What fraction of its kinetic energy is rotational?

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SUMMARY

The discussion focuses on calculating the angular velocity and the fraction of kinetic energy that is rotational for a sphere rolling down a 2.00-meter incline at a 17-degree angle. The sphere has a diameter of 8.90 cm and a mass of 310 g. Key equations utilized include the conservation of mechanical energy and the formulas for rotational kinetic energy, specifically Rotational Kinetic Energy = (1/10)mv². The analysis confirms that the total kinetic energy consists of both linear and rotational components, leading to a definitive method for solving part B of the problem.

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An 8.90-cm-diameter, 310 g sphere is released from rest at the top of a 2.00-m-long, 17 degree incline. It rolls, without slipping, to the bottom.

a) What is the sphere's angular velocity at the bottom of the incline?
b) What fraction of its kinetic energy is rotational?

If someone could help me out, it'd be great...I'm not exactly sure how to tackle this problem...
 
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To start off...

Gravitational Potential Energy = Linear Kinetic Energy + Rotational Kinetic Energy

...Rotational Kinetic Energy = 1/2 I w^2

w(its called omega) = v/r

I of Sphere = (2/5)mr^2...

..thusly...Rotational Kinetic Energy = (1/2)[(2/5)mr^2][v/r]^2...which simplifies to..
...(1/10)mv^2

Work from there...
 
anaylizing both movement rotational and linear we should consider a kinetic energy of the sum of both the linear of its center of mass and the rotational. Apply Conservation of Mechanical Energy because it's pure rolling motion (no slipping).
 
Last edited:
I kind of get what you're saying but I'm still sort of lost?
 
Actually now I got part A...I just need part B...
Thanks for the help by the way
 

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