Moment of inertia and conservation of energy problem

In summary, a uniform sphere is rolling without slipping on a horizontal surface at 2m/s. It then rolls without slipping up a hill to height h, where it briefly comes to rest. Find the vertical height h.
  • #1
knowlewj01
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0
A uniform sphere is rolling without slipping on a horizontal surface at 2m/s. It then rolls without slipping up a hill to height h, where it briefly comnes to rest.

Find the vertical height h.

[The moment of inertia, I, of a sphere is 2/5 mR^2]

where m is the mass and R is the radius of the sphere.
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I used conservation of energy, such that
1/2 I w^2 = mgh

then substituted the moment of inertia in.

1/5 m R^2 w^2 = mgh

m cancels:

1/5 R^2 w^2 = gh

after this I don't know where to go, i have to relate the angular speed, w, to the speed of the rolling sphere, v.

any ideas?
 
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  • #2
Here are some thoughts:

- w is not the angular momentum, it is the angular velocity. The [edit: magnitude of the] angular momentum is given by L = Iw, but you don't need it to solve this problem.

- For this type of motion (rolling without slipping), the velocity, v, with which the centre of the sphere translates forward is the same as the velocity, v, of any point on the great circle of the sphere that is rotating around the centre. (A great circle is any circle on the surface of a sphere that is centred on the sphere's centre. E.g. a longitude line on Earth is a great circle. In this case, due to the rolling motion, there will be a great circle that is in motion about the centre. It is the same as the set of points that is in contact with the ground over a full cycle).

- There is a very simple relationship between w and the second v I mentioned in the point above. If you don't know it, you can easily figure it out by realizing that the angular velocity is also known as the angular frequency -- it is the frequency (number of rotations per second) multiplied by 2*pi so that it measures the rate of accumulation of phase (in radians per second) in one cycle of the periodic motion. The velocity of a point on the great circle of the sphere that is rotating about the centre is just the circumference of that great circle divided by the time it takes for one rotation (the period). In other words, it is the circumference times the frequency.

[EDIT: What I'm saying, for clarity, is: distance covered by point going around in one cycle * number of cycles per second = total distance covered per second = (by definition) velocity of point.]
 
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  • #3
ok, thanks. I realized I wrote the angular momentum and edited it, I meant angular speed;)
just a typo. I'll have a go and give my solution agin in a minute.
 
  • #4
so, i got as far as:

[tex]\frac{R^2 \omega ^2}{5}[/tex] = gh

substitute in:

[tex]\omega[/tex] = [tex]\frac{v}{R}[/tex]

[tex]\frac{v^2}{5}[/tex] = gh

[tex]\frac{4}{5g}[/tex] = h

h = 0.08m

= 8cm

does this look right?
 
  • #5
No, actually, now that I think about it, I don't think that you have accounted for all of the kinetic energy of the sphere. There is kinetic energy due to the rotation about the centre of mass, but there is also kinetic energy due to the translation (i.e. forward linear motion) OF the centre of mass.

EDIT: By the way, [itex] v = 2 \pi r \cdot f = (2\pi f)r = \omega r[/itex] is the derivation I was hinting at in my previous post, but I guess you already knew the relation.
 
  • #6
so the total kinetic energy should be the sum of

[tex]\frac{1}{2}[/tex] I [tex]\omega[/tex]^2

and

[tex]\frac{1}{2}[/tex] m v^2

?

i'll try that and see what i get, 8cm does seem pretty small
 
  • #7
did it with these numbers and got 0.29m
 
  • #8
knowlewj01 said:
so the total kinetic energy should be the sum of

[tex]\frac{1}{2}[/tex] I [tex]\omega[/tex]^2

and

[tex]\frac{1}{2}[/tex] m v^2

?

i'll try that and see what i get, 8cm does seem pretty small


That is indeed what I was trying to say.

knowlewj01 said:
did it with these numbers and got 0.29m

I get the same answer.
 
  • #9
Thanks very much, this was just a section A question on a past exam paper. Understand it now, cheers :)
 

Related to Moment of inertia and conservation of energy problem

1. What is moment of inertia?

Moment of inertia is a measure of an object's resistance to rotation. It depends on the object's mass, shape, and distribution of mass.

2. How does moment of inertia affect an object's rotational motion?

The larger the moment of inertia, the more difficult it is for an object to change its rotational motion. Objects with larger moments of inertia will have slower rotational speeds compared to objects with smaller moments of inertia.

3. What is the relationship between moment of inertia and conservation of energy?

Moment of inertia is related to conservation of energy through the law of conservation of angular momentum. This law states that the total angular momentum of a system is conserved, meaning that the moment of inertia and rotational speed of an object are inversely proportional.

4. How do I calculate the moment of inertia for a given object?

The moment of inertia can be calculated using the formula I = mr^2, where m is the mass of the object and r is the distance from the axis of rotation. For more complex shapes, the moment of inertia can be found using integrals or by looking up the values in a table.

5. How can I use the concept of moment of inertia to solve conservation of energy problems?

In conservation of energy problems involving rotational motion, the moment of inertia can be used to determine the final rotational speed of an object. By setting the initial and final kinetic energies equal, the moment of inertia can be solved for and then used to find the final rotational speed.

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