Angular Velocity Of A Sphere Rotating Under Gravity

AI Thread Summary
A solid sphere of mass m and radius a rotates about a point on its surface when released from a horizontal position under gravity. The moment of inertia is calculated using the parallel axis theorem, resulting in Iz = (7/5)ma^2. The energy change equation relates the rotational kinetic energy to the potential energy, leading to the formula w = (100/(7a))^.5 for angular velocity. There is a suggestion to express the answer in terms of the radius a and keep g symbolic for unit flexibility. The discussion emphasizes the importance of correctly applying energy principles and moment of inertia in solving the problem.
Woolyabyss
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Homework Statement


A solid sphere of mass m and radius a can rotate freely about a point A on its surface. The sphere is held initially at rest with the line OA through A and the centre of the sphere O horizontal and is released under gravity. Find the angular velocity of the system when OA first becomes vertical.
(You may assume that the moment of inertia of the sphere about an axis through O is (2/5)ma^2.)

Homework Equations


energy change
.5*Iz*w^2 = mgh

where Iz is the moment of inertia about z
w is angular velocity
mgh potential energy

The Attempt at a Solution


parallel axis theorem
Iz = (2/5)ma^2 + ma^2 = (7/5)ma^2

energy change

.5*Iz*w^2 = mgh (where h is the change in height of centre of mass)

(7/10)*m*a^2*w^2 = mga

rearranging

w = (100/(7a))^.5

I'm not sure how to find a or if I even have enough information to solve for a.

Should I try angular momentum?
 
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Your work looks good. Perhaps they just want the answer to be expressed in terms of the radius a. If so, it might be preferable to also leave g in symbolic form rather than substitute a value for g. That way, your answer will be correct for any system of units (as long as you use radians for angle).
 
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thanks.
 
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