# Rotational Mechanics - Pulley and Bucket

1. Jan 19, 2007

### Kaoi

1. The problem statement, all variables and given/known data

Problem:
"A cylindrical 4.91 pulley with a radius of 0.338 m is used to lower a 3.80 kg bucket into a well. The bucket starts from rest and falls for 4.06 s. The acceleration of gravity is 9.81 m/s². What is the linear acceleration of the falling bucket? Answer in units of m/s²."

Givens:
$$\\ m_{p}=4.91 kg\\ \\ r=0.338 m\\ \\ m_{p}=3.80 kg\\ \\ v_{i}=0 m/s\\ \\ \Delta t=4.06 s\\$$

Unknown:
$$\\ a_{t}=?$$

2. Relevant equations

$$\\ a_{t} = r\alpha\\ \\ \Tau\=\I\alpha = Fd sin \theta\\ \\ I_{cylinder}\=\frac{1}{2} r^2\\$$
3. The attempt at a solution
$$\\ \tau = F_{g}r = I\alpha\\ \\ m_{b}gr = \frac{1}{2}mr^2\alpha\\ \\ \alpha = \frac{2m_{b}g}{m_{p}r}\\ \\ a_{t} = \frac{2m_{b}g}{m_{p}\\$$

Now, I thought that was a perfectly logical way to get the answer, but when I submitted it, the answer was wrong. Can anyone tell me what I'm doing wrong here?

Edit: And I seem to be having difficulties putting breaks into my TeX, heh.

Last edited: Jan 19, 2007
2. Jan 19, 2007

### Ja4Coltrane

torque is equal to the rate of change of angular momentum. That is-- T=d/dt(Lp+Lb)
I think that should work out.

3. Jan 19, 2007

### Ja4Coltrane

your approach just doesnt take into account the mass of the bucket.

4. Jan 19, 2007

### Kaoi

What do you mean by this? I used the mass of the bucket to calculate $$F_{g}$$, which I used to find the torque that I could divide by $$I$$ to find $$\alpha$$.

5. Jan 19, 2007

### Ja4Coltrane

Well that makes sense if Fg just acted on the pulley. If it acts over the whole system, there is some difference.
T=dL/dt=d(Lp+Lbucket)/dt=d(Iw+mvr)/dt
does that make sense? now you can distribute the d/dt to get your answer.

6. Jan 19, 2007

### curly_ebhc

I am going to try to put Ja4Coltrane's math into more conceptual terms.

Note: Remember those mass/pulley systems. This problem is just a rotational analogy.

1st: You are correct that the only force is the wieght (mg) of the bucket. However F=ma and there are two masses here. Now F is a net force so it is F=Fg - torque

The two masses are the rotational mass (I) of the cylinder (which you included) but also includes the mass of the bucket (which you forgot).

2nd: Treat both masses as one system and set Fg= both masses x(acceleration).

3. use some algebra and turn angular acceleration into (translational) acceleration. I have a feeling that some factors of r might cancel out.

I hope this is helpful although the mathematical explanation is suffecient for all calculus lovers.

Last edited: Jan 19, 2007