Relation between Torque and Angular acceleration and it's Application Questions in English

(B) The moment of inertia of a disc about its central axis is $I = \frac{1}{2} M R^2$.
Using the rotational analog of Newton's second law,$\tau = I \alpha$,where $\tau$ is the torque and $\alpha$ is the angular acceleration.
The tangential force $F$ applied at the rim provides a torque $\tau = F \times R$.
The angular acceleration is given by $\alpha = \frac{\omega}{t}$.
Substituting these into the torque equation: $F \times R = (\frac{1}{2} M R^2) \times (\frac{\omega}{t})$.
Solving for $F$: $F = \frac{M R \omega}{2 t}$.

(C) Given the angular velocity $\omega(t) = \alpha - \beta t$.
At time $t = 0$,$\omega = \alpha$.
The body comes to rest when $\omega(t) = 0$.
$\alpha - \beta t = 0 \implies t = \frac{\alpha}{\beta}$.
The angular displacement $\theta$ is given by the integral of angular velocity with respect to time:
$\theta = \int_{0}^{t} \omega(t) dt = \int_{0}^{\alpha/\beta} (\alpha - \beta t) dt$.
$\theta = [\alpha t - \frac{1}{2} \beta t^2]_{0}^{\alpha/\beta}$.
Substituting the limits:
$\theta = \alpha(\frac{\alpha}{\beta}) - \frac{1}{2} \beta (\frac{\alpha}{\beta})^2$.
$\theta = \frac{\alpha^2}{\beta} - \frac{1}{2} \frac{\alpha^2}{\beta} = \frac{\alpha^2}{2 \beta}$.

(A) Given: Radius $R = 0.4 \,m$,Mass $M = 1 \,kg$,Angular acceleration $\alpha = 10 \,rad \,s^{-2}$.
The moment of inertia of a disc about its central axis is $I = \frac{1}{2} MR^2$.
$I = \frac{1}{2} \times 1 \,kg \times (0.4 \,m)^2 = 0.5 \times 0.16 = 0.08 \,kg \,m^2$.
The torque $\tau$ is given by $\tau = I \alpha$.
$\tau = 0.08 \,kg \,m^2 \times 10 \,rad \,s^{-2} = 0.8 \,N \,m$.
The tangential force $F$ applied at the rim is related to torque by $\tau = F \times R$.
Therefore,$F = \frac{\tau}{R} = \frac{0.8 \,N \,m}{0.4 \,m} = 2 \,N$.

(C) Given: Radius $R = 0.4 \,m$,Mass $M = 1 \,kg$,Angular acceleration $\alpha = 10 \,rad/s^2$.
The moment of inertia $I$ of a disc about an axis passing through its center and perpendicular to its plane is given by $I = \frac{MR^2}{2}$.
Substituting the values: $I = \frac{1 \times (0.4)^2}{2} = \frac{0.16}{2} = 0.08 \,kg \cdot m^2$.
The torque $\tau$ is given by $\tau = I\alpha$. Also,for a tangential force $F$ applied at the rim,$\tau = RF$.
Equating the two: $RF = I\alpha$.
Solving for $F$: $F = \frac{I\alpha}{R} = \frac{0.08 \times 10}{0.4} = \frac{0.8}{0.4} = 2 \,N$.

(A) Given that the wheel is initially at rest,the initial angular velocity $\omega_0 = 0$.
The angular acceleration $\alpha$ produced by a constant torque $\tau$ is given by $\alpha = \frac{\tau}{I}$.
After time $t$,the angular velocity $\omega$ is given by $\omega = \omega_0 + \alpha t = 0 + \left(\frac{\tau}{I}\right)t = \frac{\tau t}{I}$.
The rotational kinetic energy $K$ is given by $K = \frac{1}{2} I \omega^2$.
Substituting the value of $\omega$,we get $K = \frac{1}{2} I \left(\frac{\tau t}{I}\right)^2 = \frac{1}{2} I \cdot \frac{\tau^2 t^2}{I^2} = \frac{\tau^2 t^2}{2 I}$.

