An athlete swings a 4.00-kg ball horizontally on the end of a rope. The ball moves in a circle of radius 0.740 m at an angular speed of 0.750 rev/s.(a)What is the tangential speed of the ball? m/s(b)What is its centripetal acceleration? Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m/s2(c)If the maximum tension the rope can withstand before breaking is 140 N, what is the maximum tangential speed the ball can have? m/s
Question
An athlete swings a 4.00-kg ball horizontally on the end of a rope. The ball moves in a circle of radius 0.740 m at an angular speed of 0.750 rev/s.(a)What is the tangential speed of the ball? m/s(b)What is its centripetal acceleration? Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m/s2(c)If the maximum tension the rope can withstand before breaking is 140 N, what is the maximum tangential speed the ball can have? m/s
Solution
(a) To find the tangential speed of the ball, we first need to convert the angular speed from revolutions per second to radians per second. We know that 1 revolution is equal to 2π radians. So, the angular speed in radians per second is 0.750 rev/s * 2π rad/rev = 1.5π rad/s.
The formula for tangential speed (v) is v = rω, where r is the radius and ω is the angular speed. Substituting the given values, we get:
v = 0.740 m * 1.5π rad/s = 3.51 m/s.
(b) The formula for centripetal acceleration (a) is a = rω². Substituting the given values, we get:
a = 0.740 m * (1.5π rad/s)² = 16.4 m/s².
(c) The maximum tension the rope can withstand before breaking is equal to the force required for the centripetal acceleration at the maximum speed. This force (F) can be calculated using the formula F = ma, where m is the mass and a is the acceleration.
Setting F equal to the maximum tension, we get:
140 N = 4.00 kg * a.
Solving for a gives a = 140 N / 4.00 kg = 35 m/s².
We can then use this acceleration to find the maximum tangential speed using the formula v = sqrt(a*r). Substituting the values, we get:
v = sqrt(35 m/s² * 0.740 m) = 5.7 m/s.
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