How Far Does a Ring Roll Up an Inclined Plane?

[lunarskull]

A ring of mass 2.4kg, inner radius 6c.0cm, and outer radius 8.0cm is rolling (without slipping) up an inclined plane that makes an angle of theta=36.9 with the horizontal. At the moment the ring is x=2.0 m up the plane its speed is 2.8 m/s. the ring continues up the plane for some additonal distance and then rolls abck down. Assuming that the plane is long enough so that the ring does not roll off the top end, how far up the plane does it go?

help by one of you good samaritans would be greatly appreciated.

so far, I've gotten :

I=.5MR1^2+.5MR2^2
I=.5(2.4)(.06^2)+.5(2.4)(.08)^2
I= .012 kg M^2

what do i do next? i am unaware of how to find distance and incorporate theta into this problem

 

[Hootenanny]

You need to use conservation of energy.

 

[lunarskull]

how?

You need to use conservation of energy.

how do you incorporate the numbers into conservation of energy?

 

[Hootenanny]

Assuming x is horizontal displacement you know that at this point the ring is traveling at 2.8m/s. From this you can work out the linear kinetic energy and the rotational kinetic energy. Also if you resolve you can work out the vertical height of the ring and from this you can calculate the potential energy. This will give you the total amount of energy is the system, which must remain constant. When the ring comes to rest (i.e. at its highest point on the plain) you know that there will be no rotational or linear kinetic energy. Therefore, all the energy must be converted into gravitational potential. This will give you a vertical height. Using trig.and resolving you can then calculate the distance up the inclined plane.

 

[AakashR]

dude the inner radius is 0.06m and outer is 0.08m. you should subtract the value of MIdisk you get for 0.08m from the one you get for 0.06m. Why did you add the two up. It's not making any sense. After you do it you can solve the problem easily by using conservation of energy.