE/ 2069 ‘A’] A ballet dancer spins with 2.4 rev/s with her arms outstretched when the moment of inertia about the axis of rotation is [latex]I[/latex]. With her arms folded; the moment of inertia about the same axis becomes [latex]0.6I[/latex]. Calculate the new rate of spin.    [3]
   Solution:
   Initial frequency (f₁) = 2.4 rev/sec
   Initial moment of inertia (I₁) = [latex]I[/latex]
   Final moment of inertia (I₂) = 0.6 [latex]I[/latex]
   Final frequency (f₂) =?
   We know, from principle of conservation of angular momentum,
   [latex]I_1\omega_1 = I_2\omega_2[/latex]
   Or, [latex]I\times 2\pi f_1 = 0.61I\times 2\pi f_2[/latex]
   Or, [latex]2.4 = 0.6\times f_2[/latex]
   Or, [latex]f_2 = \frac{2.4}{0.6}[/latex] = 4 rev/sec.
8. [2079 GIE ‘B’] A constant torque of 500 Nm turns a wheel which has a moment of inertia 20 kgm² about its centre. Find the angular velocity gained in two seconds.    [3]
   Solution:
   Torque [latex](\tau)[/latex] = 500 Nm
   Moment of inertia (I) = 20 [latex]kgm^2[/latex]
   Time (t) = 2s
   We have,
   [latex]\tau = I\alpha[/latex]
   Or, 500 = 20[latex]\alpha[/latex]
   Or, [latex]\alpha = 25\ rads^{-2}[/latex] is the angular acceleration
   Again, [latex]\omega = \omega_o + \alpha t[/latex]
   Or, [latex]\omega = 0 + 25\times 2[/latex]
   [latex]\therefore \omega = 50\ rads^{-1}[/latex] is the angular velocity after 2s.
9. [2079 ‘V’/ 2066] A disc of moment of inertia [latex]5\times 10^{-4}[/latex] kg m² is rotating freely about the axis through its centre at 40 rpm. Calculate the new revolution per minute if some wax of mass 0.02 kg dropped gently on to the disc 0.08 m from the axis.       [3]
   Solution:
   Moment of inertia (I₁) = 5 x 10^-4 kgm²
   Initial angular frequency (f₁) = 40 rpm = 0.667 revolutions/sec
   Final angular frequency (f₂) =?
   Mass of wax (m) = 0.02 kg
   Radius (r) = 0.08 m
   Moment of inertia of wax (I₂) = mr²
   = 0.02 x 0.08² = [latex]1.28\times 10^{-4}[/latex] kgm²
   Now, from principle of conservation of angular momentum,
   [latex]I_1\omega_1 = (I_1 + I_2)\omega_2[/latex]
   [latex]Or,\ (5\times 10^{-4})\times 2\pi f_1 = (5\times 10^{-4}+1.28\times 10^{-4})\times 2\pi f_2[/latex]
   Or, [latex]f_2 = \frac{(5\times 10^{-4})\times 0.667}{(6.28\times 10^{-4})}[/latex]
   = 0.531 rev/sec
   = 0.531 x 60 = 31.863 rpm.
10. [2076 ‘C’] 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.
    Solution:
    Initial angular momentum ([latex]\omega_o[/latex]) = 0 rad/s
    Final angular frequency (f) = 15 rev/s
    Final angular frequency ([latex]\omega[/latex]) = [latex]2\pi f[/latex] = [latex]2\pi \times 15 = 30\pi[/latex] rad/s
    Time (t) = 10 s
    Angular acceleration ([latex]\alpha[/latex]) =?
    We know,
    [latex]\alpha = \frac{\omega – \omega_o}{t}[/latex]
    = [latex]\frac{30\pi – 0}{10}[/latex] = [latex]3\pi = 9.42[/latex] rad/s²
    Angle ([latex]\theta[/latex]) =? rotated at the end of 2 sec
    At t = 2s
    [latex]\theta = \omega_o t + \frac{1}{2}\alpha t^2[/latex]
    = 0 + [latex]\frac{1}{2}\times 9.42 \times 2^2[/latex]
    = 18.84 rad.
11. [2073 ‘D’] An electric fan is turned off, and its angular velocity decreases uniformly 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 the 4 sec interval.
