How Is Kinetic Energy Distributed in a Rolling Sphere?

In summary, the problem asks for the kinetic energy of a sphere when it is at an initial position. The solution is found by finding the rotational inertia, mass, and the equation for rolling without slipping.
  • #1
Hindi
9
0
Hi all,

I just had a question on a homework problem I was attempting, but can't find the correct solution for it yet. The answer is in the back of the book, but I wanted to know how it is derived. Anyways, here is the question:

A hollow sphere of radius .15 with I=.040 rolls without slipping up a surface inclined at 30 degrees. At a certain initial position, the sphere's total kinetic energy is 20J.

A) How much of the totak K.E. initial is rotation?

------------------------------------------------
Variables:

Radius of sphere = .15m
Moment of Inertia = .040 kg * m^2

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Attempt:
I tried this problem with setting the initial kinetic energy of the sphere-incline system to 20J. So I did:
.5MV^2 + .5I(w)^2 = 20J

This is the step that I get lost in. Should I try to find the mass of the system first or should I try a whole different approach? Any help would be appreciated! Thanks!
 
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  • #2
Originally posted by Hindi

.5MV^2 + .5I(w)^2 = 20J

This is the step that I get lost in. Should I try to find the mass of the system first or should I try a whole different approach?
Yes, you'll need the mass of the sphere. (Hint: what's the formula for rotational inertia of a sphere?)

Also, you'll need to apply the condition for rolling without slipping: how does ω relate to V when there is no slipping?
 
  • #3
Doc Al,

Thanks for the reply. One of the biggest problems I was having with finding the mass was finding the rotational inertia of a hollow sphere. In my book (Fundamentals of Physics 6th Edition), it only gives the rotational inertia of a solid sphere.

I know the rotational inertia for a solid sphere is 2/5MR^2. Will that apply to this problem as well? If that applies, then I set:
2/5MR^2=.040 kg*m^2 and find the mass that way.

Also, you'll need to apply the condition for rolling without slipping: how does ω relate to V when there is no slipping?

v=w(r) --> w=v/r

Any hints would be greatly appreciated. Thanks again!
 
  • #4
The moment of inertia for a hollow sphere is 2/3*m*r^2.. find the mass, then with the equation you have (substitute in for v to find w), then plug w back into the rotational energy equation, and voila
 
  • #5
Hey thanks deltabourne. You gave me the answer to the missing piece of the puzzle! Found the answer to be 8.0J joules, which is correct!
 
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  • #6
Hi all again.

Now the question wants me the find the TOTAL kinetic energy of the sphere. I keep on trying to figure this out, but I keep getting 10J, while the answer is 6J. I f anyone can point what I am doing wrong, it would be greatly appreciated!

-------------------------

I found Vi=2.98m/s
Height = .75m

1/2MV^2+1/2I(W)^2 = MGH + 5/6MV^2

Subtracting 5/6MV^2 to the other side, I keep on getting 10J. What am I doing wrong?

To make things easier, I attached my work I have done so far as an image:
http://home.comcast.net/~msharma15/problem_1.jpg
 
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  • #7
Originally posted by Hindi
Now the question wants me the find the TOTAL kinetic energy of the sphere. I keep on trying to figure this out, but I keep getting 10J, while the answer is 6J. I f anyone can point what I am doing wrong, it would be greatly appreciated!
You'll have to state the problem more clearly. Obviously you are asked to find the total KE at some new point. What point? In your attachment you mention a point 1 m up the incline. Is that the point? Let me pretend that it is. You know the initial KE; when it raises up a height H, the KE will decrease by mgH. For 1 m up the plane, H = 1 sin(30) = 0.5 m.
 
  • #8
Thanks everyone...eventually solved it! Pretty easy after thinking about it.

Doc...yea it was 1m. Sorry about that!

Anyone interested in the solution, you can find it here:
http://home.comcast.net/~msharma15/problem_1.jpg
 
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FAQ: How Is Kinetic Energy Distributed in a Rolling Sphere?

What is rolling friction?

Rolling friction is the resistance that occurs when an object rolls over a surface. It is caused by the deformation of the object and the surface it is rolling on.

What is the difference between rolling friction and sliding friction?

Rolling friction occurs when an object is in motion and rolls over a surface, while sliding friction occurs when an object is in motion and slides over a surface. Rolling friction is typically lower than sliding friction, making it easier to keep an object in motion.

How does the shape of an object affect rolling friction?

The shape of an object can affect rolling friction in several ways. A round object, such as a ball or a wheel, has less surface area touching the ground, resulting in lower rolling friction. On the other hand, an object with a flat surface, such as a cube, has more surface area in contact with the ground, resulting in higher rolling friction.

What factors affect the amount of rolling friction?

The amount of rolling friction is affected by several factors, including the weight of the object, the surface it is rolling on, and the speed at which it is rolling. Heavier objects have higher rolling friction, rougher surfaces have higher rolling friction, and higher speeds can increase or decrease rolling friction depending on the object and surface.

How can rolling friction be reduced?

Rolling friction can be reduced by using rolling objects with smoother surfaces, increasing the size of the object's wheels, and using lubricants or ball bearings. Additionally, reducing the weight of the object can also decrease rolling friction.

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