(A) The torque $\tau$ applied by the force $F$ at the rim is given by $\tau = F R$.The moment of inertia $I$ of a uniform disc about an axis passing t

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(A) The torque $\tau$ applied by the force $F$ at the rim is given by $\tau = F R$.The moment of inertia $I$ of a uniform disc about an axis passing through its centre and perpendicular to its plane is $I = \frac{1}{2} M R^2$.Using the relation $\tau = I \alpha$,where $\alpha$ is the angular acceleration:$F R = (\frac{1}{2} M R^2) \alpha$$\alpha = \frac{F R}{\frac{1}{2} M R^2} = \frac{2F}{M R}$.The tangential acceleration $a_t$ at the rim is given by $a_t = \alpha R$.Substituting the value of $\alpha$:$a_t = (\frac{2F}{M R}) R = \frac{2F}{M}$.Thus,the tangential acceleration is $\frac{2F}{M}$.

Class 11 Physics System of Particles and Rotational Motion Rotation Motion Basic, Motion of Connected Mass

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$A$ uniform disc of radius $R$ and mass $M$ can rotate on a smooth axis passing through its centre and perpendicular to its plane. $A$ force $F$ is applied on its rim as shown in the figure. What is the tangential acceleration of the rim?

A
$\frac{2F}{M}$
B
$\frac{F}{M}$
C
$\frac{F}{2M}$
D
$\frac{F}{4M}$

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System of Particles and Rotational Motion

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Rotation Motion Basic, Motion of Connected Mass

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$A$ uniform disc of radius $R$ and mass $M$ can rotate on a smooth axis passing through its centre and perpendicular to its plane. $A$ force $F$ is applied on its rim as shown in the figure. What is the tangential acceleration of the rim?

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$A$ tube of length $L$ is filled completely with an incompressible liquid of mass $M$ and closed at both ends. The tube is then rotated in a horizontal plane about one of its ends with a uniform angular velocity $\omega$. The force exerted by the liquid on the tube at the other end is

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