Question

A uniform disk with mass 37.0 kg and radius 0.240 m is pivoted at its center about a horizontal, frictionless axle that is stationary. The disk is initially at rest, and then a constant force 26.5 N is applied tangent to the rim of the disk.

a) What is the magnitude v of the tangential velocity of a point on the rim of the disk after the disk has turned through 0.170 revolution?

b) What is the magnitude a of the resultant acceleration of a point on the rim of the disk after the disk has turned through 0.170 revolution?

Answer 1

Here ,

m = 37 Kg

r = 0.240 m

F = 26.5 N

a) let the angular acceleration is a

Using second law of motion

0.5 * m * r^2 * a = F * r

0.5 * 37 * 0.240 * a = 26.5

a = 5.97 rad/s^2

angular speed , w = sqrt(2 * 2pi * 0.170 * 5.97)

w = 3.57 rad/s

tangential velocity = r * w

tangential velocity = 3.57 * 0.240

tangential velocity = 0.86 m/s

b)

for the resultant acceleration

resultant acceleration = sqrt(ac^2 + at^2)

resultant acceleration = sqrt((5.97 * 0.240)^2 + (0.86^2/0.240)^2)

resultant acceleration = 3.4 m/s^2

the resultant acceleration is 3.4 m/s^2