Question 81:
easyWhat 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)
Torque is $\vec{\tau} = \vec{r} \times \vec{F}$.
$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 3 \\ 2 & -3 & 4 \end{vmatrix}$.
$= \hat{i}(8 - (-9)) - \hat{j}(12 - 6) + \hat{k}(-9 - 4) = 17\hat{i} - 6\hat{j} - 13\hat{k}$.
Question 82:
easyA 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)
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, $$Mr^2 \omega = (M + 4m)r^2 \omega' \implies \omega' = \frac{M\omega}{M + 4m}$$.
Question 83:
easyA disc is rotating with angular speed $\omega$. If a child sits on it, what is conserved:
(2002)
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.
Question 84:
easyA 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)
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 $\omega' = \frac{M\omega}{M + 2m}$.
Question 85:
easyIf a ladder is not in balance against a smooth vertical wall, then it can be made in balance by:
(1998)
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 $\theta$.
Question 86:
easyA couple produces:
(1997)
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.
Question 87:
easyFind 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)
Torque $\vec{\tau} = \vec{r} \times \vec{F}$.
Using the determinant method: $\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & 3 & 1 \\ -3 & 1 & 5 \end{vmatrix}$.
$= \hat{i}(15 - 1) - \hat{j}(35 - (-3)) + \hat{k}(7 - (-9)) = 14\hat{i} - 38\hat{j} + 16\hat{k}$.
Question 88:
easyA solid sphere is in rolling motion. In rolling motion a body possesses translational kinetic energy ($K_t$) as well as rotational kinetic energy ($K_r$) simultaneously. The ratio $K_t : (K_t + K_r)$ for the sphere is:
(2018)
For a solid sphere, $K_t = \frac{1}{2}mv^2$ and $K_r = \frac{1}{5}mv^2$. The total kinetic energy is $K_t + K_r = \frac{7}{10}mv^2$. The ratio $K_t : (K_t + K_r)$ evaluates to $5 : 7$.
Question 89:
easyA 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)
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.
Question 90:
easyA drum of radius R and mass M, rolls down without slipping along an inclined plane of angle $theta$. The frictional force:
(2005)
Static friction provides the necessary torque for rolling without slipping, converting translational kinetic energy into rotational kinetic energy without dissipating mechanical energy.