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 tr

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Solution

Q.
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]

1 $10.86 {kg m}^{2} s^{- 2}$

2 $2.86 {kg m}^{2} s^{- 2}$

3 $6.66 {kg m}^{2} s^{- 2}$

4 $8.58 {kg m}^{2} s^{- 2}$

Ans.
(3)

Here, speed of the automobile,

$v = 54 {km h}^{- 1} = 54 \times \frac{5}{18} {m s}^{- 1} = 15 {m s}^{- 1}$

Radius of the wheel of the automobile, R = 0.45 m

Moment of inertia of the wheel about its axis of rotation, $I = 3 {kg m}^{2}$

Time in which the vehicle is brought to rest, $t = 15 s$

The initial angular speed of the wheel is

$ω_{i} = \frac{v}{R} = \frac{15 {m s}^{- 1}}{0.45 m} = \frac{1500}{45} {rad s}^{- 1} = \frac{100}{3} {rad s}^{- 1}$

and its final angular speed is

$ω_{f} = 0 (as the vehicle comes to rest)$

$∴$ The angular retardation of the wheel is

$α = \frac{ω_{f} - ω_{i}}{t} = \frac{0 - \frac{100}{3}}{15 s} = - \frac{100}{45} {rad s}^{- 2}$

The magnitude of required torque is

$τ = I | α | = (3 {kg m}^{2}) (\frac{100}{45} {rad s}^{- 2})$

$= \frac{20}{3} {kg m}^{2} s^{- 2} = 6.66 {kg m}^{2} s^{- 2}$

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