The moment of inertia of a body about a given axis is $1.2 \text{ kgm}^2$. Initially, the body is at rest. In order to produce a rotational kinetic energy of $1500 \text{ joule}$, an angular acceleration of $25 \text{ rad/sec}^2$ must be applied about that axis for a duration of:
(1990)
1. $4 \text{ s}$
2. $2 \text{ s}$
3. $8 \text{ s}$
4. $10 \text{ s}$
View Answer
Using $K = \frac{1}{2} I \omega^2$, we substitute the values to get $1500 = \frac{1}{2}(1.2) \omega^2$, giving $\omega = 50 \text{ rad/s}$. Applying kinematics equation $\omega = \omega_0 + \alpha t$, we find $50 = 0 + 25t$, yielding $t = 2 \text{ s}$.
Find the torque about the origin when a force of $3 \hat{j} \text{ N}$ acts on a particle whose position vector is $2 \hat{k} \text{ m}$.
(2020)
1. $6 \hat{j} \text{ N m}$
2. $-6 \hat{i} \text{ N m}$
3. $6 \hat{k} \text{ N m}$
4. $6 \hat{i} \text{ N m}$
View Answer
Torque is given by the cross product $\vec{\tau} = \vec{r} \times \vec{F}$. Substituting the given vectors, $\vec{\tau} = (2\hat{k}) \times (3\hat{j}) = 6(\hat{k} \times \hat{j})$. Since $\hat{k} \times \hat{j} = -\hat{i}$, the torque is $-6\hat{i} \text{ N m}$.
A solid cylinder of mass $2 \text{ kg}$ and radius $4 \text{ cm}$ is rotating about its axis at the rate of $3 \text{ rpm}$. The torque required to stop after $2\pi$ revolutions is
(2019)
1. $2 \times 10^{-6} \text{ N m}$
2. $2 \times 10^{-3} \text{ N m}$
3. $12 \times 10^{-4} \text{ N m}$
4. $2 \times 10^{6} \text{ N m}$
View Answer
Here $I = \frac{1}{2}MR^2 = 1.6 \times 10^{-3} \text{ kg m}^2$, $\omega_0 = 3 \times \frac{2\pi}{60} = \frac{\pi}{10} \text{ rad/s}$, and $\theta = 4\pi^2 \text{ rad}$. Using $\omega^2 = \omega_0^2 + 2\alpha\theta$, $\alpha = -\frac{1}{800} \text{ rad/s}^2$. The required torque magnitude is $\tau = I\alpha = 2 \times 10^{-6} \text{ N m}$.
A rope is wound around a hollow cylinder of mass $3 \text{ kg}$ and radius $40 \text{ cm}$. What is the angular acceleration of the cylinder if the rope is pulled with a force of $30 \text{ N}$?
(2017-Delhi)
1. $0.25 \text{ rad/s}^2$
2. $25 \text{ rad/s}^2$
3. $5 \text{ m/s}^2$
4. $25 \text{ m/s}^2$
View Answer
For a hollow cylinder, the moment of inertia is $I = MR^2$. The torque provided by the rope is $\tau = F \times R = I\alpha$. Substituting $I$, we get $F \times R = MR^2 \alpha$, which gives $\alpha = \frac{F}{MR} = \frac{30}{3 \times 0.4} = 25 \text{ rad/s}^2$.
An automobile moves on a road with a speed of $54 \text{ km/h}$. The radius of its wheels is $0.45 \text{ m}$ and the moment of inertia of the wheel about its axis of rotation is $3 \text{ kgm}^2$. If the vehicle is brought to rest in $15 \text{ s}$, the magnitude of average torque transmitted by its brakes to wheel is:
(2015 Re)
1. $2.86 \text{ kg m}^2/\text{s}^2$
2. $6.66 \text{ kg m}^2/\text{s}^2$
3. $8.58 \text{ kg m}^2/\text{s}^2$
4. $10.86 \text{ kg m}^2/\text{s}^2$
View Answer
Initial angular velocity $\omega_0 = \frac{v}{r} = \frac{15}{0.45} = \frac{100}{3} \text{ rad/s}$. The angular acceleration is $\alpha = \frac{\omega_0}{t} = \frac{100/3}{15} = \frac{20}{9} \text{ rad/s}^2$. Torque is $\tau = I\alpha = 3 \times (\frac{20}{9}) = 6.66 \text{ kg m}^2/\text{s}^2$.
A wheel having moment of inertia $2 \text{ kg-m}^2$ about its vertical axis, rotates at the rate of $60 \text{ rpm}$ about the axis. The torque which can stop the wheel’s rotation in one minute would be:
(2004)
1. $\frac{\pi}{12} \text{ N-m}$
2. $\frac{\pi}{15} \text{ N-m}$
3. $\frac{\pi}{18} \text{ N-m}$
4. $\frac{2\pi}{15} \text{ N-m}$
View Answer
Initial angular velocity $\omega_0 = 60 \text{ rpm} = \frac{60 \times 2\pi}{60} = 2\pi \text{ rad/s}$. Final $\omega = 0$. Time $t = 60 \text{ s}$.
Angular acceleration $\alpha = \frac{\omega - \omega_0}{t} = \frac{0 - 2\pi}{60} = -\frac{\pi}{30} \text{ rad/s}^2$.
Required torque $\tau = I|\alpha| = 2 \times \frac{\pi}{30} = \frac{\pi}{15} \text{ N-m}$.
If a ladder is not in balance against a smooth vertical wall, then it can be made in balance by:
(1998)
1. Decreasing the length of ladder
2. Increasing the length of ladder
3. Increasing the angle of inclination
4. Decreasing the angle of inclination
View Answer
For equilibrium, the required frictional force at the base is $f = \frac{mg}{2} \cot\theta$, where $\theta$ is the angle of inclination with the horizontal.
To prevent slipping, $f$ must be less than or equal to the limiting friction $\mu mg$.
To decrease the required friction $f$, we must decrease $\cot\theta$, which means increasing the angle of inclination $\theta$.
A couple produces:
(1997)
1. Linear and rotational motion
2. No motion
3. Purely linear motion
4. Purely rotational motion
View Answer
A couple consists of two equal and opposite parallel forces whose lines of action do not coincide.
The net force is zero, so there is no translational (linear) acceleration.
However, there is a net torque, which produces purely rotational motion.