System of Particles and Rotational Motion - Result Question 25
25. An automobile moves on a road with a speed of 54 $km h^{-1}$. The radius of its wheels is $0.45 m$ and the moment of inertia of the wheel about its axis of rotation is $3 kg m^{2}$. If the vehicle is brought to rest in $15 s$, the magnitude of average torque transmitted by its brakes to the wheel is:
[2015 RS]
(a) $8.58 kg m^{2} s^{-2}$
(c) $2.86 kg m^{2} s^{-2}$
(b) $10.86 kg m^{2} s^{-2}$
(d) $6.66 kg m^{2} s^{-2}$
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Answer:
Correct Answer: 25. (d)
Solution:
- (d) Given : Speed $v=54 kmh^{-1}=15 ms^{-1}$
Moment of inertia, $I=3 kgm^{2}$
Time $t=15 s$
$\omega_i=\frac{v}{r}=\frac{15}{0.45}=\frac{100}{3}$
[Given $\omega_f=0$ ]
$\omega_f=\omega_i+\alpha t$
$0=\frac{100}{3}+(-\alpha)(15) \quad \Rightarrow \alpha=\frac{100}{45}$
Average torque transmitted by brakes to the wheel,
$\tau=(I)(\alpha)=3 \times \frac{100}{45}=6.66 kgm^{2} s^{-2}$
