From rest, a ball rolls from the top of a hill to thebottom, reaching a translat

Question

From rest, a ball rolls from the top of a hill to thebottom, reaching a translational speed of 6.6m/s. Ignoringfrictional losses, what is the height of the hill? Using the energy conservation, E = 0.5mv^2 + 0.5Iw^2 + mgh, Isimplified the equation down to solve for the height, h. Thisequation became as follows: h= (0.5vf^2 + 0.5Iw^2) / g Now, I'm not sure what actions to take regarding the moment ofinertia and angular velocity. I believe angular velocity isequivalent to w = vf / r and moment of inertia may be I =mr^2. However, I'm not sure if my work is correct thus far,and if it is, how to go and obtain a value for the radius, r. From rest, a ball rolls from the top of a hill to thebottom, reaching a translational speed of 6.6m/s. Ignoringfrictional losses, what is the height of the hill? Using the energy conservation, E = 0.5mv^2 + 0.5Iw^2 + mgh, Isimplified the equation down to solve for the height, h. Thisequation became as follows: h= (0.5vf^2 + 0.5Iw^2) / g Now, I'm not sure what actions to take regarding the moment ofinertia and angular velocity. I believe angular velocity isequivalent to w = vf / r and moment of inertia may be I =mr^2. However, I'm not sure if my work is correct thus far,and if it is, how to go and obtain a value for the radius, r.

Explanation / Answer

         Translational speed, V = 6.6 m/s     Height, h = ?     Mass of the ball = M     Radius of the ball = R     From law of conservation of energy,     ( KEt + KEr + PEg ) i = ( KEt + KEr +PEg ) f     Rotational Kinetic energy, KEr = ( 1/2 ) I2     0 + 0 + M g h = ( 1/2 ) M V2 + ( 1/2) [ ( 2/5 ) M R 2 ] 2 + 0     But, V = R     M g h = ( 1/2 ) M V2 + ( 1/5 ) ( M V2 )     M g h = ( 7/10 ) M V2     g h = ( 7/10 ) V 2     9.8 * h = 0.7 * 6.62     height of the hill, h = 3.11 m

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