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Homework Statement
A 0.70-kg disk with a rotational inertia given by MR^2/2 is free to rotate on a fixed horizontal
axis suspended from the ceiling. A string is wrapped around the disk and a 2.0-kg mass hangs
from the free end. If the string does not slip, then as the mass falls and the cylinder rotates,
the suspension holding the cylinder pulls up on the cylinder with a force of:
A. 6.9N
B. 9.8N
C. 16N
D. 26N
E. 29N
Homework Equations
Newton's second law for rotating bodies
The Attempt at a Solution
I was able to get the acceleration of the disk + the mass by myself following the solution below. It's probably pretty obvious but I just can't seem to figure out how he got T = mg + M(g-at), specifically the M(g-at) part assuming that the suspension holding the cylinder pulls up on the cylinder with a force equal to the sum of the weights of the pulley and the block.
This is the solution I am trying to make sense of:
The suspension holding the cylinder pulls up on the cylinder with a force of
T = mg + M(g-at), where m - disk, M - 2kg mass, at - the acceleration of a 2.0-kg mass. At the same time that is the tangential acceleration of the disk. Let's find it.
J dω/dt = M(g-at) * R,
mR²/2 * 1/R dV/dt = M(g-at) * R, where at dV/dt = at.
m at/2 = M(g-at)
at = 2g/(m/M+2) = 2*9.8/(0.7/2 + 2) = 8.34 m/s²
T = mg + M(g-at) = 0.7*9.8 + 2*(9.8-8.34) = 6.86 + 2*1.46 = 9.8 N