Two metal disks one with radius R1 235 cm and mass M1 0820

Two metal disks, one with radius R1 = 2.35 cm and mass M1 = 0.820 kg and the other with radius R2 = 5.02 cm and mass M2 = 1.62 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, suspended from the free end of the string. If the block is released from rest at a distance of 2.05 m above the floor, what is its speed just before it strikes the floor?

C: Repeat the calculation of part B, this time with the string wrapped around the edge of the larger disk.

Solution

a)

Total moment of inertia of two disks

I=(1/2)M₁R₁²+(1/2)M₂R₂²

I=(1/2)[0.82*0.0235²+1.62*0.0502²]

I=2.27*10^-3 Kg-m²

b)

By Conservation of energy

E_f-E_i =W=0

E_i=E_f

=>KE_disks,i+KE_block,i+U_block,i =KE_disks,f+KE_block,f+U_block,f

0+0+M₃gh =(1/2)IW_disk²+(1/2)M₃V_block²+0

Since W_disk=V_block/R₁

M₃gh =(1/2)I(V_block/R₁)²²+(1/2)M₃V_block²

V_block² =2gh/[1+(I/M₃R₁²)

V_block²=2*9.8*2.05/[1+(2.27*10^-3/1.5*0.0235²)]

V_block=3.28 m/s

c)

W_disk=V_block/R₂

M₃gh =(1/2)I(V_block/R₂)²²+(1/2)M₃V_block²

V_block² =2gh/[1+(I/M₃R₂²)

V_block²=2*9.8*2.05/[1+(2.27*10^-3/1.5*0.0502²)]

V_block=5.01 m/s