Rotational Motion Question 87
Question: An automobile moves on a road with a speed of $ 54km{h^{-1}} $ . The radius of its wheels is 0.45 m and the moment of inertia of the wheel about its axis of rotation is $ 3kgm^{2} $ . If the vehicle is brought to rest in 15s, the magnitude of average torque transmitted by its brakes to the wheel is
Options:
A) $ 6.66kgm^{2}{s^{-2}} $
B) $ 8.58kgm^{2}{s^{-2}} $
C) $ 10.86kgm^{2}{s^{-2}} $
D) $ 2.86kgm^{2}{s^{-2}} $
Show Answer
Answer:
Correct Answer: A
Solution:
-
As velocity of an automobile vehicle, $ v=54km/hr=54\times \frac{5}{18}=15m/s $
Angular velocity of a vehicle, $ v={\omega_0}r $
$ \Rightarrow $ $ {\omega_0}=\frac{v}{R}=\frac{15}{0.45}=\frac{100}{3}rad/s $
So, angular acceleration of an automobile,
$ \alpha =\frac{\Delta \omega }{t}=\frac{{\omega_t}-{\omega_0}}{t}=\frac{0-\frac{100}{3}}{15}=\frac{-100}{45}rad/s^{2} $
Thus, average torque transmitted by its brakes to wheel
$ \tau =I\alpha $
$ \Rightarrow $ $ 3\times \frac{100}{45}=6.66kgm^{2}{s^{-2}} $
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