Question

Two metal disks, one with radius ?1 = 2.50 cm and mass ?1 = 0.800 kg and the other with radius ?2 = 5.00 cm and mass ?2 = 1.60 kg are welded together and mounted on a frictionless axis through their common center as shown to the right. a) A light string is wrapped around the edge of the smaller disk and a 1.50 kg block is suspended from the free end of the string. How far will the mass have to descend to give the system of disks 21.0 J of rotational kinetic energy? b) How many revolutions has the system of disks made after the mass has descended a distance of 4.00 m?

Answer 1

The situation can be visualized as

The moment of inertia of the disk system can be calculated by

Putting the values in SI units

Assuming the tension in the string is T.

The torque on the disk system is

The equation of the motion of the disk system is

And the equation of the motion of the hanging block is

Putting it in the equation of motion of the disk system

The angular acceleration in terms of linear acceleration is given by

That gives us

That gives us

PART A:

The rotational kinetic energy is given by

And the linear velocity of the hanging disk corresponding to this rotational velocity is

The linear distance traveled by the hanging mass can be calculated by

So, the mass has to descend a distance of 2.025 m to give the system of disks 21.0 J of kinetic energy.

PART B:

When the mass has descended by a distance of 4 m, the total angular distance covered is

So, the total number of revolutions is

So, the number of revolutions is 25.5.