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cozzbp

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Hi, I just don't really understand rotational motion very well, and I don't know how to proceed with this problem.

You throw a 235 g ball 26 m/s at the apparatus shown in the figure.

http://volta.byu.edu/ph121/homework/hw18f3.png [Broken]

The apparatus catches the ball, causing it to rotate. There is a kinetic frictional force on the bearing of the ball of 17 Newtons. The bearing has a radius of a=2.5 cm and the apparatus has a distance of r=1.2 m between the catcher and the axis of rotation. The catcher has a mass of 408 g and the rest of the apparatus has negligible mass. How many revolutions does the apparatus rotate before it stops?

L = mvr

[tex]\Sigma[/tex][tex]\tau[/tex] = I[tex]\alpha[/tex]

[tex]\omega[/tex]=v/r

Here is what I have tried so far...

First, I need to find the velocity after the ball and the apparatus collide. So,

m

.235 * 26 = (.235 + .408)*v

solving for v

then I can find the angular velocity using [tex]\omega[/tex] = v

9.5/2.5 = 7.92 rad/s

I can also get the angular momentum from this.

My problem is, is that I have no idea how to factor gravity into the whole equation. I figure if I could find [tex]\alpha[/tex], then I could use a rotational kinematic equation to find the change in [tex]\vartheta[/tex]. My other thought was that I could somehow use the friction and rotational kinetic energy to find when it stops. Any help would be greatly appreciated.

## Homework Statement

You throw a 235 g ball 26 m/s at the apparatus shown in the figure.

http://volta.byu.edu/ph121/homework/hw18f3.png [Broken]

The apparatus catches the ball, causing it to rotate. There is a kinetic frictional force on the bearing of the ball of 17 Newtons. The bearing has a radius of a=2.5 cm and the apparatus has a distance of r=1.2 m between the catcher and the axis of rotation. The catcher has a mass of 408 g and the rest of the apparatus has negligible mass. How many revolutions does the apparatus rotate before it stops?

## Homework Equations

L = mvr

[tex]\Sigma[/tex][tex]\tau[/tex] = I[tex]\alpha[/tex]

[tex]\omega[/tex]=v/r

## The Attempt at a Solution

Here is what I have tried so far...

First, I need to find the velocity after the ball and the apparatus collide. So,

m

_{ball}*v_{ball}= (m_{ball}+ m_{catcher})*v_{f}.235 * 26 = (.235 + .408)*v

_{f}solving for v

_{f}yields 9.5 m/sthen I can find the angular velocity using [tex]\omega[/tex] = v

_{f}/r9.5/2.5 = 7.92 rad/s

I can also get the angular momentum from this.

My problem is, is that I have no idea how to factor gravity into the whole equation. I figure if I could find [tex]\alpha[/tex], then I could use a rotational kinematic equation to find the change in [tex]\vartheta[/tex]. My other thought was that I could somehow use the friction and rotational kinetic energy to find when it stops. Any help would be greatly appreciated.

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