MHT CET · Physics · Motion In Two Dimensions
A disc has mass \(M\) and radius \(R\). How much tangential force should be applied to the rim of the disc, so as to rotate with angular velocity ' \(\omega\) ' in time \(t\) ?
- A \(\frac{\mathrm{MR} \omega}{4 \mathrm{t}}\)
- B \(\frac{\mathrm{MR} \omega}{2 \mathrm{t}}\)
- C \(\frac{\mathrm{MR} \omega}{\mathrm{t}}\)
- D \(M R \omega t\)
Answer & Solution
Correct Answer
(B) \(\frac{\mathrm{MR} \omega}{2 \mathrm{t}}\)
Step-by-step Solution
Detailed explanation
Torque: \(\tau=\mathrm{I} \alpha=\frac{\mathrm{MR}^2}{2} \times \frac{\omega}{\mathrm{t}}\)
\(\therefore \tau=\frac{\mathrm{MR}^2 \omega}{2 \mathrm{t}} \)
\( \text { But } \tau=\mathrm{R} \times \mathrm{F} \)
\( \therefore \mathrm{F}=\frac{\tau}{\mathrm{R}}=\frac{\mathrm{MR} \omega}{2 \mathrm{t}}\)
\(\therefore \tau=\frac{\mathrm{MR}^2 \omega}{2 \mathrm{t}} \)
\( \text { But } \tau=\mathrm{R} \times \mathrm{F} \)
\( \therefore \mathrm{F}=\frac{\tau}{\mathrm{R}}=\frac{\mathrm{MR} \omega}{2 \mathrm{t}}\)
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