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Rolling Motion without slipping Problem

  • Thread starter adca14
  • Start date
11
0
1. Homework Statement
A small rubber wheel is used to drive a large pottery wheel, and they are mounted so that their circular edges touch. The small wheel has a radius of 2.0cm and accelerates at a rate of , and it is in contact with pottery wheel (radius 25.0 cm)without slipping. Calculate (a) the angular acceleration of the pottery wheel, and (b) the time it takes the pottery wheel to reach its required speed of 65 rpm.

r for small rubber ball = 2cm = .02m
r for large pottery ball = 25cm = .25m
[tex]\alpha[/tex] for small rubber ball = 7.2 rad/s[tex]^{}2[/tex]
[tex]\alpha[/tex] for large pottery ball = ?
t[tex]_{}1[/tex] = 0s
t[tex]_{}f[/tex] = ?
[tex]\omega[/tex] for small rubber ball = ?
[tex]\omega[/tex] for large pottery ball = ?
[tex]\ell[/tex] = ?
[tex]\theta[/tex] = ?

2. Homework Equations
v=r[tex]\omega[/tex]
[tex]\theta[/tex] = [tex]\ell[/tex]/r
[tex]\omega[/tex][tex]^{}2[/tex] = [tex]\omega_{}o[/tex][tex]^{}2[/tex] + 2[tex]\alpha[/tex][tex]\theta[/tex]
[tex]\alpha[/tex] = [tex]\omega[/tex][tex]^{}2[/tex] - [tex]\omega_{}o[/tex][tex]^{}2[/tex]/2[tex]\theta[/tex]

3. The Attempt at a Solution

My book said that under certain circumstances [tex]\ell[/tex]=2[tex]\pi[/tex]r
So i plugged 2[tex]\pi[/tex](.02m) = .12m

Than to solve for [tex]\theta[/tex], I did [tex]\ell[/tex]/r or .12/.02 = 6.28

To get [tex]\omega[/tex] I did [tex]\omega[/tex][tex]^{}2[/tex] = [tex]\omega_{}o[/tex][tex]^{}2[/tex] + 2[tex]\alpha[/tex][tex]\theta[/tex] I set the first [tex]\omega[/tex] to zero isolated [tex]\omega^{}2[/tex] by taking the square root to both sides, plugged everything in, [tex]\sqrt{}2(7.2)(6.28)[/tex] and I got 9.51 rad/s

Then to solve for v, I used v=r[tex]\omega[/tex], plugged it in, .02(9.51), and got .19

Then to get [tex]\omega[/tex] for the pottery wheel I used [tex]\omega[/tex] = v/r, plugged it in, .19/.25, and got .76

For [tex]\alpha[/tex], I used [tex]\omega^{}2[/tex]/2[tex]\theta[/tex] and got .045

this doesn't look right, I hope I did it right though, any help would be appreciated, again thanks
 

Answers and Replies

dynamicsolo
Homework Helper
1,648
4
I'm assuming that the two wheels are starting at rest, though the problem says nothing about that (but it would be typical for a problem of this sort).

The condition that the wheels are in contact and roll against each other without slipping means that the linear or tangential speeds at the edges of the two wheels are the same. So you know that [tex]\omega_{rub} = \omega_{pot} = 0[/tex] at t = 0 . You also know the relation between tangential speed at the edge of each wheel and its angular speed; what is it?

You can know set up an equation for the tangential speed of the edge of each wheel, given that each has a constant angular acceleration. You know [tex]\alpha_{rub}[/tex], so you can now find [tex]\alpha_{pot}[/tex]. Once you have that, you can find how long it will take for the pottery wheel, starting from rest, to reach 65 rpm (which is how many radians per second?).
 
Last edited:
11
0
I got it, the relation with the tangential speed at the edge of each wheel was what I needed. I set the tangential accelerations of both equal to each other and got it, .576, then I i used w = w0 + at and got t, 11.8s, thanks again, if it weren't for this forum, I'd be in a jam.
 
dynamicsolo
Homework Helper
1,648
4
I got it, the relation with the tangential speed at the edge of each wheel was what I needed. I set the tangential accelerations of both equal to each other and got it, .576, then I i used w = w0 + at and got t, 11.8s, thanks again, if it weren't for this forum, I'd be in a jam.
I think you mean that you "set the tangential velocities equal to each other".

Yes, I get 0.576 rad/(sec^2) for the pottery wheel's angular acceleration, so it runs up to speed in 11.82 seconds.
 

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