# Homework Help: Energy conversion in a closed system?

1. Dec 13, 2011

### arabianights

We wrap a light, flexible cable around a thin-walled, hollow cylinder with mass M and radius R. The cylinder is attached to the axle by spokes of a negligible moment of inertia.The cylinder rotates with negligible friction about a stationary horizontal axis. We tie the free end of the cable to a block of mass m and release the object with no initial velocity at a distance h above the floor. As the block falls, the cable unwinds without stretching or slipping, turning the cylinder. The speed of hanging mass m just as it strikes the floor is v=sqrt 2mgh/(m+M). Use energy concepts to explain why the answer to part A is different from the speed found in case of solid cylinder,which is
v= sqrt 2gh/(1+M/2m) ?????

2. Dec 13, 2011

### Simon Bridge

note: it can help to write them out to have a similar form -
$$v_{solid}^2=\frac{2gh}{1+\frac{M}{2m}}$$
$$v_{hollow}^2 = \frac{2gh}{1+\frac{M}{m}}$$

What are the energy conversions involved in the motion? (Gravitational potential energy turns into............what?) What difference does having a solid cylinder make to each energy?

3. Dec 13, 2011

### arabianights

i figure that the potential energy of small m would be converted to kinetic energy of small m and the kinetic energy of M in this system without external force being applied.

so for hollow cylinder: mgh = 1/2*m*V^2+ 1/2*M*V^2, with 1/2*m*V^2 being the kinetic energy of m and 1/2*M*V^2 being the kinetic energy of M. V being the same for both m and M since they're attached to each other.

this shall give me the v=sqrt 2mgh/(m+M)

for solid one, i think it's necessary to apply integral along the radius R to get the kinetic energy for the body, but i'm not sure how

4. Dec 13, 2011

### Simon Bridge

Careful - the cylinder is rotating on the spot.

You have implicitly used the tangential velocity of the surface of the hollow cylinder - to use the same method for the solid one, you'd have to subdivide it into hollow cylinders of different radii (r) and integrate - the tangential velocity of the surface of each cylinder depends on it's radius. But each radius will also have a different mass.

The original cylinder has mass M and radius R, so it's radial density is $M/2\pi R$, so the mass of a cylinder of a different radius r<R will be r times this.
The speed at radius R is v, then at radius r is vr/R. Thus the kinetic energy at r<R will be $$K(r)=\frac{1}{2}\frac{Mr}{2\pi R}\left ( \frac{vr}{R} \right )^2 = \left ( \frac{r^3}{2\pi R^3} \right )K(R)$$... or something.................. integrate wrt r from 0 to R. This is easy to stuff up.

OR: you could use the rotatonal form of the kinetic energy equation instead: $K_R=\frac{1}{2}I\omega^2$

For the hollow cylinder: I=MR2

$$K_{h}=\frac{1}{2}(MR^2)(\frac{v}{R})^2 =$$
... see? Now do it for $K_{s}$ - the solid cylinder.

Last edited: Dec 13, 2011
5. Dec 14, 2011

### arabianights

using K(r)=12Mr2πR(vrR)2=(r32πR3)K(R)

i came up with this:
integral K(r) |(r = 0->R) = M*v^2/2*(1/2*pi*R^3)*r^4/4|(r = 0->R)
= M*v^2/2*(1/2*pi*R^3)*R^4/4 = M*v^2*R/(8*pi)

so plug kinetic engery for solid M into formular used earlier
mgh=1/2*mv^2 + Mv^2*R/(8pi) gives me
v^2(m/2+MR/8pi)=mgh
v=sqrt(2gh/(1+RM/(4pi*m)

so the speed v is related to not only M, m, but to radius of cylinder R as well

thanks:)

6. Dec 14, 2011

### Simon Bridge

Wow - really learn to use the tex tags: I can only read the last sentence.

You are correct - sort of.
Though we'd usually think of it as the acceleration of the cylinder depending on gravity and it's moment of inertia.

If you look at the picture, the speed the weight falls is also the speed the cable unwinds - so it must be the speed of rotation of the surface the cable is wound around. That's how they all link together (via the cable tension and moment of inertia). However, I would not expect the calculation for final speed to include terms in R ... I'd expect them to cancel. Re-check your working.

I told you it was easy to stuff up the integral ... that "or something" under my example is an indication that I don't trust the way I did that: on checking, it seems I did make a mistake. The result was supposed to be (1/4)Mv^2 but it isn't. It is very very much easier to just use the rotational-K equation. I'm puzzled that you didn't.

Last edited: Dec 14, 2011
7. Dec 14, 2011

### Simon Bridge

Aside: I was sloppy setting up the integral.
I felt OK not being careful because the idea was to illustrate the process, without actually doing it for you. I hoped you'd realize that using the rotational equation was easier.

When setting up an integral, it is important to keep your goal in mind.

The strategy is to start by observing that the energy in a cylindrical shell is already known as $\frac{1}{2}Mv^2$ ... because you've already done that part. We can use this to find the energy in a solid cylinder radius R by dividing it into a very large number of cylindrical shells radius r and thickness dr ... start to see where I went wrong?

The mass in each shell will be the density of the cylinder multiplied by the volume of the cylindrical shell (which is the ratio of the volumes times the total mass).

The speed of each shell is just the angular speed of the cylinder multiplied by the radius of the shell.

Now it remains to write out $dK=\frac{1}{2}M(r)v^2(r)$ and integrate both sides.