(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A ring of mass 2.83 kg, inner radius 5.30 cm, and outer radius 6.95 cm rolls (without slipping) up an inclined plane that makes an angle of θ=37.4°, as shown in the figure below.

http://capa.physics.mcmaster.ca/figures/sb/Graph11/sb-pic1106.png [Broken]

At the moment the ring is at position x = 2.18 m up the plane, its speed is 2.69 m/s. The ring continues up the plane for some additional distance and then rolls back down. It does not roll off the top end. How much further up the plane does it go?

2. Relevant equations

I think this has something to do with Kinetic energy of the rolling object...

KE=1/2Iw^2+1/2mv^2 (where I= MR^2)

w=2pi rad/perod

delta xcm= vcm delta t (to find the period)

PE=mgh

sin 37.4=opp/hyp (which I put as the distance x)

3. The attempt at a solution

I first calculated for the period using

delta xcm= vcm delta t (to find the period)

2.18=2.69 t

t=0.810s

w=2pi rad/period

=2pi rad/0.810s

=7.76 rad/s

KE=1/2Iw^2+1/2mv^2

=1/2(2.83kg)(0.1225m)^2x(7.76rad/s)^2+1/2(2.83kg)(2.69m/s)^2

=11.52J

Sin 37.4= opp/2.18m

opp (height when ring is at x)= 1.32m

mgh at this point= (2.83kg)(10N/m^2)(1.32m)

PE=37.356

Final PE when the v of the ring=0 is 37.56+11.52

=48.88J at the highest point when all the KE would have transferred to PE

PE=mgh

48.88J=(2.83kg)(10N/m^2)(h)

h=1.73m

To find the total length of the hypotentus at this height:

sin37.4=1.73m/hyp

=2.84m (final x)

Distance that still needs to be travelled:

2.84- 2.18 (initial x)= 0.663m

But this answer is not right. I feel like that there's something that I missing or not relating properly! If anyone can provide some insight I'd really appreciate it! :)

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# Homework Help: Object Rolling up ramp

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