Two metal disks, one with radius R1 = 2.52cm and mass M1 = 0.840kg and the other

Question

Two metal disks, one with radius R1 = 2.52cm and mass M1 = 0.840kg and the other with radius R2 = 5.05cm and mass M2 = 1.59kg, are welded together and mounted on a frictionless axis through their common center. A light string is wrapped around the edge of the smaller disk and a 1.55kg block is suspended from the free end of the string.

a) What is the magnitude of the downward acceleration of the block after it is released?
Take the free fall acceleration to be 9.80 m/s^2.

Answer: ? (m/s^2)

b) Repeat the calculation of part (a), this time with the string wrapped around the edge of the larger disk.
Take the free fall acceleration to be 9.80 m/s^2.

Answer: ? (m/s^2)

Explanation / Answer

moment of inertia of smaller disc=I1=0.5*0.84*0.0252^2=2.67*10^-4 kg m^2 of larger dosc=I2=0.5*1.59*0.0505^2=2.02*10^-3 kg m^2 net inertia I=I1+I2=2.29*10^-3 kg m^2 mass of block =m mg-T=ma T*r1=I*(a/r1) a=mg/[m+I/r1^2] a=1.55*9.81/[1.55+3.6]=2.95 m/s^2 if string is wrapped around larger disc a=mg/[m+I/r2^2]=1.55*9.81/2.44=6.23 m/s^2

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