(B) The torque $\tau$ applied by the force $F$ at a distance $r$ from the axis is given by $\tau = F \times r$.
Substituting the values, $\tau = 25 \,N \times 0.4 \,m = 10 \,Nm$.
The moment of inertia $I$ of a solid cylinder about its own axis is $I = \frac{1}{2} M R^2$.
Substituting the values, $I = \frac{1}{2} \times 1 \,kg \times (0.4 \,m)^2 = 0.5 \times 0.16 = 0.08 \,kg \cdot m^2$.
Using the relation $\tau = I \alpha$, the angular acceleration $\alpha$ is $\alpha = \frac{\tau}{I}$.
$\alpha = \frac{10 \,Nm}{0.08 \,kg \cdot m^2} = 125 \,rad/s^2$.

(C) Let the initial angular velocity be $\omega_0$ and uniform angular retardation be $\alpha$.
Using the equation of rotational motion $\omega^2 = \omega_0^2 - 2\alpha\theta$,where $\theta = 2\pi n$.
At time $t$,$\omega = \frac{\omega_0}{4}$.
So,$(\frac{\omega_0}{4})^2 = \omega_0^2 - 2\alpha(2\pi n) \implies \frac{\omega_0^2}{16} = \omega_0^2 - 4\pi n\alpha$.
$4\pi n\alpha = \omega_0^2(1 - \frac{1}{16}) = \frac{15\omega_0^2}{16}$.
Thus,$2\alpha = \frac{15\omega_0^2}{32\pi n}$.
Now,for the fan to come to rest,final angular velocity $\omega_f = 0$.
Let $n'$ be the total revolutions from switch off to rest.
$0^2 = \omega_0^2 - 2\alpha(2\pi n')$.
$2\alpha(2\pi n') = \omega_0^2$.
Substituting $2\alpha$: $(\frac{15\omega_0^2}{32\pi n})(2\pi n') = \omega_0^2$.
$\frac{15n'}{16n} = 1 \implies n' = \frac{16n}{15}$.

(A) The rod is pivoted at one end. The moment of inertia of the rod about the pivot is $I = \frac{mL^2}{3}$.
The gravitational force $mg$ acts at the center of mass of the rod,which is at a distance $L/2$ from the pivot.
The torque $\tau$ about the pivot due to gravity is $\tau = mg \cdot (L/2) \sin \theta$.
Using the rotational analog of Newton's second law,$\tau = I \alpha$,where $\alpha$ is the angular acceleration:
$mg \frac{L}{2} \sin \theta = \left( \frac{mL^2}{3} \right) \alpha$
Solving for $\alpha$:
$\alpha = \frac{mg(L/2) \sin \theta}{mL^2/3} = \frac{mgL/2}{mL^2/3} \sin \theta = \frac{3g}{2L} \sin \theta$.
Thus,the angular acceleration is $\frac{3g \sin \theta}{2L}$.

(C) Given:
Mass $M = 25 \ kg$,Radius $R = 0.2 \ m$,Initial angular velocity $\omega_0 = 240 \ r.p.m. = 240 \times \frac{2\pi}{60} \ rad/s = 8\pi \ rad/s$.
Final angular velocity $\omega = 0 \ rad/s$,Time $t = 20 \ s$.
The moment of inertia of the disc is $I = \frac{1}{2}MR^2 = \frac{1}{2} \times 25 \times (0.2)^2 = 0.5 \ kg \cdot m^2$.
The angular acceleration $\alpha$ is given by $\alpha = \frac{\omega - \omega_0}{t} = \frac{0 - 8\pi}{20} = -0.4\pi \ rad/s^2$.
The retarding torque is $\tau = I|\alpha| = 0.5 \times 0.4\pi = 0.2\pi \ N \cdot m$.
Since the torque is applied tangentially,$\tau = F \times R$,so $F = \frac{\tau}{R} = \frac{0.2\pi}{0.2} = \pi \ N$.

