An automobile moves on a road with a speed of 54kmh^ - 1 . The radius of its wheels is 0.45m and the moment of inertia of the wheel about its axis of rotation is 3kgm^ 2 . If the vehicle is brought to rest in 15 s , the

Correct answer: D. 6.66kg m ^ 2 s ^ - 2 ω _ i = (15)/(0.45) = (100)/(3) ω _ f = 0 ω _ f = ω _ i + 0 = (100)/(3) + (- α) ( 15 ) α = (100)/(45) τ = (I)(α) = 3 × (100)/(45) = 6.66Nm

NEET 2015

Exam: NEET
Year: 2015
Subject: Physics
Chapter: Rotational Motion

NEET 2015 · Physics · Rotational Motion

An automobile moves on a road with a speed of 54kmh^ - 1 . The...

An automobile moves on a road with a speed of

54kmh^{- 1}

. The radius of its wheels is

0.45\text{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

\text{15 s}

, the magnitude of average torque transmitted by its brakes to the wheel is:

Options

A $8.58kg{m}^{2}{s}^{- 2}$
B $10.86kg{m}^{2}{s}^{- 2}$
C $2.86kg{m}^{2}{s}^{- 2}$
D $6.66kg{m}^{2}{s}^{- 2}$

Correct answer

D. $6.66kg{m}^{2}{s}^{- 2}$

Step-by-step solution

{\omega}_{i} = \dfrac{15}{0.45} = \dfrac{100}{3}{\omega}_{f} = 0

{\omega}_{f} = {\omega}_{i} + \alphat

0 = \dfrac{100}{3} + \left(- \alpha\right) ( 15 )

\alpha = \dfrac{100}{45}

τ = \left(I\right)\left(\alpha\right) = 3 \times \dfrac{100}{45} = 6.66Nm

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