Question

The mass of the disk is 0.35 kg with a radius of 0.18 m and is initially spinning at 5 revolutions per second. It is spinning horizontally and on a very low friction axle.

Then a uniform thin rod of length 0.40m and mass of 0.085 kg is suspended just above the spinning disk and dropped, without any rotational motion, onto the spinning disk exactly across its diameter.

After the rod is dropped, the rod and disk are moving with the same rotational speed.

a) What is the rotational speed? Show all work and explain.

b) What is the total kinetic energy of the system before and after the rod is dropped? Show all work and explain.

Answer 1

Moment of inertia of disk I_d = MR^2/2 = 0.35 times (0.18)^2/2 = 5.67 times 10^-3 kg-m^2 Moment of inertia of rod about centre of rotation I_r = mr^2 = m times 0 = 0 kg - m^2 Moment of inertia of system I = I_d + I_r = 5.67 times 10^-3 + 0 I = 5.67 times 10^-3 kg-m^2 w = 2I times 5 = 31.4I rad/s (a) since moment of inertia does not change therefore rotational speed will remain constant w = 31.42 rad/s (b) KE will also be unchanged KE = 1/2 IW^2 = 1/2 times 5.67 times 10^-3 times (31.42)^2 = 2.799 J KE = 2.80 J