An athlete swings a 4.90 kg ballhorizontally on the end of a rope. The ball move

Question

An athlete swings a 4.90 kg ballhorizontally on the end of a rope. The ball moves in a circle ofradius 0.670 m at an angular speed of0.460 rev/s. (a) What is the tangential speed of theball?
1 m/s

(b) What is its centripetal acceleration?
2 m/s2

(c) If the maximum tension the rope can withstand before breakingis 105 N, what is the maximum tangentialspeed the ball can have?
3 m/s (a) What is the tangential speed of theball?
1 m/s

(b) What is its centripetal acceleration?
2 m/s2

(c) If the maximum tension the rope can withstand before breakingis 105 N, what is the maximum tangentialspeed the ball can have?
3 m/s

Explanation / Answer

      Tagential speed       V = R          = 0.67 x0.46 rev /s ( 2 rad / 1 rev)          = 1.9364m/s Centripetal acceleraion      a = V2 /R         =(1.9364m/s)2 / 0.67         = 5.597m/s2 The angle made with the vertical is     tan = ( V2 /Rg)              = tan-1[(1.9364)2 /0.67 (9.8) ]               = 29.73o maximum tangential speed from the equation The angle made with the vertical is     tan = ( V2 /Rg)     tan = ( V2 /Rg)              = tan-1[(1.9364)2 /0.67 (9.8) ]               = 29.73o maximum tangential speed from the equation       T sin = m V2/ R        V = [ RTsin / m ]          = 2.6683m/s        

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