MHT CET · Physics · Rotational Motion
A disc of radius \(0.4 \mathrm{~m}\) and mass \(1 \mathrm{~kg}\) rotates about an axis passing through its center and perpendicular to its plane. The angular acceleration of the disc is \(10 \mathrm{rad} / \mathrm{s}^2\). The tangential force applied to the rim of the disc is
- A \(4 \mathrm{~N}\)
- B \(1 \mathrm{~N}\)
- C \(2 \mathrm{~N}\)
- D \(8 \mathrm{~N}\)
Answer & Solution
Correct Answer
(C) \(2 \mathrm{~N}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{R}=0.4 \mathrm{~m}, \mathrm{M}=1 \mathrm{~kg}, \alpha=10 \mathrm{rad} / \mathrm{s}^2\)
Moment of inertia, \(\mathrm{I}=\frac{\mathrm{MR}^2}{2}=\frac{1 \times(0.4)^2}{2}=0.08 \mathrm{~kg} \mathrm{~m}^2\)
Torque, \(\tau=\mathrm{RF}=\mathrm{I} \alpha\)
\(\therefore \mathrm{F}=\frac{\mathrm{I} \alpha}{\mathrm{R}}=\frac{0.08 \times 10}{0.4}=2 \mathrm{~N}\)
Moment of inertia, \(\mathrm{I}=\frac{\mathrm{MR}^2}{2}=\frac{1 \times(0.4)^2}{2}=0.08 \mathrm{~kg} \mathrm{~m}^2\)
Torque, \(\tau=\mathrm{RF}=\mathrm{I} \alpha\)
\(\therefore \mathrm{F}=\frac{\mathrm{I} \alpha}{\mathrm{R}}=\frac{0.08 \times 10}{0.4}=2 \mathrm{~N}\)
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