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  • A wheel on the road moves with a speed of 72km/hr. Its radius is '20cm' and the moment of inertia about the axis of rotation is '1kg<img alt="m^{2}" src="https://entrancecorner.oncodeco

A wheel on the road moves with a speed of 72km/hr. Its radius is '20cm' and the moment of inertia about the axis of rotation is '1kgm^{2}'. Find the torque applied by the breaks if the wheel stops in 10s after the application of the breaks.

Option: 1

4 \mathrm{~kg}\mathrm{m}^2s^{-2}


Option: 2

8 \mathrm{~kg}\mathrm{m}^2s^{-2}


Option: 3

12 \mathrm{~kg}\mathrm{m}^2s^{-2}


Option: 4

10 \mathrm{~kg}\mathrm{m}^2s^{-2}


Answers (1)

best_answer

\begin{aligned} & \omega_i=\frac{v}{R}=\frac{20 \mathrm{~m} / \mathrm{s}}{0.2 \mathrm{~m}}=100 \mathrm{rad} / \mathrm{s} \\ & \omega_f=0 \\ & \alpha=\frac{\omega_f-\omega_i}{t} \end{aligned}

\tau=I \alpha=I\left(\frac{\omega_f-\omega_i}{t}\right)

\tau=\frac{I(0-100)}{10} = 10 \mathrm{kgm}^2 \mathrm{~s}^{-2}

 

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