(A) The torque $\tau$ about the hinged end is given by $\tau = Mg \frac{L}{2}$.The moment of inertia $I$ of a uniform rod of mass $M$ and length $L$ a

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(A) The torque $\tau$ about the hinged end is given by $\tau = Mg \frac{L}{2}$.The moment of inertia $I$ of a uniform rod of mass $M$ and length $L$ about an axis passing through one end is $I = \frac{ML^2}{3}$.Using the rotational analogue of Newton's second law,$\tau = I \alpha$,where $\alpha$ is the angular acceleration:$Mg \frac{L}{2} = \left( \frac{ML^2}{3} \right) \alpha$.Solving for $\alpha$:$\alpha = \frac{MgL/2}{ML^2/3} = \frac{3g}{2L}$.Given the length of the metre stick is $L = 1 \ m$,we substitute $L = 1$:$\alpha = \frac{3g}{2(1)} = \frac{3}{2} g \ rad/s^2$.

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

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$A$ uniform metre stick of mass $M$ and length $L = 1 \ m$ is hinged at one end and supported in a horizontal direction by a string attached to the other end. What should be the initial angular acceleration of the free end of the stick if the string is cut? (in $rad/s^2$)

A
$\frac{3g}{2L}$
B
$g$
C
$3g$
D
$4g$

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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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$A$ uniform metre stick of mass $M$ and length $L = 1 \ m$ is hinged at one end and supported in a horizontal direction by a string attached to the other end. What should be the initial angular acceleration of the free end of the stick if the string is cut? (in $rad/s^2$)

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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 is:

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