# Moment of inertia and conservation of energy problem

#### knowlewj01

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 dont 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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#### cepheid

Staff Emeritus
Gold Member
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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#### knowlewj01

ok, thanks. I realised 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.

#### knowlewj01

so, i got as far as:

$$\frac{R^2 \omega ^2}{5}$$ = gh

substitute in:

$$\omega$$ = $$\frac{v}{R}$$

$$\frac{v^2}{5}$$ = gh

$$\frac{4}{5g}$$ = h

h = 0.08m

= 8cm

does this look right?

#### cepheid

Staff Emeritus
Gold Member
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, $v = 2 \pi r \cdot f = (2\pi f)r = \omega r$ is the derivation I was hinting at in my previous post, but I guess you already knew the relation.

#### knowlewj01

so the total kinetic energy should be the sum of

$$\frac{1}{2}$$ I $$\omega$$^2

and

$$\frac{1}{2}$$ m v^2

?

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

#### knowlewj01

did it with these numbers and got 0.29m

#### cepheid

Staff Emeritus
Gold Member
so the total kinetic energy should be the sum of

$$\frac{1}{2}$$ I $$\omega$$^2

and

$$\frac{1}{2}$$ 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.

did it with these numbers and got 0.29m
I get the same answer.

#### knowlewj01

Thanks very much, this was just a section A question on a past exam paper. Understand it now, cheers :)

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