A wheel is rotating at a rate of 120 revolutions per minute. It slows down at a

Question

A wheel is rotating at a rate of 120 revolutions per minute. It slows down at a rate of 2.00 revolutions per minute per minute; in other words, at any given time it is going two revolutions per minute slower than it was a minute earlier. How many revolutions does it make before coming to a stop? Suppose the length of a simple pendulum increases by one-tenth of one percent as a result of thermal expansion. By what percent does the period of the pendulum increase? Use the equation t = 2PI SquarerootL/g.

Explanation / Answer

Here,

angular speed , wi = 120 rev/min

angular acceleration , a = - 2 rev/min^2

let the number of revolutions is theta

Using third equation of motion

wf^2 - wi^2 = 2 * a * theta

0 - 120^2 = -2 * 2 * theta

theta = 3600 rev

the wheel makes 3600 revs before stopping.

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as T = 2pi * sqrt(L/g)

for increasing length by 10 percent , L1 = 1.1 * L

T1 = 2pi * sqrt(L1/g)

T1 = 2pi * sqrt(1.1) * sqrt(L/g)

T1 = 1.049 * 2pi * sqrt(L/g)

T1 = 1.049 * T

increase in period of pendulum = 4.9 %

the increase in period of pendulum is 4.9 %

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