Torque

Question

A uniform disc of radius ‘R” and mass ‘M’ can rotate without friction on an axis passing through its centre and perpendicular to its plane. A cord is wound at the rim of the disc and a uniform force of F is applied on the cord. The tangential acceleration of a point on the rim of the disc is

Easy

Solution

$\mathrm{FR}=\mathrm{I\alpha}=\frac{{\mathrm{MR}}^{2}}{2}\frac{\mathrm{a}}{\mathrm{R}}$

$\mathrm{a}=\frac{2\mathrm{F}}{\mathrm{m}}$

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