COMEDK · Physics · 8. Rotational Motion
A motor bike is moving with a speed of \(36 \mathrm{~km} / \mathrm{h}\) along a straight road. The radius of its wheel and its moment of inertia about the axis of rotation are 40 cm and \(5 \mathrm{~kg} \mathrm{~m}^2\) respectively. What is the magnitude of the torque applied by the brake to the wheel so as to stop the bike in 10 s?
- A \(12.5 \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-2}\)
- B \(0.125 \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-2}\)
- C \(125 \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-2}\)
- D \(1.25 \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-2}\)
Answer & Solution
Correct Answer
(A) \(12.5 \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-2}\)
Step-by-step Solution
Detailed explanation
The speed of the motorbike is \(v = 36 \text{ km/h} = 36 \times \dfrac{5}{18} \text{ m/s} = 10 \text{ m/s}\). The radius of the wheel is \(r = 40 \text{ cm} = 0.4 \text{ m}\). The angular velocity of the wheel is \(\omega = \dfrac{v}{r} = \dfrac{10}{0.4} = 25 \text{ rad/s}\).…
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