AP EAMCET · PHYSICS · Rotational Motion
A circular disc of diameter 0.8 m and mass 4 kg is rolling on a smooth horizontal plane. If 2.56 N m torque is acting on the disc, then its angular acceleration is
- A \(8 \mathrm{rad} \mathrm{s}^{-2}\)
- B \(4 \mathrm{rad} \mathrm{s}^{-2}\)
- C \(2 \mathrm{rad} \mathrm{s}^{-2}\)
- D \(16 \mathrm{rad} \mathrm{s}^{-2}\)
Answer & Solution
Correct Answer
(A) \(8 \mathrm{rad} \mathrm{s}^{-2}\)
Step-by-step Solution
Detailed explanation
\(r = D/2 = 0.8/2 = 0.4 \text{ m}\) \(I = \frac{1}{2} m r^2 = \frac{1}{2} (4)(0.4)^2 = 2(0.16) = 0.32 \text{ kg m}^2\) \(\alpha = \tau/I = 2.56/0.32 = 8 \text{ rad s}^{-2}\)
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