# Homework Help: Help with the questions!

1. Nov 27, 2012

### NasuSama

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

In each question: you place mass m on the mass holder, and released the system from rest. The system includes two pulleys and a long rod with two point masses on it.

a) In procedure 1: if you increase m, the mass on the mass holder, what happens to the following quantities:
- α, the angular accelertion of the pulley.
- I, the moment of inertia of the pulleys-plus-rod system.

In procedure 2: if you move the two masses attached to the rotating rod away from the center of mass of the rod, what happens to the following quantities:
- α, the angular accelertion of the pulley.
- I, the moment of inertia of the pulleys-plus-rod system.

2. The attempt at a solution

For part (a), I thought the answer is α increases, I increases, but it's wrong.
For part (b), I thought the answer is α decreases, I decreases, but it's also wrong.

2. Nov 27, 2012

### Simon Bridge

A lot depends on the arrangement of the pulleys - do you have a diagram?
You should also show your reasoning ... in each case the answer could be "increases", "decreases" or "stays the same" ... so why would you expect the quantity to "increase"? Under what circumstances would each quantity remain the same or decrease?

3. Nov 29, 2012

### NasuSama

Oh!

The first image I uploaded is the diagram for the first part.

The second part? I can't picture the diagram. Here is the procedure...

Procedure 2
1. Taking Data
a) Trial 0:
•You did procedure 1 with the two masses on the rodat the ends of the rod.
Measure d, the distance from the centerof each mass to the center of the
axis of rotation. Record this value on your data sheet.
•Fill in your data sheet with the values you calculated in procedure 1.
b) Trial 1
•Slide the two mases on the rod inward so d (the distance of each mass to
the axis of rotation) is about 3 cm less than in trial 0. Tighten the set
screws..
•Use the vernier calipers to measure d, the distance from the center of each
mass to the center of the axis of rotation. BE SURE THE DISTANCE d IS
THE SAME FOR BOTH MASSES. Record this value on your data sheet.
•Now repeat the steps from procedure 1 trial 1 (wrapping the string around
the smallest wheel and using the emptymass holder). Record all data and
calculated values on your data table.
c) Trial 2: Slide the two mases on the rod inward so d is about 3 cm less than in
trial 1. BE SURE THE DISTANCE d IS THE SAME FOR BOTH MASSES.
Now, repeat all the steps from trial 1, and record all data and calculations on
d) Trial 3: Slide the two mases on the rod inward so d about 3 cm less than in
trial 2. BE SURE THE DISTANCE d IS THE SAME FOR BOTH MASSES.
Now, repeat the steps from trial 1, and record alldata and calculated values
e) Trial 4: Slide the two mases on the rod inward so d about 3 cm less than in
trial 3. BE SURE THE DISTANCE d IS THE SAME FOR BOTH MASSES.
Now, repeat the steps from trial 1, and record alldata and calculated values
2.Bring up Excel:
•Create two graphs: Ivs. dand Ivs. d
2
.
•Use a linear fit for the second graph.
•Put a title on the graph that includes your name, and print the graph.

#### Attached Files:

• ###### here.JPG
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4. Nov 29, 2012

### NasuSama

For both parts of the question, why aren't my answers correct?

5. Nov 29, 2012

### NasuSama

Sorry for another post.

I am not sure why the first and second answers are incorrect.

1. α should increase if you add more mass, but I would thought that inertia increases too.

My second chance is that α increases while inertia stays the same.

2. α should decrease if both masses are added to both sides. Then, inertia stays the same as always?

6. Nov 29, 2012

### Simon Bridge

Focussing on part 1 for now:
In your attached diagram, part 1 only has one pulley. Is not part of a system involving more pulleys?
I notice there is no rod in the diagram you supplied - but there is one in the question. What happened to it?

from the free body diagrams:

for the pulley: $rT=I\alpha=\frac{1}{2}MR^2\alpha$ (taking anticlockwise as positive)
for the mass: $mg-T=ma$ (taking down as positive, so that $a=R\alpha$)

r= radius of where the string pulls
M=mass of pulley
m=mass of weight

eliminate T from both equations and solve for $\alpha$.
How does $\alpha$ depend on the mass?

Last edited: Nov 29, 2012
7. Nov 29, 2012

### NasuSama

That is just a pulley for the first part only. For the second part, there are two masses on the pulleys...

Oh! I mean for the diagram I just posted, I treat the circle as the rod.

#### Attached Files:

• ###### here.JPG
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Views:
131
8. Nov 29, 2012

### Simon Bridge

You mean the rod is an axle?
Doesn't it have two point masses on it?
Anyway - I added a bit to post #6 while you were replying.

9. Nov 29, 2012

### Simon Bridge

Do the questions in reverse:

In Part 1.
You added the mass to the falling weight - what does that do to the
(a) moment of inertia of the pulley (you said in increased - why? what is your reasoning?)
(b) angular acceleration (you said it increased - what is your reasoning?)
... the previous discussion of a free-body diagram should help here.

In Part 2.
The falling weight stays the same - but the small masses are extended farther from the axle.
(a) moment of inertia of the pulley (you said it decreased - why? what is your reasoning?)
(b) angular acceleration (you said it decreased - what is your reasoning?)

Looking at them in that order should help.
Telling me your reasoning will help me explain it to you if you still don't see it ;)

Last edited: Nov 29, 2012