Question #e264b

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Question #e264b

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Answer 1 · 2017-01-21T12:23:40

Velocity of the automobile $v = 80 km/h = \frac{80 \times 10^{3} m}{60 \times 60 s} = \frac{200}{9} m/s$ $v=80"km/h"=(80xx10^3m)/(60xx60s)=200/9"m/s"$

Radius of Tyre $r = 80 c m = 0.8 m$ $r=80cm =0.8m$

Initial angular velocity of wheels

$ω_{0} = \frac{v}{r} = \frac{200}{0.8 \times 9} rad/s = \frac{250}{9} rad/s$ $omega_0=v/r=200/(0.8xx9)"rad/s"=250/9"rad/s"$

On application of brakes the car stops in 30 complete turns. So the angular displacement of wheels $θ = 30 \times 2 π = 60 π$ $theta =30xx2pi=60pi$ radian.
Final angular velocity of wheel $ω = 0$ $omega =0$

If angular acceleration be $α$ $alpha$ then we can write from equation of rotational motion

$ω^{2} = ω_{0}^{2} + 2 α \times θ$ $omega^2=omega_0^2+2alphaxxtheta$

Inserting values we get

$0^{2} = {(\frac{250}{9})}^{2} + 2 \times α \times 60 π$ $0^2=(250/9)^2+2xxalphaxx60pi$

$\Rightarrow α = - \frac{250 \times 250}{9 \times 9 \times 120 π} \approx - 2 rad/ s^{2}$ $=>alpha=-(250xx250)/(9xx9xx120pi)~~-2"rad/"s^2$

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