Find Angular Speed: Urgent Help Needed - Can't Get 32.9

AI Thread Summary
The discussion revolves around finding the angular speed of a sphere rolling down a slope, with the target answer being 32.9. The user initially struggles with the lack of time data and attempts to apply energy conservation principles, realizing that potential energy at the top converts to kinetic energy at the bottom. They derive the equation mgh = (1/2)Iω² + (1/2)mv², leading to the conclusion that the total kinetic energy includes both rotational and linear components. Ultimately, they confirm that using the relationship ω = v/r is appropriate for calculating angular speed in this context, clarifying that the force applied is not relevant to the final answer. The discussion emphasizes the importance of understanding energy conservation and the definitions of angular velocity in rotational motion.
riseofphoenix
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Urgent help needed with finding angular speed? This is impossible!

I've even asked on Y!A and THEY all got it WRONG.

Webassign says the answer is 32.9 but how?
This is so freaking frustrating...anlfdsfjlkdjf

IMPOSSIBLEQUESTION.png


Ok...so what I tried doing was ask myself what the equation for angular speed was.ω = at

And then I realized, they don't give me time. So I MUST use some OTHER equation to find angular speed. And then I realized that the key word in the problem was that sphere STARTED FROM REST (INITIAL POSITION) and ENDED AT THE BOTTOM OF THE SLOPE (FINAL.) And then I was like OHHHHH... ok this should be a freaking walk in the park...

-.-

PEinitial + [STRIKE]KEinitial[/STRIKE] = [STRIKE]PEfinal[/STRIKE] + KEfinal

In other words, the ball has PE at the top when it's at rest and no KE, and when it rolls to the bottom it has no PE but it has KE.

So...

1) PEinitial = KEfinal
mgh = (1/2)mω2

2) Find mass and plug it in equation above

F = ma
(300 N) = m(9.8)
300/9.8 = m
30.61 kg = m

And then here I said to myself "for the love of God...mass cancels out anyways -.- what a huge waste of time"

3) gh = (1/2)ω2
(9.8)(6 sin 31) = (0.5)ω2
30.28 = (0.5)ω2

30.28/0.5 = ω2
60.56 = ω2
7.78 = ω

?
Help?

Can someone please SHOW me and explain to me how they got 32.9?
 
Last edited:
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The motion of the sphere combines translation and rotation. Total KE is the sum of the respective KE's. What are they?
 


voko said:
The motion of the sphere combines translation and rotation. Total KE is the sum of the respective KE's. What are they?

1) PE = Total KE (Rotational KE + Linear KE) since the solid sphere is rotating while it rolls down the slope.

2) mgh = (1/2)Iω2 + (1/2)mv2
m(9.8)(3.09) = (1/2)[(2/5)mr2](v/r)2 + (1/2)mv2
(30.282)m = (1/5)mr2(v2/r2) + (1/2)mv2
(30.282)m = (1/5)mv2 + (1/2)mv2
30.282 = (1/5)v2 + (1/2)v2
30.282 = (0.2)v2 + (0.5)v2
30.282 = 0.7v2
43.26 = v2
6.57 m/s = v

3) Angular velocity (ω) = v/r

ω = 6.57/0.20
ω = 32.85
ω = 32.9

So ultimately, it didn't matter what the Force (F) was in this particular problem right?
Also, I thought angular velocity/speed (ω) was ω = at?
Do I use (ω) = v/r when a solid sphere is in motion/rolling?
 
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That seems good.
 


It's not a case of using only one or the other...

Also, I thought angular velocity/speed (ω) was ω = at?

That equation is an equation of motion similar to Velocity = acceleration/time eg...

Angular Velocity = Angular acceleration * time

it's fine to use that to solve a problem if you know the Angular acceleration and time or can calculate them from the data given.

Do I use (ω) = v/r when a solid sphere is in motion/rolling?

That equation comes from the properties of a circle and the definition of angular velocity...

velocity = distance / time
v = circumference / time for one revolution
v = 2∏r / T

angular velocity ω = angle / time
ω = 2∏ (radians) / time for one revolution

so

v = ωr and
ω = v/r

It's fine to use that to solve a problem if you know or can calculate v.
 
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