A uniform disk with mass 38.7 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 30.0 N is applied tangent to the rim of the disk. What is the magnitude v of the tangential velocity of a point on the rim of the disk after the disk has turned through 0.320 revolution?

Explanation:

Given:

Moment of inertia of the given disk:

Also the torque due to the force is given as:

Torque in terms of angular acceleration:

where:

angular acceleration

Now using the equation of motion:

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y

1.195000

explanation:

total mass=0.0117+0.0137+0.0881+0.0111

total mass= 0.01172kg

total velocity = 0.283-0.743+0.871-0.799+0.155+0.637+0.301=0.102

momentum=mv=0.1172*0.102=1.195000//