Rotational motion homework help

[dobiegerl]

Homework Statement

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

Homework Equations

Conservation of energy

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.

 

[Doc Al]

I started off trying to use KEri + KEti + GPEi = KErf + KEtf + GPEf

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

After substituting and eliminating I got to gh = 1/2Vf^2 + 1/4Vf^2

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?

 

[dobiegerl]

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?

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

 

[Doc Al]

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

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

 

[dobiegerl]

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

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

 

[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 90^o (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.

 

[SammyS]

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 90^o (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.

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.

 

[dancingsquirrel]

How can π/2 radians be a height?

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 K_T. After fixing this I finally got the correct answer of 1.87 m/s.

 

[SammyS]

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 K_T. After fixing this I finally got the correct answer of 1.87 m/s.

You may want to show more details of your work. It appears that the given answer may be in error.

 

[haruspex]

You may want to show more details of your work. It appears that the given answer may be in error.

Yes, 1.87m/s is the speed of the mass centre.