Hollow sphere, angular momentum, torque problem? help?

AI Thread Summary
A hollow sphere with a radius of 0.15m and rotational inertia of 0.040 kg m^2 rolls up a 30-degree incline with an initial total kinetic energy of 20J. The discussion focuses on calculating the rotational kinetic energy, using the formulas for rotational inertia and kinetic energy. The confusion arises from combining translational and rotational kinetic energy, specifically how to express the rotational term in relation to the translational term. The key realization is that angular velocity (ω) can be expressed in terms of linear velocity (v), allowing the terms to be combined correctly. Ultimately, the participant resolves the confusion by recognizing the relationship between linear and angular motion.
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Hollow sphere, angular momentum, torque problem? help??

a hollow sphere or radius 0.15m with rotational inertia = 0.040 kg m^2 about a line through its center of mass, rolls without slipping up a surface inclined 30 degree to the horizontal. at a certain initial position,the sphere's total kinetic energy is 20J.

a) How much of this initial KE is rotational?

what i did

Hollow sphere = 2/3MR^(2)
KErot = 1/2 Iω^2

KEtotal = KEtranslational + KErotational

a) 20 = 1/2mv^2 + 1/2 x 2/3 mr^2
20 = 1/2mv^2 + 1/3 mr^2

this part confuses me
1/2mv^2 + 1/3 mr^2 = 5/6mv^2
20 = 5/6mv^2

How was 1/2mv^2 + 1/3 MR^2 = 5/6mv^2?? mv and MR are different yet how were they both added together??
 
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When dimensions don't match up, backtrack to the first instance. It's here:
nchin said:
a) 20 = 1/2mv^2 + 1/2 x 2/3 mr^2
ω is missing.
 


haruspex said:
When dimensions don't match up, backtrack to the first instance. It's here:

ω is missing.

1/2mv^2 + 1/2 x 2/3 mr^2ω = 5/6mv^2

ok so if i put a ω there, how would it still = 5/6mv^2? in other words how can you change mr^2ω into mv^2? help!
 


oh nvm i got it. w = V^2/R^2. r cancel out. thanks!
 
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