A wheel of diameter 68.0 cm slows down uniformly from 8.4 0M/s to rest over a di

Question

A wheel of diameter 68.0 cm slows down uniformly from 8.4 0M/s to rest over a distance of 115 m. A) what is the total number of revolutions at the wheel rotates before coming to rest B) what is the angular acceleration C) how long does it take for the wheel to come to a stop. A wheel of diameter 68.0 cm slows down uniformly from 8.4 0M/s to rest over a distance of 115 m. A) what is the total number of revolutions at the wheel rotates before coming to rest B) what is the angular acceleration C) how long does it take for the wheel to come to a stop. B) what is the angular acceleration C) how long does it take for the wheel to come to a stop.

Explanation / Answer

A)

Tricky question, as you don't need to know the rate of deceleration, which they give you. You need only the distance, 115 meters.

Find the circumference of the wheel, pi times the diameter, then divide 115 by that number.

The distance a wheel travels in one revolution is equal to its circumference

So,

No. of revolutions at the wheel rotates before coming to rest = 115 / (pi*d)

= 53.83181899

= 54 revolutions

B)

First find the linear acceleration using

v^2 = v0^2 + 2*a*x

v = 0

so a = -v0^2/(2*x) = -(8.40m/s)^2/(2*115m) = -0.307m/s^2

Now = a/R = -0.307/0.34 = -0.902rad/s^2

C)

As we know that

v= u + at

0 = -8.40 - 0.307*t

So,

t = 27.36156352 seconds

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