# Find inner radius of hollow cylinder given inertia

## Homework Statement

A roll of toilet paper ( a partially hollow cylinder with R2=7.0 cm, M=320 g, I=6.0x 10 ^(-4) kg m is mounted on an axle. initially at rest, until a child grabs the end and starts running at a constant linear acceleration.

part a) what is the inner radius (R1)

## Homework Equations

I know the equation for a hollow cylinder's moment of inertia is I=1/2 M (R1^2 + R2^2)

## The Attempt at a Solution

I converted all my units to S.I. units

7 cm= 0.07m
320g= 0.320kg

I rearranged the equation as
R= ({(2*I)/M} -R2^2) ^(1/2)
2 (6.0x 10^-4)=0.0012 kg m
0.0012 kg m/.320 kg= .00375m
0.00375 m-(0.07m)^2= -0.0015 m^2
I would now need to take the square-root of this number, but its a negative number. I cannot seem to find what I did wrong. and I need part a) to find the other four parts of this question. Thanks to anyone who takes the time to help.

## The Attempt at a Solution

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PhanthomJay