A wheel having moment of inertia 2 kg-m² about its vertical axis, rotates at the rate of 60 rpm about this axis. The torque which can stop the wheel’s rotation in one minute would be
ω = ω0 + α.t
0= (60 × 2π /60) + α.60
⇒ α = - π/30
Torque = I α = (2× π/30) = (π/15) N-m
The figure shows a horizontal block of mass M suspended by two wires A and B. The centre of mass of the block is closer to B than A. (i) Is the magnitude of the torque due to wire A is greater, less or equal to that due to B w.r.t. centre of mass ? (ii) Which wire A or B exerts more force on the block ?
As the object is in rotational equilibrium, Net torque acting on the object is zero.
so, Torque of TA = Torque of TB
TAXA= TBXB
\[ \frac{T_{A}}{T_{B}}=\frac{X_{B}}{X_{A}} \]
\[ X_{B} < X_{A} \]
\[ T_{A} < T_{B} \]
A ladder of length \(\ell\) and mass m is placed against a smooth vertical wall but the ground is not smooth. Coefficient of friction between the ground and the ladder is \(\mu\). The minimum angle \(\theta\) with ground at which the ladder will stay in equilibrium is :
For translational and rotational equilibrium of the ladder, taking torque about the base gives \(N_{\text{wall}} \ell \sin\theta = mg \frac{\ell}{2} \cos\theta\). With \(N_{\text{wall}} = f \le \mu mg\), we get the minimum angle for equilibrium to be \(\tan\theta = \frac{1}{2\mu}\).
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)
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)
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 $\alpha = \frac{F}{MR} = \frac{30}{3 \times 0.4} = 25 \text{ rad/s}^2$.
A solid cylinder of mass $50\text{ kg}$ and radius $0.5\text{ m}$ is free to rotate about the horizontal axis. A massless string is wound round the cylinder with one end attached to it and other hanging freely. Tension in the string required to produce an angular acceleration of $2\text{ rev/s}^{2}$ is:
(2014)
Moment of inertia $I = \frac{1}{2}MR^{2} = 6.25\text{ kg m}^{2}$. Angular acceleration $\alpha = 2\text{ rev/s}^{2} = 4\pi\text{ rad/s}^{2}$. Torque $\tau = I\alpha = 25\pi = 78.5\text{ N m}$. Tension $T = \frac{\tau}{R} = \frac{78.5}{0.5} = 157\text{ N}$.
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)
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 $\tau = I|\alpha| = 2 \times \frac{\pi}{30} = \frac{\pi}{15} \text{ N-m}$.