# Calculate angular velocity of a ball rolling down incline?

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1. Apr 28, 2017

### Dinkleberg

1. The problem statement, all variables and given/known data
An 7.80-cm-diameter, 400 g solid sphere is released from rest at the top of a 1.70-m-long, 20.0 degree incline. It rolls, without slipping, to the bottom

2. Relevant equations
I=2/5 Mr^2
K = 1/2 m*v^2
Kroll = 1/2 Iw2
mgh=K+Kroll
3. The attempt at a solution
Using the above energy equation and the fact that the ball is rolling without slipping I replaced v with 'rw' and then solved for w which gets me:
w2 = (10gh)/(7r2)
Where h is the height of the ramp. But I domt know what to do next. I tried finding h with trig and I got h=0.69
but when I subbed those values in it was incorrect apparently. So the question is a little ambiguous when it says 'long' so I thought what if it means that is the length of the ramp itself which gets h=0.58 which is still wrong.

I am taking the units to be rad/s and they actually matter for the answer if that is the problem.

2. Apr 29, 2017

### TSny

Hello. Welcome to PF!
OK
I think this is the intended interpretation.
Your outline of the solution looks correct to me. Sometimes there can be a careless error in substituting the numbers. You did not state the numerical value of your answer.

3. Apr 29, 2017

### Dinkleberg

So using that formula I get w=36.53rad/s

4. Apr 29, 2017

### Dinkleberg

Woops actually I get 79.69 because I forgot it is radius not diameter but that is still incorrect apparently

5. Apr 29, 2017

### Staff: Mentor

You might be running into rounding or truncation issues affecting your significant figures. For example, if you truncate the change in height to two figures: h = 0.58 m, does it make any sense that anything calculated from that value could have more than two significant figures?

Re-run your calculation keeping extra digits for all intermediate values (don't round anything until the very end) and see what you get. Use a value for g that has at least three significant figures.

6. Apr 29, 2017

### TSny

This answer corresponds to h = 0.69 m. What if you use h = .58 m? As gneill pointed out, your data is given to 3 significant figures, so you want your calculation to be accurate to the same number of significant figures.

7. Apr 30, 2017

### TomHart

I think you should just try the math again. I ended up with the same final equation you did, plugged in the numbers, and got a different answer.
I used
g = 9.81m/s^2
h = 0.5814 m
r = 0.039 m

Edit: Based on your previous posts, I suspect your radius may be wrong.
Edit2: @TSny, you are probably right in post #6. I missed that when I posted.

Last edited: Apr 30, 2017
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