Homework Help: Rotational motion homework help

1. Dec 14, 2015

dobiegerl

1. The problem statement, all variables and given/known data
A uniform, solid sphere of mass 350 grams and radius 25.0 cm has an axle attached to it tangent to its surface. The axle is oriented horizontally, causing the sphere to be suspended below it, its center of mass directly below the axle. Someone lifts the sphere until its center of mass is at the same vertical level as the axle. Upon releasing it, the sphere begins to rotate counterclockwise about the axle. ) Calculate the linear speed of a point on its outer edge (farthest from the axle) when the sphere’s center of mass returns to its position directly below the axle

2. Relevant equations
Conservation of energy

3. The attempt at a solution
I started off trying to use KEri + KEti + GPEi = KErf + KEtf + GPEf
After substituting and eliminating I got to gh = 1/2Vf^2 + 1/4Vf^2
And then I got 2.55 m/s but according to the answer key, I should be getting 1.87 m/s. I not sure where I'm going wrong.

2. Dec 14, 2015

Staff: Mentor

It might be easier to consider the sphere as purely rotating.

Not sure how you got that. What's Vf? The velocity of the center of mass? What's the moment of inertia of the sphere about the given axis?

Why not compute the angular velocity first?

3. Dec 14, 2015

dobiegerl

That's what I'm trying to do, but I don't know how to without acceleration or time

4. Dec 14, 2015

Staff: Mentor

You don't need acceleration or time, just conservation of energy.

5. Dec 14, 2015

dobiegerl

Would I just do KEri = KErf ? Or do I need to include GPE. Because I tried other formulas with conservation of energy and I'm still not getting the correct answer

6. Dec 14, 2015

dancingsquirrel

I am also working on this problem. I've been using conservation of energy and set up the equation:

mgh = ½(2/5mr2)(ω2)

The sphere rotates 90o (based on the diagram we have) and I converted that to (π/2) radians and used that as my height.

After plugging in values for variables I got:

5.393 = 0.004375ω2
after dividing by 0.004375 and taking the square root of that number I got the ω value to be 35.11 rad/s

Using the formula v = rω, I plugged in v = .25(35.11) which gave me the linear velocity of 8.778 m/s.

I'm not quite sure where I went wrong, but it clearly isn't the right answer based on our answer key.

7. Dec 14, 2015

SammyS

Staff Emeritus
You went wrong in a number of places.

How can π/2 radians be a height?

You need the moment of inertia about an axis through the edge of the sphere, not through its center. That is unless you use a slightly different method.

8. Dec 14, 2015

dancingsquirrel

I realized that I should have used 0.25m as the height, not π/2 radians. As for the rest of the problem, I needed to include KT. After fixing this I finally got the correct answer of 1.87 m/s.

9. Dec 14, 2015

SammyS

Staff Emeritus