Solid sphere Kinetic Energy problem

AI Thread Summary
A solid sphere rolls down an incline from a height H at a 30° angle, reaching a speed of 65 cm/s at the bottom. The problem is approached using conservation of energy, equating potential energy (PE) at the top to kinetic energy (KE) at the bottom. The relevant equations include PE = mgh, KE = 1/2 mv^2, and rotational energy terms. The key challenge is to solve the equations without needing the mass or radius of the sphere. The final calculated height H is 3.02 cm, which appears to be correct.
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Homework Statement


A solid sphere is released from rest at the top of an incline of height H and angle 30°. The sphere then rolls down the incline without slipping until it reaches the bottom of the incline, at which point the speed of the center-of-mass of the sphere is found to be 65 cm/s.

What is the value of the height H?

Homework Equations



conservation of energy

KEi + REi + PEi = KEf + REi +PEf

PE=mgh
RE= 1/2*I*w^2
KE= 1/2*m*v^2
where I= 2/5*m*r^2
and v=w*r
w=omega
r=radius
m=mass
v=velocity

The Attempt at a Solution



apparently this is supposed to be solved without mass or the radius of the sphere.
but i can't get all those variables to cancel any help would be awesome.

mgh=1/2*m*(w*r)^2 + 1/2*(2/5mr^2)*(v/r)^2
 
Last edited:
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Firstly, how did you set up the conservation of energy?
Secondly, did you use the fourth and fifth relevant equation you listed in order to substitute?
 
As

PEi = KE(linear)f + KE(rotational)f

and ended up with

mgh=1/2*m*(w*r)^2 + 1/2*(2/5mr^2)*(v/r)^2
 
The kinetic energy term lies the problem. You don't want to end up with neither w nor r, so don't replace v!
 
Thank you so much..this was really bugging me...ended up with h= 3.02cm.
 
Seems right to me.
 

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