A 0.50-kg ball that is tied to the end of a 1.9-m light cord is revolved in a ho

Question

A 0.50-kg ball that is tied to the end of a 1.9-m light cord is revolved in a horizontal plane, with the cord making a ? = 30 angle with the vertical. (See the figure below.)

(a) Determine the ball's speed.
m/s

(b) If, instead, the ball is revolved so that its speed is 3.7 m/s, what angle does the cord make with the vertical?


(c) If the cord can withstand a maximum tension of 9.9 N, what is the highest speed at which the ball can move?
m/s

A 0.50-kg ball that is tied to the end of a 1.9-m light cord is revolved in a horizontal plane, with the cord making a ? = 30½ angle with the vertical. (See the figure below.) (a) Determine the ball's speed. m/s (b) If, instead, the ball is revolved so that its speed is 3.7 m/s, what angle does the cord make with the vertical? ½ (c) If the cord can withstand a maximum tension of 9.9 N, what is the highest speed at which the ball can move? m/s

Explanation / Answer

A. If g is the acceleration of gravity then v^2/R = g tan(A). R = L sin(A)
So, v^2 = Rg tan(A) = gL sin(A) tan(A) = 9.8*1.9*sin 30 * tan 30 = 5.375

v = 2.318 m/s
B. The formula is v^2 = gL sin(A) tan(A) = gL (1-x^2)/x where x is cos(A).
x^2 + (v^2/gL) x -1 = 0

x^2 + (3.7*3.7/(9.8*1.9)) x -1 = 0
Letting v be 3.7 m/s and solving for a positive value of x

x = 0.697

Angle is cos A = x

Angle = 45.81 degree

C. The tension of the rope is mg / cos(A) = mg / x =9.9

x = 0.5*9.8/9.9 = 0.49 m

v^2 = gL (1-x^2)/x = 9.8*1.9*(1 - 0.49^2)/0.49 = 28.87

v = 5.37 m/s

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