Question

An athlete swings a 4.00 kg ball horizontally on the end of a rope. The ball moves in a circle of radius 0.860 m at an angular speed of 0.660 rev/s. (a) What is the tangential speed of the ball? m/s m/s2 m/s (b) What is its centripetal acceleration? (c) If the maximum tension the rope can withstand before breaking is 136 N, what is the maximum tangential speed the ball can have?

Answer 1

Answer:

a)  Tangential speed = 3.57 m/s

b) Centripetal acceleration = 17.79 m/s²

c) Maximum tangential speed the ball can have = 5.41 m/s

Explanation:

a) Tangential speed, v = rω

   Radius, r = 0.86 m

   Angular speed, ω = 0.660 rev/s = 0.66 x 2 x π = 4.15 rad/s

  Tangential speed, v = rω = 0.86 x 4.15 = 3.57 m/s

b) Centripetal acceleration,

          $a=\frac{v^2}{r}=\frac{3.57^2}{0.86}=14.79m/s^2$

c) Maximum tension in horizontal circular motion

           $T=\frac{mv^2}{r}\\\\136=\frac{4\times v^2}{0.86}\\\\v=5.41m/s$

  Maximum tangential speed the ball can have = 5.41 m/s