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 m2. If the vehicle is brought to rest in 15 s, the magnitude of average torque transmitted by its brakes to the wheel is :

  • Option 1)

    8.58 kg m2 s-2

  • Option 2)

    10.86 kg m2 s-2

  • Option 3)

    2.86 kg m2 s-2

  • Option 4)

    6.66 kg m2 s-2

 

Answers (1)
P Prateek Shrivastava

As we learned in concept

Analogue of second law of motion for pure rotation -

vec{	au }=I, alpha

- wherein

Torque equation can be applied only about two point

(i) centre of motion.

(ii) point which has zero velocity/acceleration.

 

 V=54 Km/hr = 15m/s

 

Angular velocity of vehicle V=W0r

W_{0} = \frac{V}{R} = \frac{15}{0.45} = \frac{100}{3} rad/s

Angular Accn          \propto = \frac{\Delta W}{t} = \frac{-100}{45} \:rad/s^{2}

\vec{\tau }=I\propto

=> 3\times\frac{100}{45}=6.66\:Kg\:m^{2}s^{-2}

 

 


Option 1)

8.58 kg m2 s-2

This option is incorrect.

Option 2)

10.86 kg m2 s-2

This option is incorrect.

Option 3)

2.86 kg m2 s-2

This option is incorrect.

Option 4)

6.66 kg m2 s-2

This option is correct.

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