At that speed, what is the radial acceleration of the ball?

Question

A ball of mass 5.0 kg is suspended by two wires from a horizontal arm that is attached to a vertical shaft, as shown in the figure. The shaft is in uniform rotation about its axis. The rate of rotation is adjusted until the tensions in the two wires are EQUAL. At that speed, what is the radial acceleration of the ball?

The length of the horizontal arm is 0.8 m.

Wire 1 is attached to the end of the horizontal arm and measures 0.6 m.

Wire 2 connects the ball to the vertical shaft and measures 1m.

The two wires and horizontal arm form a right-angled triangle with the vertical shaft.

The final answer is 4.9 m/s^2. How is this calculated?

Diagram: https://www.chegg.com/homework-help/questions-and-answers/ball-mass-50-kg-suspended-two-wires-horizontal-arm-attached-vertical-shaft-shown-figure-sh-q55510590

Answer 1

Let Dv=.6 and Dr=.8 ... Make a FBD showing a vertical tension (Tv), a radial tension (Tr) and a weight (5g). Sum forces in the horizontal = ma =Tr*.8. Sum forces in the vertical =5g=.6Tr+Tv. Given that Tr=Tv the second equation reduces to 1.6Tv=5g and the first reduces to Tv=5a/.8. Solving the remaining two equations yields 1.6*5a/.8=5g or a = g/2=4.9m/s²