(C) The torque $\tau$ is related to the moment of inertia $I$ and angular acceleration $\alpha$ by the equation $\tau = I\alpha$.
Since the torque $\tau$ is the same for both,the angular acceleration is given by $\alpha = \frac{\tau}{I}$.
The moment of inertia of a ring is $I_{\text{ring}} = MR^2$ and for a disc is $I_{\text{disc}} = \frac{1}{2}MR^2$.
Since $I_{\text{ring}} > I_{\text{disc}}$,the angular acceleration of the disc will be greater than that of the ring $(\alpha_{\text{disc}} > \alpha_{\text{ring}})$.
Consequently,the disc will rotate with a greater angular frequency (or angular velocity) compared to the ring after the same time interval.

(A) Power $(P)$ is defined as the product of torque $(\tau)$ and angular velocity $(\omega)$.
$P = \tau \times \omega$
We know that torque is given by the product of moment of inertia $(I)$ and angular acceleration $(\alpha)$,i.e.,$\tau = I \alpha$.
Substituting this into the power equation:
$P = (I \alpha) \times \omega$
Rearranging the equation to solve for angular velocity $(\omega)$:
$\omega = \frac{P}{I \alpha}$
$\omega = P(I \alpha)^{-1}$

(D) The initial angular velocity is $\omega_0 = 240 \ r.p.m. = \frac{240 \times 2\pi}{60} \ rad/s = 8\pi \ rad/s$.
Using the rotational kinematic equation $\omega = \omega_0 - \alpha t$,where the final angular velocity $\omega = 0$ at $t = 20 \ s$:
$0 = 8\pi - \alpha(20) \Rightarrow \alpha = \frac{8\pi}{20} = 0.4\pi \ rad/s^2$.
The torque $\tau$ is given by $\tau = I\alpha$,where $I = \frac{1}{2}MR^2$ for a disc.
Given $M = 25 \ kg$ and $R = 0.2 \ m$,$I = \frac{1}{2} \times 25 \times (0.2)^2 = 0.5 \ kg \cdot m^2$.
The torque is also $\tau = F \cdot R$,where $F$ is the tangential force.
Equating the two expressions for torque: $F \cdot R = I \alpha
\Rightarrow F = \frac{I \alpha}{R} = \frac{0.5 \times 0.4\pi}{0.2} = \frac{0.2\pi}{0.2} = \pi \ N$.

(A) Initial angular velocity $\omega_i = 120 \ rpm = \frac{120 \times 2\pi}{60} = 4\pi \ rad/s$.
Final angular velocity $\omega_f = 0 \ rad/s$.
Time taken $t = 10 \ s$.
Angular acceleration $\alpha = \frac{\omega_f - \omega_i}{t} = \frac{0 - 4\pi}{10} = -0.4\pi \ rad/s^2$.
Moment of inertia of the disc $I = \frac{1}{2}MR^2 = \frac{1}{2} \times 10 \times (0.1)^2 = 0.05 \ kg \ m^2$.
Retarding torque $\tau = I|\alpha| = 0.05 \times 0.4\pi = 0.02\pi \ Nm$.
Since $\tau = F \times R$, the force $F = \frac{\tau}{R} = \frac{0.02\pi}{0.1} = 0.2\pi \ N$.

(A) The relationship between torque $\tau$ and the rate of change of angular momentum $L$ is given by Newton's second law for rotation: $\tau = \frac{dL}{dt}$.
Given initial angular momentum $L_1 = 1 \, J \cdot s$ and final angular momentum $L_2 = 4 \, J \cdot s$.
The change in angular momentum is $\Delta L = L_2 - L_1 = 4 - 1 = 3 \, J \cdot s$.
The time interval is $\Delta t = 4 \, s$.
Therefore, the torque is $\tau = \frac{\Delta L}{\Delta t} = \frac{3}{4} = 0.75 \, N \cdot m$.

(C) $1$. Option $A$ is correct: $\tau = I \alpha$,which is the rotational analogue of Newton's second law $(F = ma)$.
$2$. Option $B$ is correct: $\tau = p \times B$ (or $\mu \times B$),representing the torque on a dipole in a magnetic field.
$3$. Option $C$ is incorrect: The correct relation is $I = \frac{\tau}{\alpha}$. The given relation $I = \tau \times \alpha$ is dimensionally and physically wrong.
$4$. Option $D$ is incorrect: The correct relation for angular momentum is $L = I \omega$. Linear momentum is $p = mv$. The relation $p = I \omega$ is physically incorrect.
Note: Since the question asks for the incorrect relation and both $C$ and $D$ are incorrect,$C$ is the most fundamental error in rotational dynamics definitions.

