Question

001 10.0 points A car accelerates uniformly from rest and reaches a speed of 25.5 m/s in 8.7 s. The diameter of a tire is 83.6 cm. Find the number of revolutions the tire makes during this motion, assuming no slip- ping. Answer in units of rev. 002 10.0 points A figure skater begins spinning counterclock- wise at an angular speed of 3.6 π rad/s. Dur- ing a 5.0 s interval, she slowly pulls her arms inward and finally spins at 8.0 π rad/s. What is her average angular acceleration during this time interval? Answer in umits of rad/s.

Answer 1

1.

Number of revolution during this motion will be given by:

Number of revolution = total displacement/distance traveled in one revolution

N = d/C

Using kinematic equations:

V = U + a*t

d = U*t + (1/2)*a*t^2

U = 0 m/sec

V = 25.5 m/sec

t = 8.7 sec

So,

a = (V - U)/t

d = U*t + (1/2)*[(V - U)/t]*t^2

d = [U + V]*t/2

d = [0 + 25.5]*8.7/2 = 110.925 m

Now, C is circumference of tire

C = 2*pi*r = pi*D

D = diameter of tire = 83.6 cm = 0.836 m

C = pi*0.836 m

So, Number of revolutions will be

N = 110.925/(pi*0.836) = 42.23 rev

N = 42 rev

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