    Solution:
    Initial rate of revolution [latex](f_1)[/latex] = 500 rev/min = 500/60 = [latex]\frac{25}{3}[/latex] rev/s
    Final rate of revolution [latex](f_2)[/latex] = 200 rev/min = 200/60 = [latex]\frac{10}{3}[/latex] rev/s
    Initial angular velocity [latex](\omega_1) = 2\pi f_1 = 2\pi \times \frac{25}{3}[/latex] = 52.3 rad/s
    Final angular velocity [latex](\omega_2) = 2\pi f_2 = 2\pi \times \frac{10}{3}[/latex] = 20.9 rad/s
    Time (t) = 4.5 s
    Now, angular acceleration, [latex]\alpha = \frac{\Delta \omega}{\Delta t}[/latex]
    = [latex]\frac{\omega_2 – \omega_1}{t} = \frac{20.9-52.3}{4} = – 7.85\ rad/s^2[/latex]
    Also, angular displacement, [latex]\theta = \omega_1 t + \frac{1}{2}\alpha t^2[/latex]
    = [latex]52.3\times 4 + \frac{1}{2}(-7.85)\times 4^2[/latex] = 146.4 radian
    [latex]\therefore[/latex] No. of revolutions made in 4 s intervals = [latex]\frac{146.4}{2\pi} [\because \theta = 2\pi n][/latex]
    = 23.3 revolutions.
12. [2072 ‘C’ / 2070 ‘S’] A constant torque of 500 Nm turns 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.
    Solution:
    Torque [latex](\tau)[/latex] = 500 Nm
    Moment of inertia (I) = 20 [latex]kgm^2[/latex]
    Time (t) = 2s
    We have,
    [latex]\tau = I\alpha[/latex]
    Or, 500 = 20[latex]\alpha[/latex]
    Or, [latex]\alpha = 25\ rads^{-2}[/latex] is the angular acceleration
    Again, [latex]\omega = \omega_o + \alpha t[/latex]
    Or, [latex]\omega = 0 + 25\times 2[/latex]
    [latex]\therefore \omega = 50\ rads^{-1}[/latex] is the angular velocity after 2s
    Also, Rotational kinetic energy = [latex]\frac{1}{2}I\omega^2 = \frac{1}{2}\times 20\times (50)^2[/latex] = 25000 J.
13. [2072 ‘E’] A disc of radius 1 m and mass 5 kg is rolling along a horizontal plane, its moment of inertia about its centre is 2.5 kgm². If its velocity along the plane is 2ms^-1, find its angular velocity and the total energy.
    Solution:
    Radius (r) = 1m
    Mass (m) = 5 kg
    Moment of inertia (I) = [latex]2.5\ kgm^2[/latex]
    Velocity (v) = 2 m/s
    Angular velocity [latex](\omega)[/latex] =?
    Total energy (E) =?
    We know that,
    v = [latex]\omega r[/latex]
    Or, [latex]\omega = \frac{v}{r} = \frac{2}{1} = 2\ rads^{-1}[/latex]
    Now, Total energy (E) = [latex]\frac{1}{2}\times 5\times 2^2 + \frac{1}{2}\times 2.5\times 2^2[/latex]
    = 15 J.
14. [2071 ‘S’ / 2070 ‘S’] A constant torque of 500 Nm turns a wheel about its centre. The moment of inertia about this axis if 100 kgm². Find the angular velocity gained in 4 seconds and kinetic energy gained after 20 revolutions.
    Solution:
    Torque [latex](\tau)[/latex] = 500 Nm
    Moment of inertia (I) = 100 [latex]kgm^2[/latex]
    Angular velocity [latex](\omega)[/latex] =?
    Kinetic energy (K.E.) =?
    No. of revolutions (n) = 20
    Time (t) = 4s
    We have,
    [latex]\tau = I\alpha[/latex]
    Or, 500 = 100[latex]\alpha[/latex]
    Or, [latex]\alpha = 5\ rads^{-2}[/latex] is the angular acceleration
    Now, [latex]\alpha = \frac{\omega}{t}[/latex]
    Or, [latex]\omega = \alpha t = 5\times 4[/latex] = 20 rad/s
    Again, after 20 revolutions, angle displaced [latex](\theta) = 2\pi n[/latex]
    = [latex]2\pi \times 20 = 40\pi[/latex] rad
    Again, [latex]\omega^2 = \omega_o^2 + 2\alpha \theta[/latex]
    = [latex]0 + 2\times 5\times 40\pi[/latex]
    Or, [latex]\omega^2 = 400\pi[/latex]
    Now, Rotational kinetic energy = [latex]\frac{1}{2}I\omega^2 = \frac{1}{2}\times 100\times 400\pi[/latex] = 62831.853 J.
15. [2070 ‘S’] A computer disk drive is turned on starting from the rest and has constant angular acceleration, (a) how long did it take to make the first complete rotation, and (b) what is its angular acceleration? Given that the disk took 0.750 sec for the drive to make its second complete revolution.
    Solution:
    Initial angular velocity [latex](\omega)[/latex] = 0
    Let, it takes ‘t’ seconds to make the first complete rotation.
    Angular acceleration [latex](\alpha)[/latex] =?
