Question

A uniform disk with mass 38.9 kgkg and radius 0.300 mm is pivoted at its center about a horizontal, frictionless axle that is stationary. The disk is initially at rest, and then a constant force 31.5 NN is applied tangent to the rim of the disk.

1. What is the magnitude v of the tangential velocity of a point on the rim of the disk after the disk has turned through 0.260 revolution?

Express your answer with the appropriate units.

2. What is the magnitude aa of the resultant acceleration of a point on the rim of the disk after the disk has turned through 0.260 revolution?

Express your answer with the appropriate units.

Answer 1

answer) from angular kinematics we have

F*r=I*

=F*r/I

I=moment of inertia=1/2mr²=0.5*38.9*0.3m²=1.75 kgm²

=31.5*0.3/1.75=5.40 rad/s²

a) again from kinematics we have

w²=wo²+2........1)

wo=0

=0.260 rev=2pi*0.260=1.63 rad

w²=2*5.40*1.63

w=17.6=4.20

we know that angular velocity

v=r*w=0.3*4.20=1.26 m/s

so answer to part a) 1.26 m/s

b) for this part we have

=x²+y²

x=tangential acceleration=r*=0.3*5.40=1.62 m/s²

y=radial=r*w²=0.3*4.20²=5.29 m/s²

so we have

=1.62²+5.29²=5.53 m/s²

so answer is 5.53 m/s²