A car moves forward 10.0 m in 1.5 s. Each tire rotates through an arc length of

Question

A car moves forward 10.0 m in 1.5 s. Each tire rotates through an arc length of 10.0 m, and each car tire has a radius of 3.5 times 10^-1 m. Find the angular displacement of one of the tires. Find the average angular speed of the tire. Assume the tire starts from rest and accelerates uniformly. Find the angular acceleration of the tire. What is the instantaneous angular speed of the tire after 1.5 s? The period, T, of rotational motion is the time required for one complete revolution, or the time for the object to rotate through 2 pi rad. 2m Starting with delta theta = omega Delta t, show that T = 2 pi r/v.

Explanation / Answer

(3)

(a)formula related angualr displacment and arc is

s= r theta

theta = s/r = 10 m/3.5 * 10^-1 m = 29 rad

(b)

w = del theta/ dt = 29/1.5 = 19 rad/s

(c)

theta = wi t + 1/2 * alpha t^2

29 = 0(1.5) + 1/2 * alpha * 1.5^2

alpha = 25 rad/s^2

(d)

Apply rotational kinematic equation

wf = wi + alpha t

= 0 + 25(1.5)

=38 rad/s

4 )

w = v/r

del theta/ del t = v/r

del theta = v( del t)/ r

if del t = T

del theta = 2 pi

2 pi = vT/r

2 pi r/v= T

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