MHT CET · Physics · Rotational Motion
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 \(\frac{\mathrm{mR} \omega}{2 \mathrm{t}}\)
- B \(\mathrm{mR} \omega \mathrm{t}\)
- C \(\frac{\mathrm{mR} \omega}{4 \mathrm{t}}\)
- D \(\frac{\mathrm{mR} \omega}{\mathrm{t}}\)
Answer & Solution
Correct Answer
(A) \(\frac{\mathrm{mR} \omega}{2 \mathrm{t}}\)
Step-by-step Solution
Detailed explanation
Angular acceleration, \(\alpha=\frac{\omega}{\mathrm{t}}\)
And, moment of inertia of disc \(\mathrm{I}=\frac{1}{2} \mathrm{MR}^2\)
Hence, torque,
\(\begin{aligned} & \tau=I \cdot \alpha=\frac{1}{2} M^2 \frac{\omega}{t} c \\ & \tau=F \cdot R \\ & \Rightarrow F=\frac{\tau}{R} \\ & \Rightarrow F=\frac{M R \omega}{2 t}\end{aligned}\)
And, moment of inertia of disc \(\mathrm{I}=\frac{1}{2} \mathrm{MR}^2\)
Hence, torque,
\(\begin{aligned} & \tau=I \cdot \alpha=\frac{1}{2} M^2 \frac{\omega}{t} c \\ & \tau=F \cdot R \\ & \Rightarrow F=\frac{\tau}{R} \\ & \Rightarrow F=\frac{M R \omega}{2 t}\end{aligned}\)
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