Two metal disks, one with radius R_1 = 2.39 cm and mass M_1 = 0.770 kg and the o

Question

Two metal disks, one with radius R_1 = 2.39 cm and mass M_1 = 0.770 kg and the other with radius R_2 = 4.97 cm and mass M_2 = 1.63 kg, are welded together and mounted on a frictionless axis through their common center.

Part c) Repeat the calculation of part B, this time with the string wrapped around the edge of the larger disk. (answer must be in m/s)

I've tried to get part b and c and cant. I got 3.95m/s but it was wrong. Please help!

Two metal disks, one with radius R_1 = 2.39 cm and mass M_1 = 0.770 kg and the other with radius R_2 = 4.97 cm and mass M_2 = 1.63 kg, are welded together and mounted on a frictionless axis through their common center. uploaded image Part a) What is the total moment of inertia of the two disks? The answer was 2.2310^-3 kg. m^2 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 2.09 m above the floor, what is its speed just before it strikes the floor? (answer must be in m/s) Part c) Repeat the calculation of part B, this time with the string wrapped around the edge of the larger disk. (answer must be in m/s)

Explanation / Answer

(a) Total MOI I = m1r12/2 + m2r22/2 = 22.33 kg-cm2 = 2.23*10-3 kg-m2

(b) We have mg-T = ma and T*r1 = I

But = a/r1

Thus, T*r1 = I*a/r1

Thus, T = I*a/r12

Thus, mg-I*a/r12 = ma

So, a = mg/(m+I/r12)

Putting values, a = 1.5*9.81/(1.5+2.23*10-3/(2.39*10-2)2) = 2.72 m/s2

Speed just before it strikes the floor is given by v2 = u2 + 2ah

So, v2 = 0 + 2*2.72*2.09

This gives v = 3.37 m/s

(c) Now, a = mg/(m+I/r22)

Putting values, a = 1.5*9.81/(1.5+2.23*10-3/(4.97*10-2)2) = 6.12 m/s2

Speed just before it strikes the floor is given by v2 = u2 + 2ah

So, v2 = 0 + 2*6.12*2.09

This gives v = 5.06 m/s

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