Answered

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An athlete swings a 6.90-kg ball horizontally on the end of a rope. The ball moves in a circle of radius 0.830 m at an angular speed of 0.680 rev/s.
(a) What is the tangential speed of the ball?
m/s
(b) What is its centripetal acceleration?
m/s2
(c) If the maximum tension the rope can withstand before breaking is 110 N, what is the maximum tangential speed the ball can have?
m/s

Sagot :

Pipsqeak1

Answer:

a) 0.5644 m/s

b) ~0.384 m/s^2

c) ~3.638 m/s

Explanation:

a) Tangential speed is found be the radius*rotational speed, so it is 0.83*0.68 = 0.5644 m/s

b) Centripetal acceleration is found by v^2/r, so it is (0.5644^2)/0.83 = ~0.384 m/s^2

c) Let the tangential speed be v. The maximum centripetal force 110 N (as given). Centripetal force = mass*centripetal acceleration = mass*v^2/r (because centripetal acceleration is found by v^2/r). Inputting the values from the problem and solving for v, we get:

110 = 6.9*v^2/0.83

v = sqrt(110*0.83/6.9) = ~3.638 m/s

I hope this helps! :)

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