Part A.) If the pulley has radius R and moment of inertia I about its axle, dete

Question

Part A.) If the pulley has radius

R and moment of inertia I about its axle, determine the acceleration of the masses m1 and m2. [Hint: The tensions FT1 and FT2 are not equal.] Express answers in terms of m1, m2, I, R.

Part B.) Compare to the situation in which the moment of inertia of the pulley is ignored. Express answer in terms of m1, m2, and R.

m2 just before it strikes the ground. Assume the pulley is frictionless.

Use this figure for all 3 parts

R and moment of inertia I about its axle, determine the acceleration of the masses m1 and m2. [Hint: The tensions FT1 and FT2 are not equal.] Express answers in terms of m1, m2, I, R. Compare to the situation in which the moment of inertia of the pulley is ignored. Express answer in terms of m1, m2, and R. M1= 14kg, m2=29kg,Pulley is a uniform cylendar and has a radius of r=0.2 and a mass of 10.3 kg. Initially m1 is on the ground and m2 is 2.9m above the ground. If the system is now released, use conservation of energy to determine the speed of

Explanation / Answer

1) let 'a' be the acceleration

m2g - FT2 = m2a

FT1 - m1g = m1a

total torque on the pulley = R(FT2 - FT1) = Ia/R

from above 3 eqautions we get

a = g(m2 - m1)/(I/R2 + m1 + m2)

2) Without the moment of inertia of the pulley it is just the two mass equations we have to solve:
FT1 - m1g = m1a
m2g - FT2 = m2a
If the pulley has no moment of inertia, then FT1 and FT2 are the same

therefore the equations become

FT - m1g = m1a

m2g - FT = m2a

by solving above 2 equations we get

a = (m2 - m1)g/(m2 + m1)

3) here I = (1/2)MR2

where M = 10.3 kg R = 0.2 m

m1 = 14kg m2 = 29 kg

therefore we get I = 0.206 kgm2

by subtituting these values in equation from '1)' we get

a = 3.053 m/s2

velocity just before it strikes the ground be 'v' then

v = ?(2ah) where h = 2.9 m

v = 4.208 m/s

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