Calculate angular velocity of a ball rolling down incline?

[Dinkleberg]

Homework Statement

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

Homework Equations

I=2/5 Mr^2
K = 1/2 m*v^2
K_roll = 1/2 Iw²
mgh=K+K_roll

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.

 

[TSny]

Hello. Welcome to PF!

w² = (10gh)/(7r²)

OK

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 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.

 

[Dinkleberg]

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

 

[Dinkleberg]

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.

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

 

[gneill]

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

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.

 

[TSny]

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

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.

 

[TomHart]

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

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.