(A) The net torque $\tau$ about the pivot point $P$ is given by the difference in torques due to the two masses.$\tau = (5M_0g)(l) - (2M_0g)(2l) = 5M_

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(A) The net torque $\tau$ about the pivot point $P$ is given by the difference in torques due to the two masses.$\tau = (5M_0g)(l) - (2M_0g)(2l) = 5M_0gl - 4M_0gl = M_0gl$.The moment of inertia $I$ about the pivot point $P$ is the sum of the moments of inertia of the two masses.$I = (5M_0)(l)^2 + (2M_0)(2l)^2 = 5M_0l^2 + 8M_0l^2 = 13M_0l^2$.The angular acceleration $\alpha$ is given by $\alpha = \frac{\tau}{I}$.Substituting the values,$\alpha = \frac{M_0gl}{13M_0l^2} = \frac{g}{13l}$.

Class 11 Physics System of Particles and Rotational Motion Relation between Torque and Angular acceleration and it's Application

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$A$ rigid massless rod of length $3l$ has two masses attached at each end as shown in the figure. The rod is pivoted at point $P$ on the horizontal axis (see figure). When released from the initial horizontal position,its instantaneous angular acceleration will be

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A
$\frac{g}{13l}$
B
$\frac{g}{3l}$
C
$\frac{g}{2l}$
D
$\frac{7g}{3l}$

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Class 11

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System of Particles and Rotational Motion

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Relation between Torque and Angular acceleration and it's Application

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DifficultMHT CET 2022

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$A$ wheel having a moment of inertia $2 \; kg \cdot m^2$ about its vertical axis rotates at the rate of $60 \; rpm$ about this axis. The torque required to stop the wheel's rotation in one minute is:

MediumAIPMT 2004

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The moment of inertia of a wheel about an axis passing through its center is $200 \, kg \cdot m^2$. $A$ constant torque of $1000 \, N \cdot m$ is applied to rotate the wheel. After $3 \, s$,the angular velocity of the wheel will be ........ $rad/s$.

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$A$ rigid massless rod of length $3l$ has two masses attached at each end as shown in the figure. The rod is pivoted at point $P$ on the horizontal axis (see figure). When released from the initial horizontal position,its instantaneous angular acceleration will be

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$A$ disc of mass $25 \ kg$ and radius $0.2 \ m$ is rotating at $240 \ r.p.m.$ $A$ retarding torque brings it to rest in $20 \ s$. If the torque is due to a force applied tangentially on the rim of the disc,then the magnitude of the force in newton is

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