Disk A and B are identical and roll across a floor with equalspeeds. Then disk A

Question

Disk A and B are identical and roll across a floor with equalspeeds. Then disk A rolls up an incline, reaching a maximum heighth, and disk B moves upl an incline that is identical except that itis frictionless. Is the Maximum height reached by disk B greaterthan, less than, or equal to h?

I figured out that disk B would reach a lesser height than disk Adue to the fact that its deceleration is greater than disk A'sdeceleration. (Used Newton's second law)

How would I explain the difference in heights using conservation ofenergy? I keep thinking that disk B would not stop because it hasKE at the bottom - no friction - so it keeps going. But then wheredid the rotational energy go? What am I missing?

Explanation / Answer

Disk B will only go up the hill to a point where it loses itstranslational energy, ie. h = 1/2 mv2/mg h = v2 /2g This will be less than the height reached by Disk A, whichwill be h = (1/2 m v2 + 1/2 I 2)/mg . But Disk B will continue to have its rotational energy -- itwill keep spinning at the same rate. So it will continue tohave E = 1/2 I 2 . So, at the top, Disk B will be spinning forward, but slidingbackwards. ******* I'm not sure how far along you are in your physics education,but the rotational kinetic energy of an object is: E = 1/2 I2 . And total kinetic energy is E = 1/2 mv2 + 1/2 I 2 . So, at the top, Disk B will be spinning forward, but slidingbackwards. ******* I'm not sure how far along you are in your physics education,but the rotational kinetic energy of an object is: E = 1/2 I2 . And total kinetic energy is E = 1/2 mv2 + 1/2 I 2

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