Two metal disks, one with radius R 1 = 2.53 cm and mass M 1 = 0.790 kg and the o

Question

Two metal disks, one with radius R1 = 2.53 cm and mass M1 = 0.790 kg and the other with radius R2 = 4.99 cm and mass M2 = 1.52 kg , are welded together and mounted on a frictionless axis through their common center. (Figure 1) .

Part A

What is the total moment of inertia of the two disks?

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Part 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 2.01 m above the floor, what is its speed just before it strikes the floor?

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Part C

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

Two metal disks, one with radius R1 = 2.53 cm and mass M1 = 0.790 kg and the other with radius R2 = 4.99 cm and mass M2 = 1.52 kg , are welded together and mounted on a frictionless axis through their common center. (Figure 1) .

Part A

What is the total moment of inertia of the two disks?

I =   kgm2  

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Part 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 2.01 m above the floor, what is its speed just before it strikes the floor?

v =   m/s  

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Part C

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

v =   m/s  

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Explanation / Answer

R1=2.53
M1=0.790
R2=4.99cm
M2=1.52kg
m=1.50
d=2.01

If the discs have a common axis, then the moments can be directly summed. The moment of a single disc is m*r²/2,
so the moment of the two discs is 0.790*0.0253²/2 + 1.52*0.0499²/2; I = 0.00215 kg-m²

Let Ft be the tension in the spring. The force balance on the hanging block (mass mb) is

mb*g - Ft = mb*a

The angular acceleration of the discs is = T/I where T = torque on the discs. T = Ft*r so = Ft*r/I but a = *r a = Ft*r²/I. Solve the above eq for Ft and insert; Ft = mb*g = mb*a

a = (mb*g - mb*a)*r²/I = g*mb*r²/I - mb*a*r²/I

a*(1 + mb*r²/I) = mb*r²/I

a = mb*r²/I / (1 + mb*r²/I) = 2.50 m/s²

h = 0.5*a*t² so the time to reach the floor is [2*h/a]. The speed is a*t, so v - [2*h*a] = 3.12 m/s

Alternate (and simpler) approach:

Initial energy of the system is mb*g*h

Final energy is 0.5*mb*v² + 0.5*I*² = v/r so

mb*g*h = 0.5*mb*v² + 0.5*I*v²/r² = v²*( 0.5*mb + 0.5*I/r²)

v = [2*mb*g*h/(mb + I/r²)]

v = 3.12m/s

calculation may be wrong but concept is right

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