COMEDK2025PhysicsRotational MotionActual
A motor bike is moving with a speed of 36 ~km / 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 ~kg ~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?
Options
- A12.5 ~kg ~m ^2 ~s ⁻²
- B0.125 ~kg ~m ^2 ~s ⁻²
- C125 ~kg ~m ^2 ~s ⁻²
- D1.25 ~kg ~m ^2 ~s ⁻²
Correct answer
A. 12.5 ~kg ~m ^2 ~s ⁻²
Step-by-step solution
The speed of the motorbike is v = 36 km/h = 36 5 18 m/s = 10 m/s . The radius of the wheel is r = 40 cm = 0.4 m . The angular velocity of the wheel is = v r = 10 0.4 = 25 rad/s . The wheel is to be stopped in t = 10 s , so the final angular velocity is _f = 0 . The angular acceleration is given by = _f - t = 0 - 25 10 = -2.5 rad/s ^2 . The magnitude of the angular acceleration is | | = 2.5 rad/s ^2 . The torque applied by the brake is given by = I | | , where I = 5 kg m ^2 . Substituting the values, = 5 2.5 = 12.5