# Physichs

Two metal disks, one with radius R1 = 2.40 cm and mass M1 = 0.900 kg and the other with radius R2 = 4.95 cm and mass M2 = 1.50 kg , 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 1.55 kg block is suspended from the free end of the string.

What is the magnitude of the downward acceleration of the block after it is released?
Take the free fall acceleration to be 9.80m/s2.

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 9.80m/s2.

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1. mass of first disk = M1 = 0.750 kg

radius of first disk = R1= 2.45 cm = 0.0245 m

mass of second disk = M2 = 1.55 kg

radius of second disk = R2= 5.10 cm = 0.0510 m

moment of inertia of first disk = I1= (0.5) M1 R12 = (0.5) (0.750) (0.0245)2 = 2.251 x 10-4 kgm2

moment of inertia of second disk = I2= (0.5) M2 R22 = (0.5) (1.55) (0.0510)2 = 2.016 x 10-3 kgm2

Total moment of inertia of the combination = I = I1 + I2 = 2.251 x 10-4+ 2.016 x 10-3= 2.241x 10-3 kgm2

mass of the block hanging = m = 1.60 kg

part a)

for the hanging block , the force equation is given as ::

mg - T = ma Eq-1

for the disk system , torque is given as

Torque = T R1

but torque = I \alpha

so I \alpha = T R1

T = Ia/R21 Eq-2

from equation 1 and 2

mg - Ia/R21 = ma

a = mg/ (m + I/R21)

inserting the value

a = (1.6 x 9.8) / (1.6 + (2.241 x 10-3)/(0.0245)2)

a = 2.94 m/s2

part b)

for the hanging block , the force equation is given as ::

mg - T = ma Eq-1

for the disk system , torque is given as

Torque = T R2

but torque = I \alpha

so I \alpha = T R2

T = Ia/R22 Eq-2

from equation 1 and 2

mg - Ia/R22 = ma

a = mg/ (m + I/R22)

inserting the value

a = (1.6 x 9.8) / (1.6 + (2.241 x 10-3)/(0.051)2)

a = 6.37 m/s2

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