Sphere rolling down incline then up incline?

In summary, the solution to this problem involves setting up a conservation of energy equation for both halves of the ramp. The equation for the first half is mgH = 1/2mv^2 + 1/2Iω^2. Solving for v^2, we get (gH)/c. Then, for the second half of the ramp, we have 1/2mv^2 + 1/2Iω^2 = mgh + 1/2Iω^2. Simplifying and plugging in the value for v^2 from the first equation, we get 2gH/c+1 as the final answer for the height reached by the sphere.
  • #1
Watney
7
0

Homework Statement


A sphere of mass M and radius R is not necessarily solid or hollow. It has moment of inertia I = cMR^2 . The sphere starts from rest and rolls without slipping down a ramp from height H. It then moves back up the other side with height h, but now with no friction at all between the sphere and the ramp.
What height h does the sphere reach?
(Answer should only include H and c.)

Homework Equations


I = cMR^2
KE = 1/2Iω^2
PEg = mgH

The Attempt at a Solution


So I tried setting up a conservation of energy for both sides of the ramp. For the first half I got
mgH = 1/2mv^2 + 1/2Iω^2. I then solved for v^2.
For the second half of the ramp (one without friction), I got
1/2mv^2 + 1/2Iω^2 = mgh + 1/2Iω^2.
I solved for h then plugged in v^2 from the first equation and got g^2H/c but my answer can't have g so I know this is wrong. Any pointers? I've never done a problem like this so I'm pretty sure I'm way off.
 
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  • #2
In this first part, what did you get when you solved for v2?
 
  • #3
TomHart said:
In this first part, what did you get when you solved for v2?
]

I got (gH)/c but I'm not entirely sure if it's correct.
 
  • #4
Yeah, I got something different for v2.

In your equation for the second half of the ramp, you have 1/2Iω2 on both sides of the equation. I assume those are the same and they cancel out, true? At least that's how I see the problem. When the sphere starts up the other side of the ramp, none of the rotational kinetic energy converts back to potential energy. So it looks like your method is good. I think you are just having algebra issues. I was able to come up with a solution in terms of only H and c.

Edit: I feel obligated to put in my standard disclaimer. I am prone to making mistakes. And even though I found a solution in terms of only H and c, it could still be incorrect. :)
 
  • #5
TomHart said:
Yeah, I got something different for v2.

In your equation for the second half of the ramp, you have 1/2Iω2 on both sides of the equation. I assume those are the same and they cancel out, true? At least that's how I see the problem. When the sphere starts up the other side of the ramp, none of the rotational kinetic energy converts back to potential energy. So it looks like your method is good. I think you are just having algebra issues. I was able to come up with a solution in terms of only H and c.

Edit: I feel obligated to put in my standard disclaimer. I am prone to making mistakes. And even though I found a solution in terms of only H and c, it could still be incorrect. :)

Haha no worries! So I tried it again and got 2gH/c+1. Is this correct?
 
  • #6
Plugged it in and got the right answer! Thanks Tom!
 

1. What factors affect the speed of a sphere rolling down an incline?

The speed of a sphere rolling down an incline is affected by several factors, including the angle of the incline, the mass and size of the sphere, and the surface of the incline. A steeper incline will result in a faster speed, while a larger and heavier sphere will have more momentum and roll faster. Additionally, a smooth and frictionless surface will allow the sphere to roll faster compared to a rough or sticky surface.

2. How does the angle of the incline affect the path of the sphere?

The angle of the incline plays a significant role in determining the path of a sphere rolling down and then up. A steeper incline will result in a faster and more direct path, while a shallower incline will cause the sphere to roll slower and follow a longer and curved path. At a certain angle, the sphere will reach its maximum height before rolling back down.

3. What causes a sphere to roll back up an incline?

The force of gravity is what causes a sphere to roll down an incline. However, when the sphere reaches the bottom of the incline, its momentum and kinetic energy are conserved and allow it to roll back up the incline. This is due to the fact that the sphere's center of mass is higher than its point of contact with the incline, creating a torque that causes it to roll upwards.

4. How does the mass of the sphere affect its motion on an incline?

The mass of the sphere affects its motion on an incline in two ways. Firstly, a heavier sphere will have more momentum and therefore roll faster down an incline. Secondly, the mass of the sphere affects its rotational inertia, which is the resistance to change in rotational motion. A sphere with a larger mass will have a higher rotational inertia and may roll slower compared to a smaller mass sphere.

5. What is the relationship between the height of the incline and the distance the sphere will roll?

The relationship between the height of the incline and the distance the sphere will roll is directly proportional. This means that as the height of the incline increases, the distance the sphere will roll also increases. This is because a higher incline will result in a faster speed and more momentum, allowing the sphere to travel a greater distance before coming to a stop.

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