(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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$A$ 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

Similar Questions

$A$ solid disc has a diameter of $0.5 \ m$ and a mass of $16 \ kg$. What torque must be applied to increase its angular velocity from $0$ to $120 \ rpm$ in $8 \ s$?

$A$ uniform disc rotates at $10$ revolutions per second. $A$ torque is applied to it, producing an angular acceleration of $5 \text{ rad s}^{-2}$. After $2 \text{ s}$, its angular velocity is ...... $\text{rad s}^{-1}$ and the number of revolutions completed by the disc in $2 \text{ s}$ is ...... .

The instantaneous angular position of a point on a rotating wheel is given by the equation $\theta(t) = 2t^3 - 6t^2$. The torque on the wheel becomes zero at $t = $ ...... $s$.

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