Help with Past Exam Question on Rolling Disk on Hemisphere

[Ichigo449]

Homework Statement

Hoping to outdo his physics professor, Doofus wants to dramatically demonstrate parabolic motion by throwing a cheese wheel of massm and radius r off of the top of the UW observatory, which is at a height 3R above the roof of the physics building, as shown in the diagram. Just as he is finishing his climb, he accidentally let's go of the cheese wheel at the top of the hemispherical dome of radius R. It begins rolling without slipping from rest down the edge of the dome.

a) Draw a free-body diagram of the cheese when it has rolled a distance s along the dome

b) What is the value of s when the cheese leaves the dome

c) Diligent is standing on the roof of the physics building and catches the cheese just as it hits the ground. How fast is the cheese moving when he catches it

Homework Equations

Conservation of Energy , s = r*a, Newton's Second Law

The Attempt at a Solution

a) Forces are static friction, normal force and gravity
b) Applying conservation of energy gives v^2 = 4/3 ((R+r)(1-cos(a))g. Since the wheel falls off when N =0, a = arccos(4/7). Therefore s = r*arccos(4/7).
c) Conservation of energy gives 0.5m (v_f^2 - v_c^2) = mg(s+3R). Therefore v_f^2 = 2g(s+3R) +4/3 (R+r)(1-4/7)g.

 

[Ichigo449]

Ooops, the change in gravitational potential energy is mg(-25/7 R).

 

[ehild]

Homework Statement

Hoping to outdo his physics professor, Doofus wants to dramatically demonstrate parabolic motion by throwing a cheese wheel of massm and radius r off of the top of the UW observatory, which is at a height 3R above the roof of the physics building, as shown in the diagram. Just as he is finishing his climb, he accidentally let's go of the cheese wheel at the top of the hemispherical dome of radius R. It begins rolling without slipping from rest down the edge of the dome.

a) Draw a free-body diagram of the cheese when it has rolled a distance s along the dome

b) What is the value of s when the cheese leaves the dome

c) Diligent is standing on the roof of the physics building and catches the cheese just as it hits the ground. How fast is the cheese moving when he catches it

Homework Equations

Conservation of Energy , s = r*a,

s is the distance traveled along the dome, and r is the radius of the cheese.

 

[ehild]

Ooops, the change in gravitational potential energy is mg(-25/7 R).

That is correct. What is the final velocity of the cheese then? Note that the rotation of the cheese does not change after leaving the dome. Velocity means the velocity of its center of mass.

 

[Ichigo449]

It would be v_f^2 = 2g(-25/7 R) + 4/3 (R+r) (3/7)g = 2/7 (2r-23R)g.

 

[haruspex]

There are several steps I'm not following.
Where does s+3R come from? s is a distance around a curve. What has it to do with GPE change?
What phase of the motion does the GPE change of -25/7 mgR refer to?
Seems to me the final answer has to be of the form v²=g[4R-(R+r)c] for some constant c.

 

[Ichigo449]

Firstly s+3R was a stupid mistake. The -25/7 mgR refers to the parabolic motion of the wheel when it falls off of the dome. It falls of the dome at a height of Rcos(a), or 4/7 R and then falls the 3R for a total change in height of 25/7 R.

 

[haruspex]

It falls of the dome at a height of Rcos(a)

Are you sure about that? That's the height of what part of the cheese relative to what? Doesn't r feature?

 

[Ichigo449]

Oh, thank you. That would only be the height of the bottom of the cheese, but we need the height of the center of mass of the cheese, so it's (R+r)cos(a). Thanks :)

 

[haruspex]

Oh, thank you. That would only be the height of the bottom of the cheese, but we need the height of the center of mass of the cheese, so it's (R+r)cos(a). Thanks :)

Still not quite right. The mass centre does not reach the ground.

 

[ehild]

It would be v_f^2 = 2g(-25/7 R) + 4/3 (R+r) (3/7)g = 2/7 (2r-23R)g.

The cheese is certainly smaller then the dome, so r<R. That means you get negative value for v_f^2.

 

[Ichigo449]

Yes, I see that the velocity ends up imaginary but am not sure how to get a positive value. Am I applying energy conservation correctly for part c)?

 

[ehild]

Yes, I see that the velocity ends up imaginary but am not sure how to get a positive value. Am I applying energy conservation correctly for part c)?

Yes, you should apply conservation of energy, but you did not show your work.

 

[Ichigo449]

Ok, so for c) . The velocity when the cheese wheel falls off is v^2 = g(R+r)cos(a) from the condition that the normal force goes to zero. Since cos(a) was found to be 4/7 in part b) this is v^2 = 4/7 (R+r)g. Conservation of energy tells us that 0.5m(v_f^2 -v^2) = mg(0-h), where h is the height the cheese falls through to reach the ground. Since the cheese falls off the dome at a height of 4/7 R and then falls an additional 3R to the roof h = 25/7 R. Using the result for v^2 I get the negative answer for v_f^2.

 

[ehild]

Ok, so for c) . The velocity when the cheese wheel falls off is v^2 = g(R+r)cos(a) from the condition that the normal force goes to zero. Since cos(a) was found to be 4/7 in part b) this is v^2 = 4/7 (R+r)g. Conservation of energy tells us that 0.5m(v_f^2 -v^2) = mg(0-h), where h is the height the cheese falls through to reach the ground. Since the cheese falls off the dome at a height of 4/7 R and then falls an additional 3R to the roof h = 25/7 R. Using the result for v^2 I get the negative answer for v_f^2.

You have a sign error (and some other mistakes). Write out the energy as PE + KE both at the initial and final positions.You will find the sign error. Include also r. what is h?

 

[Ichigo449]

I found the sign error but still can't figure out what h should be.

 

[ehild]

I found the sign error but still can't figure out what h should be.

At what height above the ground is the center of mass of the cheese when it leaves the dome? And at what height it is when the cheese reaches the ground?

 

[Ichigo449]

Oh, it's at height r when the cheese hits the ground and 4/7 (R+r) + 3R when it leaves the dome.

 

[ehild]

Oh, it's at height r when the cheese hits the ground and 4/7 (R+r) + 3R when it leaves the dome.

Yes.

 

[Ichigo449]

Ok, thank you so much for the patient help.

 

[ehild]

What result did you get?

 

[Ichigo449]

v_f^2 = 2/7 (27R -r) *g

 

[Ichigo449]

My work for that is 0.5m(v_f^2 -v^2) = -mg(r-(4/7(R+r) +3R)) = mg/7 (25R -3r).
Since v^2 = 4/7(R+r)g, v_f^2 = 2/7g (25R -3r) + 4/7(R+r)g = 2/7 (27R -r)g.

 

[ehild]

I got the same.

 

[Ichigo449]

Ok, thank you very much for the help and now I have a much better understanding of how to deal with these types of questions.