Question

An athlete swings a 3.10-kg ball horizontally on the end of a rope. The ball moves in a circle of radius 0.660 m at an angular speed of 0.520 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 136 N, what is the maximum tangential speed the ball can have? m/s

Answer 1

(a) Radius of the circle, r = 0.660 m

So, circumference of the circle = 2*pi*r = 2*3.141*0.660 = 4.146 m

Angular speed of the ball = 0.520 rev/s

And in one revolution, linear distance covered by the ball = circumference of the circle.

Therefore, tangential speed of the ball = 0.520 * 4.146 m/s = 2.16 m/s

(b) Centripetal acceleration, a = v^2 / r = 2.16^2 / 0.660 = 7.04 m/s^2

(c) Tension in the rope, T = m*v^2/r

Put the maximum value of tension (T) and then calculate the maximum corresponding value of tangential speed.

=> 136 = 3.10*v^2 / 0.660

=> v^2 = (136*0.66) / 3.10

=> v = 5.38 m/s

So, maximum tangential speed = 5.38 m/s (Answer)

answered by: anonymous