PHYSICS Problem?

Question

Case 1 and case 2 show two situations of a block hanging
from a string wrapped around the outside of a pulley. The
blocks are identical, and the pulleys are uniform solid
disks of the same radius, but the pulley in case 2 has twice
the mass (and therefore twice the moment of inertia) of
the pulley in case 1. The systems are released from rest at
the same time from the same height above the ground, and
in both cases the block accelerates down. Neglect friction.

(a) In which case is the acceleration of the block
larger?

(b) In which case is the tension in the string, while the block is falling, larger?

(c) In which case is the net torque on the pulley, while the block is falling, larger?

(d) In which case does the block have a higher speed after falling through a distance h?

(e) In which case does the pulley have a larger rotational kinetic energy, measuring the
kinetic energy at the instant the block reaches the ground in each case?

Choices : Case 1, Case 2, or Equal in both

Please Explain Reasoning.

Answer 1

Let's consider FBDs of the system:
The mass of the block is m
the mass of the pulley is M
for case 2 it is 2*M
T1 is the tension for case 1 and T2 is the tension for case 2
the moment of inertia for the pulley in case 1 is M*r^2/2
and case 2 is M*r^2

case 1 the block
m*g-T1=m*a1
T1=m*(g-a1)

the pulley
T1*r=M*r^2/2*a1/r
simplify
T1=M*a1/2

Ok, let's solve for a1
M*a1/2=m*(g-a1)
M*a1/2+m*a1=m*g
a1=m*g/(M/2+m)

now solve for T1
T1=M*m*g/(2*(M/2+m))

Now consider Case 2

the block
m*g-T2=m*a2
T2=m*(g-a2)

T2*r=M*r^2*a2/r
simplify
T2=M*a2

solve for a2
M*a2=m*(g-a2)
a2=m*g/(M+m)

and T2
T2=M*m*g/(M+m)

a) Compare a1 versus a2
a1=m*g/(M/2+m)
a2=m*g/(M+m)

a1/a2=(M+m)/(M/2+m)

The denominator is smaller, therefore a1>a2, which makes sense since there is less mass to spin on the pulley.

b) Compare T1 and T2
T1=M*m*g/(2*(M/2+m))
T2=M*m*g/(M+m)
divide T1/T2
T1/T2=(M+m)/(2*(M/2+m))
simplify
T1/T2=(M+m)/(M+2*m)

The denominator is larger, so T2 is larger than T1, again it makes sense.


c) The torque is the tension times the radius, so case 2 has the larger torque.

d) Case 1 will achieve higher speeds since the acceleration is higher.

e) Case 2 will have more energy in the spinning pulley since there was more work done by a greater torque thorough the same displacement.

The symmetry of the cases means that case 1 will have more kinetic energy in the falling block on impact since it will have a higher velocity.

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