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**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