Posted by **john** on Saturday, March 17, 2012 at 11:20pm.

In each of the following cases, the mass is 7.99 kg and the radius is 7.50 cm.

(a) What is the moment of inertia (kg/m2

) of a solid cylinder rotating about an axis parallel to the symmetry axis but passing through the edge of the cylinder?

(b) What is the moment of inertia (kg/m2

) of a solid sphere rotating about an axis tangent to its surface?

The wording in (a) is confusing me. I tried but wasn't even close on both.

- Physics Please help! -
**Adam**, Tuesday, March 25, 2014 at 9:27pm
Using the parallel axis theorem, we know that I = I_cm + mr^2, where r is the amount of displacement from the symmetry axis.

And we know that for a solid cylinder, I_cm = 1/2(mr^2); Further, for a solid sphere I_cm = 2/5(mr^2)

(a)

==> I = 1/2(mr^2) + mr^2 = 3/2(mr^2) = 3/2(7.99)(.075^2) = 6.7416E-2 kg

(b)

==> I = 2/5(mr^2) + mr^2 = 7/5(mr^2) = 7/5(7.99)(.075^2) = 6.2921E-2 kg

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