Question

Two wires are tied to the 2.0 kg sphere shown in FIGURE P8.45. The sphere revolves in a horizontal circle at constant speed.
a. For what speed is the tension the same in both wires?
b. What is the tension?

Short answer

a) The tension in both wire should be same for the speed of 2.91 m/sec

b) Tension in wires at that speed is T =14.34 N

Step-by-step solution

Part(a) Step1 : Given Information

Mass of the ball =2 kg

The tension in each string is same.

The diagram shows the arrangement

Part(a) Step2 : Explanation

Lets first draw the free body diagram and then equate horizontal and vertical forces.

Horizontal components:

$T_{1}cos30°+T_{2}cos60°=\left(m{v}^2\right)/r....................\left(1\right)$

Vertical components:

$T_{1}sin30°+T_{2}sin60°=mg...............\left(2\right)$

Given T₁=T₂ , So after solving equation (1) and(2), we get

$v=\sqrt{rg}...................\left(3\right)$

Find the r using the triangle M L N and L N Q, we get

r= 0.866 m

Upon substituting in equation (3) we get

$$\begin{gathered} v=\sqrt{(0.866m)\times (9.8m/s^2)} \\ v=2.91m/s \end{gathered}$$

Part(b) Step1: Given information

Tension in both string are equal

Mass of the ball is 2 kg

Part(b) Step2: Explanation

Substitute T1=T2=T and m= 2kg and g=9.8m/s² in equation (2)

$T_{1}sin30°+T_{2}sin60°=mg$

we get

$$\begin{gathered} Tsin30°+Tsin60°=2kg(9.8m/s^2) \\ 1.366T=2kg(9.8m/s^2) \\ T=\frac{\left(2kg\right)(9.8m/s^2)}{1.366} \\ T=14.34N \end{gathered}$$