Question

Chapter 10, Problem 010

Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 6.00 s, it rotates 39.3 rad. During that time, what are the magnitudes of (a) the angular acceleration and (b) the average angular velocity? (c) What is the instantaneous angular velocity of the disk at the end of the 6.00 s? (d) With the angular acceleration unchanged, through what additional angle (rad) will the disk turn during the next 6.00 s?

Answer 1

here,

the time interval , t = 6 s

angular displacement , theta = 39.3 rad

a)

let the angular acceleration be alpha

using second equation of motion

theta = w0 * t + 0.5 * alpha * t^2

39.3 = 0 + 0.5 * alpha * 6^2

solving for alpha

alpha = 2.18 rad/s^2

the angular acceleration is 2.18 rad/s^2

b)

the final angular velocity , w = w0 + alpha * t

w = 0 + 2.18 * 6 rad/s = 13.1 rad/s

c)

the instantaneous angular velocity of the disk at the end of the 6 s is 13.1 rad/s

d)

for the next 6 s

initial angular velocity , w0' = w = 13.1 rad/s

the new angle covered , theta' = w0 * t + 0.5 * alpha * t^2

theta' = 13.1 * 6 + 0.5 * 2.18 * 6^2 rad = 117.84 rad

the addittional angle , dtheta = theta' - theta

dtheta = 78.54 rad