(A) Given,initial angular frequency of the wheel,$f_0 = 1200 rpm = \frac{1200}{60} rps = 20 rps$.
Initial angular velocity,$\omega_0 = 2 \pi f_0 = 2 \pi \times 20 = 40 \pi rad/s$.
Final angular frequency,$f = 3120 rpm = \frac{3120}{60} rps = 52 rps$.
Final angular velocity,$\omega = 2 \pi f = 2 \pi \times 52 = 104 \pi rad/s$.
Time taken,$t = 16 s$.
Using the equation of rotational motion,$\omega = \omega_0 + \alpha t$,where $\alpha$ is the angular acceleration.
$\alpha = \frac{\omega - \omega_0}{t} = \frac{104 \pi - 40 \pi}{16} = \frac{64 \pi}{16} = 4 \pi rad/s^2$.

(B) Given: Mass $M = 2 \,kg$, Radius $R = 1 \,m$, Final angular velocity $\omega = 10 \,rad/s$, Time $t = 2 \,s$, Initial angular velocity $\omega_0 = 0$.
First, calculate the angular acceleration $\alpha$ using $\omega = \omega_0 + \alpha t$:
$10 = 0 + \alpha(2) \implies \alpha = 5 \,rad/s^2$.
The moment of inertia $I$ of a solid sphere about its axis is $I = \frac{2}{5}MR^2 = \frac{2}{5}(2)(1)^2 = 0.8 \,kg \cdot m^2$.
The torque $\tau$ is given by $\tau = I\alpha = 0.8 \times 5 = 4 \,N \cdot m$.
Since the force $F$ is applied tangentially, $\tau = F \times R$.
Therefore, $F = \frac{\tau}{R} = \frac{4 \,N \cdot m}{1 \,m} = 4 \,N$.

(B) Initial angular velocity of the flywheel, $\omega_0 = 150 \text{ rev/minute}$.
Converting to $\text{rad/s}$:
$\omega_0 = \frac{150 \times 2\pi}{60} \text{ rad/s} = 5\pi \text{ rad/s}$.
Final angular velocity when the wheel comes to rest, $\omega = 0 \text{ rad/s}$.
Constant retardation, $\alpha = -\pi \text{ rad/s}^2$ (negative sign indicates slowing down).
Using the first equation of rotational motion, $\omega = \omega_0 + \alpha t$:
$0 = 5\pi + (-\pi)t$.
$t = \frac{5\pi}{\pi} = 5 \text{ s}$.

(A) Given: Diameter $D = 0.8 \ m$,so radius $R = 0.4 \ m$. Mass $M = 4 \ kg$. Torque $\tau = 2.56 \ N \ m$.
For a circular disc,the moment of inertia about its central axis is $I = \frac{1}{2} M R^2$.
$I = \frac{1}{2} \times 4 \times (0.4)^2 = 2 \times 0.16 = 0.32 \ kg \ m^2$.
Using the relation $\tau = I \alpha$,where $\alpha$ is the angular acceleration:
$\alpha = \frac{\tau}{I} = \frac{2.56}{0.32}$.
$\alpha = 8 \ rad \ s^{-2}$.

(C) The rate of change of angular momentum is equal to the external torque applied,given by $\tau = \frac{dL}{dt}$.
Since $L = I\omega$,we have $\frac{dL}{dt} = I \frac{d\omega}{dt} = I \alpha$.
Given:
Initial angular velocity $\omega_i = 6 \ rad \ s^{-1}$
Final angular velocity $\omega_f = 21 \ rad \ s^{-1}$
Time interval $\Delta t = 1.5 \ s$
Moment of inertia $I = 100 \ g \ m^2 = 0.1 \ kg \ m^2$.
First,calculate the angular acceleration $\alpha = \frac{\omega_f - \omega_i}{\Delta t} = \frac{21 - 6}{1.5} = \frac{15}{1.5} = 10 \ rad \ s^{-2}$.
Now,calculate the rate of change of angular momentum: $\frac{dL}{dt} = I \alpha = 0.1 \ kg \ m^2 \times 10 \ rad \ s^{-2} = 1 \ N \ m$ (or $kg \ m^2 \ s^{-2}$).
Thus,the correct option is $C$.

