Question

Two metal disks, one with radius R1 = 2.60 cm and mass M1 = 0.900 kg and the other with radius R2 = 5.00 cm and mass M2 = 1.50 kg, are welded together and mounted on a frictionless axis through their common center

A. What is the total moment of inertia of the two disks?
B. 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. If the block is released from rest at a distance of 1.60 m above the floor, what is its speed just before it strikes the floor?
C. Repeat part B, this time with the string wrapped around the edge of the larger disk.

Answer 1

To calculate the total moment of inertia of the two disks, sum the moments of inertia of each disk. For the speed of the block before hitting the floor, apply energy conservation to account for both rotational and linear kinetic energy considering which disk the string is wrapped around.

Step-by-step explanation:

The student asks about the moment of inertia of two metal disks welded together and investigates dynamics of rotational motion when a mass is attached to one of the disks. Part A involves combining individual moments of inertia, Part B is an application of conservation of energy in rotational and linear motion, and Part C repeats Part B with the string attached to the other disk. To calculate the moment of inertia in each case, we use the formula for a solid disk, I = (1/2)MR², where M is the mass of the disk and R is the radius.

Answer for Part A

The total moment of inertia Itotal would be the sum of the moments of inertia of the two disks. For Disk 1 with mass M1 and radius R1, the moment of inertia is I1 = (1/2)M1R1². For Disk 2 with mass M2 and radius R2, the moment of inertia is I2 = (1/2)M2R2². Therefore, Itotal = I1 + I2.

Answer for Parts B & C

In Part B and C, the student is expected to use energy conservation principles to determine the speed of the falling block. The gravitational potential energy of the block converts into both rotational kinetic energy of the disks and linear kinetic energy of the block itself. For Part B, the smaller disk is involved, and in Part C, it's the larger disk. The equations for rotational and translational kinetic energy are applied to solve for the block's speed just before it hits the floor.