Rolling Motion without slipping Problem

In summary, the problem involves a small rubber wheel driving a large pottery wheel through contact at their edges. The small wheel has a radius of 2.0cm and accelerates at a rate of 7.2 rad/s^2. The large pottery wheel has a radius of 25.0cm and the goal is to calculate its angular acceleration and the time it takes to reach a speed of 65 rpm. Using equations for tangential speed and angular acceleration, the angular acceleration of the pottery wheel is found to be 0.576 rad/s^2 and it takes 11.82 seconds for the pottery wheel to reach 65 rpm.
  • #1
adca14
11
0

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] = ?

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]

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
 
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  • #2
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:
  • #3
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.
 
  • #4
adca14 said:
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.
 

1. What is rolling motion without slipping?

Rolling motion without slipping is a type of motion in which a body, such as a wheel or a ball, rotates and translates at the same time without any slipping or skidding.

2. What is the difference between rolling motion with and without slipping?

In rolling motion with slipping, the body rotates and translates while also experiencing slipping or skidding. In rolling motion without slipping, the body rotates and translates without any slipping or skidding.

3. How do you calculate the velocity of a rolling object without slipping?

The velocity of a rolling object without slipping can be calculated by using the equation v = ωr, where v is the linear velocity, ω is the angular velocity, and r is the radius of the object.

4. What is the role of friction in rolling motion without slipping?

Friction plays a crucial role in rolling motion without slipping, as it provides the necessary force to prevent slipping or skidding. Without friction, the rolling object would slide instead of roll.

5. Can an object have both rolling and sliding motion at the same time?

No, an object cannot have both rolling and sliding motion at the same time. In order for an object to roll without slipping, the point of contact between the object and the surface must remain stationary, and this cannot occur if there is also sliding motion.

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