# HW 9 Problem 6

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You have a pulley 10.7 cm in diameter and with a mass of 1.5 kg. You get to wondering whether the pulley is uniform. That is, is the mass evenly distributed, or is it concentrated toward the center or the rim? To find out, you hang the pulley on a hook, wrap a string around it several times, and suspend your 2.0 kg physics book 0.9 m above the floor. With your stopwatch, you can find that it takes 0.63 s for your book to hit the floor.

What is the moment of inertia if the pulley is uniform?

What is the actual moment of inertia of the pulley based on your measurement?

+1 vote

Part1: If the pulley is uniform, then it can be considered as a uniform "disk", which has $I=\frac {1}{2}mR^2$.

Part2: Based on the time and distance it falls, you can find its actual acceleration as: $$\Delta y=V_{y0} \Delta t+\frac 1 2 a\Delta t^2.$$

Plug in the given distance, time and mass, etc., one can find $a$.

With the rotational and linear motion, one can also find the linear acceleration in terms of the moment of inertia -- here you would leave $I$ in the equation as an unknown variable.

Solving the equations, you will get $$a=\frac {m_1}{\frac {I}{R^2}+m_1}.$$

Plug in the $a$ found with the falling distance and time, you can find the numerical value of $I$. Compare this $I$ with the $I$ calculated in the first part to determine if the disk is uniform or not.

answered Oct 20, 2013 by (21,750 points)
I also solved for I using a system of equations but found something that seems to be notably different.

I=(m1(g-a)R^2)/a

I found this by the following:

a = alpha*R

TR=I*alpha

a=((TR/I)R)          simplifiy

a=((TR^2)/I)        now substitute for T

T= m1g-m1a

a=((m1(g-a)R^2)/I)      now solve for I

I= ((m1(g-a)R^2)/a)   finally substitute for a

I= (m1(g-(2y/t^2))/(2y/t^2))

Is this the equivalent to the equation posted above?
I am not sure if this will work 100% of the time but I do know that it worked for myself. Some feedback would be appreciated.