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 $\vec{\tau} = \vec{r} \times \vec{F}$.
$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 3 \\ 2 & -3 & 4 \end{vmatrix}$.
$= \hat{i}(8 - (-9)) - \hat{j}(12 - 6) + \hat{k}(-9 - 4) = 17\hat{i} - 6\hat{j} - 13\hat{k}$.
A thin circular ring of mass $M$ and radius ‘$r$’ is rotating about its axis with a constant angular velocity $\omega$. Four objects each of mass $m$, are kept gently to the opposite ends of two perpendicular diameters of the ring. The angular velocity of the ring will be:
(2003)
1. $\frac{M\omega}{4m}$
2. $\frac{M\omega}{M + 4m}$
3. $\frac{(M + 4m)\omega}{M}$
4. $\frac{(M + 4m)\omega}{M + 4m}$
View Answer
By conservation of angular momentum, $I_{initial}\omega_{initial} = I_{final}\omega_{final}$.
Initial moment of inertia $I_i = Mr^2$. Final moment of inertia $I_f = Mr^2 + 4mr^2$.
Thus, $$Mr^2 \omega = (M + 4m)r^2 \omega' \implies \omega' = \frac{M\omega}{M + 4m}$$.
A disc is rotating with angular speed $\omega$. If a child sits on it, what is conserved:
(2002)
1. Linear momentum
2. Angular momentum
3. Kinetic energy
4. Potential energy
View Answer
When the child sits on the rotating disc gently, no external torque acts on the system.
According to Newton's second law for rotation, if net external torque is zero, the total angular momentum of the system remains conserved.
A circular ring of mass $M$ and radius $R$ is rotating about its axis with constant angular velocity $\omega$. Two particles each of mass $m$ are attached gently to the opposite ends of a diameter of the ring. The angular velocity of the ring will now become:
(1998)
1. $\frac{M - 2m}{m\omega}$
2. $\frac{m\omega}{M - 2m}$
3. $\frac{M\omega}{M + 2m}$
4. $\frac{M - 2m}{m}$
View Answer
Since the particles are attached gently, external torque is zero, meaning angular momentum is conserved.
$I_1\omega_1 = I_2\omega_2 \implies (MR^2)\omega = (MR^2 + 2mR^2)\omega'$.
Solving for the new angular velocity yields $\omega' = \frac{M\omega}{M + 2m}$.
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: $\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & 3 & 1 \\ -3 & 1 & 5 \end{vmatrix}$.
$= \hat{i}(15 - 1) - \hat{j}(35 - (-3)) + \hat{k}(7 - (-9)) = 14\hat{i} - 38\hat{j} + 16\hat{k}$.
A solid cylinder and a hollow cylinder, both of the same mass and same external diameter are released from the same height at the same time on an inclined plane. Both roll down without slipping. Which one will reach the bottom first?
(2010 Mains)
1. Both together only when angle of inclination of plane is $45^{\circ}$
2. Both together
3. Hollow cylinder
4. Solid cylinder
View Answer
The acceleration of a rolling body depends on its moment of inertia ratio $I/mR^2$. The solid cylinder has a smaller moment of inertia ratio ($1/2$) compared to the hollow cylinder ($1$), giving it a higher acceleration and causing it to reach the bottom first.
A drum of radius R and mass M, rolls down without slipping along an inclined plane of angle $theta$. The frictional force:
(2005)
1. Converts translational energy to rotational energy
2. Dissipates energy as heat
3. Decreases the rotational motion
4. Decreases the rotational and translational motion
View Answer
Static friction provides the necessary torque for rolling without slipping, converting translational kinetic energy into rotational kinetic energy without dissipating mechanical energy.