A 0.50 kg ball that is tied to the end of a 1.5 m light cord is revolved in a ho
ID: 2244109 • Letter: A
Question
A 0.50 kg ball that is tied to the end of a 1.5 m light cord is revolved in a horizontal plane with the cord making a 30
A 0.50 kg ball that is tied to the end of a 1.5 m light cord is revolved in a horizontal plane with the cord making a 30 degree angle, with the vertical Determine the ball's speed. If, instead, the ball is revolved so that its speed is 4.0 m/s, what angle does the cord make with the vertical? Your response differs from the correct answer by more than 10%. Double check your calculations. degree If the cord can withstand a maximum tension of 9.3 N, what is the highest speed at which the ball can move?Explanation / Answer
The trajectory of the ball is an horizontal circle of radius R = L sin(A) where L (= 1.7 m) is the length of the string and A is the angle (called "theta" in the picture) which it makes with the vertical.
If the ball travels at speed v, it undergoes a centripetal (horizontal) acceleration v^2/R (see first link below for the general theorem--where R is the radius of curvature of the trajectory-- which says that the above formula always give the component of the acceleration which is perpendicular to the trajectory, even when the speed is not constant and/or when the trajectory is not a circle). This is the basis for the following answers to your questions:
(a) If g is the acceleration of gravity then v^2/R = g tan(A).
So, v^2 is Rg tan(A) = gL sin(A) tan(A)
With A = 30
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