(A) Given:
Radius of the wheel,$r = 0.4 \,m$
Angular acceleration,$\alpha = 8 \,rad \,s^{-2}$
Mass hung,$m = 4 \,kg$
Acceleration due to gravity,$g = 10 \,m \,s^{-2}$
The torque $\tau$ produced by the weight of the mass hanging from the rim is given by:
$\tau = m \cdot g \cdot r$
Also,the torque is related to the moment of inertia $I$ and angular acceleration $\alpha$ by the equation:
$\tau = I \cdot \alpha$
Equating the two expressions for torque:
$I \cdot \alpha = m \cdot g \cdot r$
Substituting the given values:
$I \cdot 8 = 4 \times 10 \times 0.4$
$I \cdot 8 = 16$
$I = \frac{16}{8} = 2 \,kg \,m^2$
Therefore,the moment of inertia of the wheel is $2 \,kg \,m^2$.

(A) The relationship between torque $\tau$ and angular momentum $L$ is given by the equation $\tau = \frac{dL}{dt}$.
Given the initial angular momentum $L_i = A_0$ and the final angular momentum $L_f = 4 \,A_0$.
The time interval $\Delta t = 4 \,s$.
The change in angular momentum is $\Delta L = L_f - L_i = 4 \,A_0 - A_0 = 3 \,A_0$.
Therefore, the magnitude of the torque is $\tau = \frac{\Delta L}{\Delta t} = \frac{3 \,A_0}{4}$.

(A) Given: Mass of wheel $m = 20 \,kg$, Radius $R = 30 \,cm = 0.3 \,m$.
Initial angular speed $\omega_0 = 80 \,rpm = \frac{80 \times 2 \pi}{60} = \frac{8 \pi}{3} \,rad/s$.
Angular displacement $\theta = 5 \,rev = 5 \times 2 \pi = 10 \pi \,rad$.
Final angular speed $\omega = 0$.
Using the rotational kinematic equation $\omega^2 = \omega_0^2 + 2 \alpha \theta$, we find angular acceleration $\alpha$:
$\alpha = \frac{\omega^2 - \omega_0^2}{2 \theta} = \frac{0 - (8 \pi / 3)^2}{2 \times 10 \pi} = -\frac{64 \pi^2 / 9}{20 \pi} = -\frac{16 \pi}{45} \,rad/s^2$.
The retarding torque $\tau$ is provided by the tangential force $F$:
$\tau = I \alpha = F R$.
For a disc, $I = \frac{1}{2} m R^2$.
Thus, $F = \frac{I \alpha}{R} = \frac{1}{2} m R \alpha$.
Substituting the values: $F = \frac{1}{2} \times 20 \times 0.3 \times \left| -\frac{16 \pi}{45} \right| = 10 \times 0.3 \times \frac{16 \pi}{45} = 3 \times \frac{16 \pi}{45} = \frac{16 \pi}{15} \approx 1.06 \pi \,N$.

(B) The moment of inertia of a disc about its central axis is given by $I = \frac{1}{2} M R^2$.
Given the angular acceleration is $\alpha = a$.
The torque $\tau$ acting on a rotating body is given by the formula $\tau = I \alpha$.
Substituting the values,we get $\tau = (\frac{1}{2} M R^2) \times a$.
Therefore,the torque acting on the disc is $\tau = \frac{M R^2 a}{2}$.

(D) Given: Moment of inertia $I = 4 \,kg-m^2$,Torque $\tau = 8 \,N-m$,and time $t = 20 \,s$. The body starts from rest,so initial angular velocity $\omega_0 = 0$.
Using the relation $\tau = I \alpha$,the angular acceleration $\alpha$ is:
$\alpha = \frac{\tau}{I} = \frac{8}{4} = 2 \,rad/s^2$.
The angular displacement $\theta$ in time $t$ is given by:
$\theta = \omega_0 t + \frac{1}{2} \alpha t^2 = 0 + \frac{1}{2} \times 2 \times (20)^2 = 400 \,rad$.
The work done $W$ by the torque is given by:
$W = \tau \theta = 8 \,N-m \times 400 \,rad = 3200 \,J$.

