A uniform disc of radius R and mass M can rot | vedclass

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

Vedclass · Accepted Answer

$\frac{2F}{M}$

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(A) The torque $	au$ applied by the force $F$ at the rim is given by $	au = 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 $	au = 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 $\frac{2F}{M}$.

$A$ 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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