A 68-kg gymnast is swinging on a high bar. The distance between his waist and th

Question

A 68-kg gymnast is swinging on a high bar. The distance between his waist and the bar is 1.0m. At the top of the swing his speed is momentarily zero. Ignore friction and treat the gymnast as if all his mass is located at his waist.
a) Using the work-energy theorem find his speed at the bottom of the swing.
b) Find the centripetal acceleration of the gymnast at the top and at the bottom of the swing.
c) If friction were not negligible, his kinetic energy at the bottom would be reduced by 10%. How would the work of friction be, while the gymnast moves from the top to the bottom of the swing?

Explanation / Answer

This represents motion of a body in a vertical plane
(a)
mgh = 68*9.8*1
= 666.4
1/2 mv^2 = 666.4
v= sqrt(666.4*2/68)
= sqrt(19.6) = 4.42 m/s _____________________________________________________________ (b) a = v^2/r at the bottom a = (4.42)^2/1                         = 19.6 m/s^2 at the top speed is zero so acceleration will lbe zero. ____________________________________________________________ kinetic energy = 1/2mv^2                      = 1/2*68*(4.42)^2                      = 664 10% of K.E = 66 J change in K.E. = 664-66                        = 598 J

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