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 \(\tan\theta = \frac{1}{2\mu}\).