    Time Taken to make 2 complete rotation (t’) = (t+0.75)s
    First Case:
    [latex]\theta = \omega_o t + \frac{1}{2}\alpha t^2[/latex]
    Or, [latex]2\pi n = 0 + \frac{1}{2}\alpha t^2[/latex]
    Or, [latex]2\pi = \frac{1}{2}\alpha t^2[/latex] ……………… (i)
    Second Case:
    We have,
    [latex]\theta’ = \omega_o t + \frac{1}{2}\alpha t^2[/latex]
    Or, [latex]2n\pi = 0 + \frac{1}{2}\alpha(t+0.75)^2[/latex] ……………… (ii)
    Dividing [latex]eq^{n.}[/latex] (ii) by [latex]eq^{n.}[/latex] (i), we get,
    2 = [latex]\frac{(t+0.75)^2}{t^2}[/latex]
    Or, [latex]2t^2 = (t+0.75)^2[/latex]
    Taking root on both sides,
    [latex]\sqrt{2}t = t + 0.75[/latex]
    Or, [latex]\sqrt{2}t – t = 0.75[/latex]
    Or, 0.414t = 0.75
    Or, t = 1.81 s.
    Again, putting the value of t in [latex]eq^{n.}[/latex] (i), we get,
    [latex]2\pi = \frac{1}{2}\alpha \times (1.81)^2[/latex]
    Or, [latex]4\pi = \alpha \times 3.278[/latex]
    Or, [latex]\alpha = \frac{4\times 3.14}{3.278} = 3.833\ rad/s^2[/latex]
16. [2069 ‘S’] A constant torque of 200 Nm turns a wheel about its centre. The moment of inertia about the axis is 100 kgm^-2. Find the angular velocity gained in 4 seconds and the kinetic energy gained after 10 revolutions.
    Solution:
    Torque [latex](\tau)[/latex] = 200 Nm
    Moment of inertia (I) = 100 [latex]kgm^2[/latex]
    Angular velocity [latex](\omega)[/latex] =?
    Kinetic energy (K.E.) =?
    No. of revolutions (n) = 10
    Time (t) = 4s
    We have,
    [latex]\tau = I\alpha[/latex]
    Or, 200 = 100[latex]\alpha[/latex]
    Or, [latex]\alpha = 2\ rads^{-2}[/latex] is the angular acceleration
    Now, [latex]\alpha = \frac{\omega}{t}[/latex]
    Or, [latex]\omega = \alpha t = 2\times 4[/latex] = 8 rad/s
    Again, after 10 revolutions, angle displaced [latex](\theta) = 2\pi n[/latex]
    = [latex]2\pi \times 10 = 20\pi[/latex] rad
    Again, [latex]\omega^2 = \omega_o^2 + 2\alpha \theta[/latex]
    = [latex]0 + 2\times 2\times 20\pi[/latex]
    Or, [latex]\omega^2 = 80\pi[/latex]
    Now, Rotational kinetic energy = [latex]\frac{1}{2}I\omega^2 = \frac{1}{2}\times 100\times 80\pi[/latex] = 12566.37 J.
17. [2069 ‘A’] A constant torque of 200 Nm turns a wheel about its centre. The moment of inertia about this axis is 100 kgm². Find the kinetic energy gained after 20 revolutions.
    Solution:
    Torque [latex](\tau)[/latex] = 200 Nm
    Moment of inertia (I) = 100 [latex]kgm^2[/latex]
    Kinetic energy (K.E.) =?
    No. of revolutions (n) = 20
    We have,
    [latex]\tau = I\alpha[/latex]
    Or, 200 = 100[latex]\alpha[/latex]
    Or, [latex]\alpha = 2\ rads^{-2}[/latex] is the angular acceleration
    Again, after 20 revolutions, angle displaced [latex](\theta) = 2\pi n[/latex]
    = [latex]2\pi \times 20 = 40\pi[/latex] rad
    Again, [latex]\omega^2 = \omega_o^2 + 2\alpha \theta[/latex]
    = [latex]0 + 2\times 2\times 40\pi[/latex]
    Or, [latex]\omega^2 = 160\pi[/latex]
    Now, Rotational kinetic energy = [latex]\frac{1}{2}I\omega^2 = \frac{1}{2}\times 100\times 160\pi[/latex] = 25132.74 J.
18. [2067 ‘S’] A ballet dancer spins about a vertical axis at 1 revolution per second with her arms stretched. With her arms folded, her moment of inertia about the axis decreases by 40%, calculate the new rate of revolution.
    Solution:
    Initial frequency (f₁) = 1 rev/sec
    Initial moment of inertia (I₁) = [latex]I[/latex]
    Final moment of inertia (I₂) = I – 40% of [latex]I[/latex]
    = I – 0.4I = 0.6I
    Final frequency (f₂) =?
    We know, from principle of conservation of angular momentum,
    [latex]I_1\omega_1 = I_2\omega_2[/latex]
    Or, [latex]I\times 2\pi f_1 = 0.6I\times 2\pi f_2[/latex]
    Or, [latex]1 = 0.6\times f_2[/latex]
    Or, [latex]f_2 = \frac{1}{0.6}[/latex] = 1.667 rev/sec.