Question

A weightlifter’s barbell consists of two 25-kg masses on the ends of a 15-kg rod 1.6 m long. The weightlifter holds the rod at its center and spins it at 10 rpm about an axis perpendicular to the rod. What’s the magnitude of the barbell’s angular momentum?

Answer 1

The angular momentum is

`\vec{L} = I\vec{\omega}\\`

So, we need to find the moment of inertia of the barbell.

The barbell consist of three parts, two masses of 25 kg and a rod of 15 kg.

The moment of inertia of the barbell can be found by summing the moment of inertia of the rod and two masses.

`I = I_(rod) + 2I_(mass) = (1)/(12)ML^2 + 2*mr^2 \\I = (1)/(12)15(1.6)^2 + 2(25)(0.8)^2 = 35.2`

We need to convert 10 rpm to rad/s.

`10 ~rpm = (10)/(60) ~rev/sec = (10*2\pi)/(60)~rad/sec = 1.047 ~ rad/s`

The angular momentum can now be calculated.

`L = I\omega = 35.2* 1.047 = 36.87~kg* m^2/s`