(B) Given: Mass $M = 25 \,kg$,Radius $R = 0.2 \,m$,Initial angular velocity $\omega_i = 240 \,rpm = \frac{240 \times 2 \pi}{60} \,rad/s = 8 \pi \,rad/s$,Time $t = 20 \,s$,Final angular velocity $\omega_f = 0 \,rad/s$.
First,calculate the angular deceleration $\alpha$:
$\alpha = \frac{\omega_i - \omega_f}{t} = \frac{8 \pi - 0}{20} = \frac{8 \pi}{20} = 0.4 \pi \,rad/s^2$.
Next,calculate the moment of inertia $I$ of the flywheel (assuming it is a disc):
$I = M R^2 = 25 \times (0.2)^2 = 25 \times 0.04 = 1 \,kg \cdot m^2$.
Finally,calculate the torque $\tau$:
$\tau = I \alpha = 1 \times 0.4 \pi = 0.4 \pi \,Nm$.

(D) When an impulse $P$ is applied at one end of the rod,the linear momentum of the center of mass is $P = mv$,so the velocity of the center of mass is $v = \frac{P}{m}$.
The angular impulse about the center of mass is $J = P \times \frac{L}{2}$. Since $J = I\omega$,where $I = \frac{mL^2}{12}$ is the moment of inertia about the center of mass,we have $\omega = \frac{P(L/2)}{mL^2/12} = \frac{6P}{mL}$.
The total kinetic energy $K$ is the sum of translational and rotational kinetic energy: $K = K_{trans} + K_{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.
Substituting the values: $K = \frac{1}{2}m\left(\frac{P}{m}\right)^2 + \frac{1}{2}\left(\frac{mL^2}{12}\right)\left(\frac{6P}{mL}\right)^2$.
$K = \frac{P^2}{2m} + \frac{1}{2}\left(\frac{mL^2}{12}\right)\left(\frac{36P^2}{m^2L^2}\right) = \frac{P^2}{2m} + \frac{3P^2}{2m} = \frac{4P^2}{2m} = \frac{2P^2}{m}$.

(B) The moment of inertia $I$ of the system about the axis $AB$ is the sum of the moments of inertia of the two discs and the rod.
For each disc,the axis $AB$ is at a distance of $10 \ \text{cm}$ $(R)$ from the centre of the left disc and $20 \ \text{cm}$ $(2R)$ from the centre of the right disc.
Using the parallel axis theorem,$I_{disc} = I_{cm} + md^2$.
For the left disc: $I_1 = \frac{1}{4}mR^2 + mR^2 = \frac{5}{4}mR^2$.
For the right disc: $I_2 = \frac{1}{4}mR^2 + m(2R)^2 = \frac{17}{4}mR^2$.
For the rod of length $L = 3R = 30 \ \text{cm}$,the axis $AB$ passes through a point $10 \ \text{cm}$ from one end. The distance of the centre of the rod from $AB$ is $d = 5 \ \text{cm} = R/2$.
$I_{rod} = I_{cm} + md^2 = \frac{mL^2}{12} + m(R/2)^2 = \frac{m(3R)^2}{12} + \frac{mR^2}{4} = \frac{9mR^2}{12} + \frac{mR^2}{4} = \frac{3mR^2}{4} + \frac{mR^2}{4} = mR^2$.
Total $I = I_1 + I_2 + I_{rod} = \frac{5}{4}mR^2 + \frac{17}{4}mR^2 + mR^2 = \frac{22}{4}mR^2 + mR^2 = 5.5mR^2 + mR^2 = 6.5mR^2$.
Given $m = 600 \ \text{g}$,$R = 10 \ \text{cm}$.
$I = 6.5 \times 600 \times (10)^2 = 6.5 \times 60000 = 390000 \ \text{g cm}^2 = 39 \times 10^4 \ \text{g cm}^2$.
Angular acceleration $\alpha = \frac{\tau}{I} = \frac{43 \times 10^5}{39 \times 10^4} = \frac{430}{39} \approx 11.02 \ \text{rad/s}^2$.
Thus,the angular acceleration is approximately $11 \ \text{rad/s}^2$.

