Hollow sphere, angular momentum, torque problem? help?

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Homework Help Overview

The discussion revolves around a problem involving a hollow sphere rolling up an incline, focusing on the concepts of angular momentum, torque, and kinetic energy. The original poster presents a scenario where the sphere has a specific rotational inertia and total kinetic energy, and seeks to understand the distribution of this energy between translational and rotational forms.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to relate the total kinetic energy to its translational and rotational components, expressing confusion over the addition of terms with different dimensions. Some participants question the inclusion of angular velocity in the equations and how it relates to the overall kinetic energy.

Discussion Status

The conversation has seen participants exploring the relationships between translational and rotational kinetic energy, with some guidance provided regarding the need for angular velocity in the calculations. The original poster appears to have resolved their confusion regarding the relationship between angular and linear quantities.

Contextual Notes

Participants are navigating the complexities of rotational dynamics, particularly in the context of rolling motion and the conservation of energy. There is an emphasis on ensuring dimensional consistency in the equations used.

nchin
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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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