Question

A uniform disk with mass 8.5 kg and radius 8 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 31.5N is applied to the rim of the disk. The force direction makes an angle of 35 degrees with the tangent to the rim. What is the magnitude v of the tangential velocity of a point on the rim of the disk after the disk has turned through 8.1 revolutions? The unit of the tangential velocity is m/s.

Answer 1

Mass of the disk = M = 8.5 kg

Radius of the disk = R = 8 m

Moment of inertia of the disk = I

I = MR²/2

I = 272 kg.m²

Force applied to the rim of the disk = F = 31.5 N

Angle the force makes with the tangent = = 35^o

Torque due to the force =

= FRCos

= (31.5)(8)Cos(35)

= 206.426 Nm

Angular acceleration of the disk =

I =

(272) = 206.426

= 0.759 rad/s²

Number of revolutions made by the disk = n = 8.1 rev

Angular displacement of the disk =

= 2n

= 2(8.1)

= 50.894 rad

Initial angular speed of the disk = ₁ = 0 rad/s (initially at rest)

Angular speed of the disk after 8.1 revolutions = ₂

₂² = ₁² + 2

₂² = (0)² + (2)(0.759)(50.894)

₂ = 8.79 rad/s

Tangential velocity of a point on the rim after 8.1 revolutions = V

V = ₂R

V = (8.79)(8)

V = 70.3 m/s

Magnitude of tangential velocity of a point on the rim of the disk after 8.1 revolutions = 70.3 m/s