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.
Find the torque of a force $\vec{F} = -3\hat{i} + \hat{j} + 5\hat{k}$ acting at the point $\vec{r} = 7\hat{i} + 3\hat{j} + \hat{k}$
(1997)
1. $-21\hat{i} + 4\hat{j} + 4\hat{k}$
2. $-14\hat{i} + 34\hat{j} - 16\hat{k}$
3. $14\hat{i} - 38\hat{j} + 16\hat{k}$
4. $4\hat{i} + 4\hat{j} + 6\hat{k}$
View Answer
Torque $\vec{\tau} = \vec{r} \times \vec{F}$.
Using the determinant method: $\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & 3 & 1 \\ -3 & 1 & 5 \end{vmatrix}$.
$= \hat{i}(15 - 1) - \hat{j}(35 - (-3)) + \hat{k}(7 - (-9)) = 14\hat{i} - 38\hat{j} + 16\hat{k}$.
What is torque of the force $\vec{F} = 2\hat{i} – 3\hat{j} + 4\hat{k}$ acting at the point $\vec{r} = 3\hat{i} + 2\hat{j} + 3\hat{k}$ about origin?
(1995)
1. $-6\hat{i} + 6\hat{j} - 12\hat{k}$
2. $-17\hat{i} + 6\hat{j} + 13\hat{k}$
3. $6\hat{i} - 6\hat{j} + 12\hat{k}$
4. $17\hat{i} - 6\hat{j} - 13\hat{k}$
View Answer
Torque is $\vec{\tau} = \vec{r} \times \vec{F}$.
$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 3 \\ 2 & -3 & 4 \end{vmatrix}$.
$= \hat{i}(8 - (-9)) - \hat{j}(12 - 6) + \hat{k}(-9 - 4) = 17\hat{i} - 6\hat{j} - 13\hat{k}$.