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wendo
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Homework Statement
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
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?
Homework 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)
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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