MHT CET · Physics · Rotational Motion
A rope is wound around a solid cylinder of mass \(1 \mathrm{~kg}\) and radius \(0 \cdot 4 \mathrm{~m}\). What is the angular acceleration of cylinder, if the rope is pulled with a force of \(25 \mathrm{~N} ?\)
(cylinder is rotating about its own axis)
- A \(50 \mathrm{rad} / \mathrm{s}^{2}\)
- B \(125 \mathrm{rad} / \mathrm{s}^{2}\)
- C \(10 \mathrm{rad} / \mathrm{s}^{2}\)
- D \(1 \mathrm{rad} / \mathrm{s}^{2}\)
Answer & Solution
Correct Answer
(B) \(125 \mathrm{rad} / \mathrm{s}^{2}\)
Step-by-step Solution
Detailed explanation
Torque \(\tau=F r=25 \times 0.4=10 \mathrm{Nm}\)
\(\begin{array}{l}I=\frac{M R^{2}}{2}=\frac{1 \times(0.4)^{2}}{2}=0.08 \mathrm{~kg}-\mathrm{m}^{2} \\ \alpha=\frac{\tau}{I}=\frac{10}{0.08}=125 \mathrm{rad} / \mathrm{s}^{2}\end{array}\)
\(\begin{array}{l}I=\frac{M R^{2}}{2}=\frac{1 \times(0.4)^{2}}{2}=0.08 \mathrm{~kg}-\mathrm{m}^{2} \\ \alpha=\frac{\tau}{I}=\frac{10}{0.08}=125 \mathrm{rad} / \mathrm{s}^{2}\end{array}\)
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