# Angular Displacement + Torque

1. Aug 10, 2008

### hot2moli

Assume that the Titanic is drifting in Southampton harbor before its fateful journey and the captain wishes to stop the drift by dropping an anchor. The iron anchor has a mass 1000 kg. It is attached to a massless rope. The rope is wrapped around a reel in the form of a solid disk of radius 0.350 m and mass 300 kg, that rotates on a frictionless axle.
(a) Find the angular displacement of the reel when the anchor moves down 15.0 m.
(b) Find the acceleration of the anchor as it falls through the air, which offers negligible resistance.
____m/s2
(c) While the anchor continues to drop through the water, the water exerts a drag force of 2500 N on it. With what acceleration does the anchor move through the water?
____m/s2
(d) While the anchor drops through the water, what torque is exerted on the reel?
____N·m

I haven't done this kind of material for a while now, and as I look back at this problem I am unsure how to attempt it...

2. Aug 10, 2008

### Chrisas

Draw a free body diagram for the anchor and put in the two forces acting on it. Use sum of forces = m*a to setup your first differential equation. You have all the initial conditions needed to integrate it.

Draw a diagram for the wheel. The same force acting against gravity on the anchor is acting to accelerate the wheel. Use Torque = I * dw/dt to set up another differential equation. This also can be integrated from initial conditions. Remember Torque comes from linear force as force * distance from spin axis.

I can be calculated from thw wheel geometry and mass.

When the anchor is in the water, then you have the same problem but with a third force, water drag, acting on the anchor.

3. Aug 10, 2008

### hot2moli

How do I find Tension.. assuming I can solve for acceleration using T-mg = ma?

4. Aug 10, 2008

### Chrisas

Ok, I threw that off the top of my head without working it through on paper first. So it is a bit harder than I first made it seem. Sorry.

You need to eliminate T. You have one equation with T from the anchor equation and you have another equation with T from the torque equation. Combine the two to eliminate T. Now it's not so easy to integrate as I initially led you to think since the result has y_doubledot (downward accel of anchor) and w_dot in it.

There is another equation you can use to convert w_dot and y_doubledot into each other. Look up what is called the "rolling" condition. Basically the fact that the rope comes off the wheel without slipping gives you a relation between w and y_dot.

5. Aug 10, 2008

### hot2moli

torque = -T = 1/2Ma
is that right?
but is that M different than the m in T-mg = ma?

6. Aug 10, 2008

### Chrisas

You are given two masses. Mass of the anchor is used in the anchor free body diagram. Mass of the wheel is used in the wheel/torque diagram. Also don't confuse the variable "T" which is tension (force) with Torque (force*distance). If you use "T" for tension, then use "N" or something else to represent torque to keep them straight.

Remember that Torque, N = I * w_dot. It is also the force causing the rotation multiplied by the distance from the spin axis.

You can look up or calculate I, moment of inertial, of a thin disk.

7. Aug 10, 2008

### LowlyPion

You have 4 parts to the problem. Do you understand how you solve each of the parts? For instance what formula do you think you would use for part (a)?

8. Aug 11, 2008

### hot2moli

Displacement = S/r

But then I get lost in the equations for the rest.
I understand T - mg = ma
and Torque I believe is = Tension= (1/2)Ma
is that correct? and then I can solve for a by substituting the 2 equations

9. Aug 11, 2008

### LowlyPion

That looks right for (a).
As to the next part. Careful there. The torque is going to be the moment of inertia of the spool times the angular acceleration. It's not quite as simple as 1/2 m a.
Do you know what the moment of inertia of a cylinder would be? Because you will want to use that to relate to the Tension in the line running to the anchor.