Simple energy question with rotation

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SUMMARY

The discussion focuses on determining the minimum height \( h \) required for a small solid sphere of mass \( m \) and radius \( r \) to roll down a hill and remain on track throughout a loop of radius \( R \), where \( R >> r \). The key equation derived is \( h = \frac{11}{5} R \), which accounts for both translational and rotational kinetic energy. The moment of inertia used is \( I = \frac{2}{5} m r^2 \), and the analysis confirms that at the top of the loop, the normal force is zero, leading to the condition \( v^2 = gR \).

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Homework Statement


a small solid sphere of mass m and radius r starts from rest and rolls down a hill without slipping. the sphere encounters a loop of radius R where R >> r.

Given R determine the min height h such that the ball remains on the track throughout the loop


Homework Equations



I = 2/5 mR^2

ω = v/r

The Attempt at a Solution



highest point requiring most energy is at top of loop:

mgh = mg2R + 1/2 Iω^2

mgh = mg2R + 1/2 * 2/5 m r^2 *v^2/r^2

gh = 2Rg + 1/5*v^2



at top of loop, normal force is 0,

Fy = N + mg
ma = mg

m*v^2/R = mg
v^2 = gR


...

gh = 2Rg + 1/5 Rg

h = 11/5 R

this is correct?
 
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To check your results - it is useful to compare to something:

i.e. If this were a block sliding around the track without friction, then what would the minimum h have to be?
(for the ball, some energy is stored in rotation - so you need to start higher than this.)

How fast is the ball rolling at the top of the hoop?
(i.e. is the kinetic energy a lot or a little more than the potential energy there? says if you should start a little higher or a lot higher.)
 
oreosama said:
mgh = mg2R + 1/2 Iω^2

Have you included all of the kinetic energy at the top?
 

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