Question

Consider an aluminum annular disk with an outer radius 63.2 mm and inner radius 7.9 mm. The mass of the disk is 464 grams.

The disk is allowed to rotate on a frictionless table with the rotation axis at its center. The disk has a small pulley rigidly mounted at the top concentrically. The pulley's radius is 12.1 mm, and the mass of the pulley is negligible. A string is wrapped around the pulley, and a hanging mass of 19.3 g is tied at the other end of the string. When the mass falls under gravity, it causes the aluminum annular disk to rotate. Ignore the string's mass, and assume that the string's motion is frictionless.

What is the angular speed of the aluminum disk when the mass has fallen 9.1 cm?

What is the angular acceleration of the aluminum disk?

How long does it take for the mass to reach this point?

Answer 1

here,

outer radius , ro = 63.2 mm = 0.0632 m

inner radius , ri = 7.9 mm = 0.0079 m

mass of the disk , m1 = 464 g = 0.464 kg

radius of pulley r = 9.1 cm = 0.091 m

hanging mass , m2 = 19.3 g = 0.0193 kg

the accelration , a = net force /effective mass

a = ( m2 * g)/(m2 + 0.5 * m1 * (ro^2 - ri^2))

a = (0.0193 * 9.81)/(0.0193 + 0.5 * 0.464 * ( 0.0632^2 - 0.0079^2))

a = 9.37 m/s^2

the angular speed of the aluminum disk when the mass has fallen 9.1 cm , w = sqrt(2*9.37*0.091) = 1.31 rad/s

the angular acceleration of the aluminum disk, alpha = a/r = 9.37 /0.091 = 103 rad/s^2

let the time taken be t

s = 0 + 0.5 * a * t^2

0.091 = 0 + 0.5 * 9.37 * t^2

soving for t

t = 0.14 s

the time taken to reach this point is 0.14 s