A 350 gram ball is swung at the end of a string that is 82 cm long in a vertical

Question

A 350 gram ball is swung at the end of a string that is 82 cm long in a vertical circle at a uniform speed. If it completes 35.5 rpm calculate the tension in the string when the ball is at the top of it's path.

Explanation / Answer

A 350 gram ball is swung at the end of a string that is 82 cm long in a vertical circle at uniform speed. If it completes 35.5 rpm, calculate the tension in the string when the ball is at the top of it's path. solution: i) for vertical circle Fy = 0 , T+mg = -m*?^2*r (here r = l) => T = m*( (?^2)*r -g) = 0.5317 N for details view the horizantal circle part :) i) for horizantal circle In this problem , the centripetal force is crucial, splitting the force's acting on the ball in x and y components and balancing them Fx = 0 =T*sin(?) -m*v^2/r [Fcent=m*v^2/r => centripetal force] Fy =0 =T*cos(?) -mg r = radius of the circular path l = length of the string. h = height from the ceiling to the horizantal postion of the ball. sin(?) = l/r , where ? , is angle by the string while in motion , with the vertical. cos(?) = h/r v = ?*r so taking Fx components T *r/l = m*(?*r)^2/r T = m*?^2*l now m = 350 gram = 0.35 kg ? = 35.5 rpm = 3.717 rad/s l = 82 cm =0.82 m T = (0.35)*(3.717)^2*0.82 = 3.966 N so tension in the string is 3.996 N

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