A group of students performed the same \"Conservation of Mechanical Energy\" exp

Question

A group of students performed the same "Conservation of Mechanical Energy" experiment that you performed in lab. However, instead of keeping the horizontal portion of the ramp at a constant height h2 from the tabletop, they varied this height and released the ball from the same height h1 each time. They then entered their results into the first two columns of the table shown in part (a).

(a) Complete the table below. ******completed below, with v2kin avgerage = 236.6 cm/s*********

(b) Use the average value of v2kin to determine the height h1 through which the ball falls. ?????????

****I used v2kin^2 = sqrt ((10*981*h1)/7) and solved for h1, however, am only getting partial credit. I'm not sure what I am doing wrong? ************

# Height of ramp from table top h2 (cm) Horizontal distance d (cm) v2kin (cm/s) 1 10.60 34.81 2 23.17 50.99 3 31.20 61.01 4 41.22 67.67 5 49.10 74.77

Explanation / Answer

). If the object is moving at linear speed "v" and rotating at rate "?" at the bottom of the ramp, then it has this much KE: Total KE = linear KE + rotational KE = ½mv² + ½I?² Substituting the value of "I" from above: KE = ½mv² + ½(ßmR²)?² If the object is not slipping, then "v" and "?" are related as follows: ? = v/R Substiting into the KE formula: KE = ½mv² + ½(ßmR²)(v/R)² Simplifying: KE = ½mv²(1 + ß) If mechanical energy is conserved, then: PE+KE at top = PE+KE at bottom; mgh = ½mv²(1 + ß) Simplifying (divide by "m"): gh = ½v²(1+ß) Thats a general formula that doesn't depend on the object's shape or mass distribution. For the specific case of the solid sphere: gh = ½(v_sphere)²(1 + ß_sphere) For the specific case of the solid cylinder: gh = ½(v_cylinder)²(1 + ß_cylinder) Since these two are both equal to "gh", they're both equal to each other: ½(v_sphere)²(1 + ß_sphere) = ½(v_cylinder)²(1 + ß_cylinder) Solve for v_cylinder: v_cylinder = (v_sphere)sqrt((1 + ß_sphere)/(1 + ß_cylinder))

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