(D) The rod is pivoted at a distance $\frac{L}{3}$ from the lower end. The center of mass of the rod is at a distance $\frac{L}{2}$ from the lower end.
Initially,the center of mass is at a height $h_i = \frac{L}{2} - \frac{L}{3} = \frac{L}{6}$ above the pivot point.
When the rod hits the table,the center of mass is at the same level as the pivot point,so the final height $h_f = 0$.
The change in potential energy is $\Delta PE = Mg(h_f - h_i) = -Mg\frac{L}{6}$.
By the principle of conservation of energy,the loss in potential energy equals the gain in rotational kinetic energy: $Mg\frac{L}{6} = \frac{1}{2} I \omega^2$.
The moment of inertia $I$ about the pivot point (which is at distance $\frac{L}{3}$ from the end) is calculated using the parallel axis theorem: $I = I_{cm} + M d^2 = \frac{ML^2}{12} + M(\frac{L}{2} - \frac{L}{3})^2 = \frac{ML^2}{12} + M(\frac{L}{6})^2 = \frac{ML^2}{12} + \frac{ML^2}{36} = \frac{3ML^2 + ML^2}{36} = \frac{4ML^2}{36} = \frac{ML^2}{9}$.
Substituting $I$ into the energy equation: $Mg\frac{L}{6} = \frac{1}{2} (\frac{ML^2}{9}) \omega^2$.
$Mg\frac{L}{6} = \frac{ML^2}{18} \omega^2$.
$\omega^2 = \frac{MgL}{6} \cdot \frac{18}{ML^2} = \frac{3g}{L}$.
$\omega = \sqrt{\frac{3g}{L}}$.

(A) The angular momentum $L$ is given by $L = I\omega$,where $I$ is the moment of inertia and $\omega$ is the angular velocity. Thus,$\omega = \frac{L}{I}$.
First,locate the centre of mass $(CM)$ of the system. Let the mass $m$ be at $x = 0$ and $2m$ be at $x = d$. The position of the $CM$ is $x_{cm} = \frac{m(0) + 2m(d)}{m + 2m} = \frac{2md}{3m} = \frac{2d}{3}$.
The distance of mass $m$ from the $CM$ is $r_1 = \frac{2d}{3}$,and the distance of mass $2m$ from the $CM$ is $r_2 = d - \frac{2d}{3} = \frac{d}{3}$.
The moment of inertia $I$ about the axis passing through the $CM$ is $I = m(r_1)^2 + 2m(r_2)^2 = m(\frac{2d}{3})^2 + 2m(\frac{d}{3})^2$.
$I = m(\frac{4d^2}{9}) + 2m(\frac{d^2}{9}) = \frac{4md^2 + 2md^2}{9} = \frac{6md^2}{9} = \frac{2}{3}md^2$.
Substituting $I$ into the formula for angular velocity: $\omega = \frac{L}{\frac{2}{3}md^2} = \frac{3L}{2md^2}$.

(D) Initial angular velocity $\omega_0 = 1200\text{ rpm} = \frac{1200 \times 2\pi}{60} = 40\pi\text{ rad/s}$.
Angular acceleration $\alpha = \frac{\Delta\omega}{\Delta t} = \frac{0 - 40\pi}{10} = -4\pi\text{ rad/s}^2$.
Moment of inertia of a solid sphere $I = \frac{2}{5}mR^2 = \frac{2}{5} \times 5 \times (0.04\text{ m})^2 = 2 \times 0.0016 = 0.0032\text{ kg m}^2$.
Torque applied $\tau = I|\alpha| = 0.0032 \times 4\pi = 0.0128\pi\text{ Nm}$.
Angular displacement $\theta = \omega_0 t + \frac{1}{2}\alpha t^2 = 40\pi(10) + \frac{1}{2}(-4\pi)(10)^2 = 400\pi - 200\pi = 200\pi\text{ rad}$.
Number of rotations $N = \frac{\theta}{2\pi} = \frac{200\pi}{2\pi} = 100$.