Question

Two wires are tied to the 300 g sphere shown in FIGURE P8.46. The sphere revolves in a horizontal circle at a constant speed of 7.5 m/s. What is the tension in each of the wires?

Short answer

Tension in each wire is T₁ = 14.17 N and T₂=8.30 N

Step-by-step solution

Given information

Mass of sphere= 300 g = 0.3 kg

Length of wire is 1 m

Speed v= 7.5 m/s.

Explanation

First draw the free body diagram and then equate forces

Horizontal Components:

$T_{1}cos\theta +T_{2}cos\theta =\frac{m{v}^2}{R}....................................\left(1\right)$

Vertical components

$T_{1}sin\theta -T_{2}sin\theta -mg=0..................................\left(2\right)$

From trigonometric ratio we get

R=0.866 m

Solve equation(1) and (2)

Multiply by sinθ to equation(1) and by cosθ to equation (2), we get

$T_{1}cos\theta sin\theta +T_{2}cos\theta sin\theta =\frac{m{v}^2}{R}sin\theta ..........\left(3\right)$

$T_{1}sin\theta cos\theta -T_{2}sin\theta cos\theta -mgcos\theta =0........\left(4\right)$

By adding equation (3) and (4) , we get

$T_1=\left(\frac{m}{sin2\theta }\right)\left(\right(\frac{v^2\sin \theta }{R})+gcos\theta )$

Substitute the given values we get

$$\begin{gathered} T_1=\left(\frac{0.3kg}{sin2\left({30}^o\right)}\right)\left(\right(\frac{{(7.5m/s)}^2\left(\sin {30}^o\right)}{0.866m})+((9.8m/s^2)cos\theta ) \\ {T}_1=14.17N \end{gathered}$$

Substitute the value of T₁ in the equation (2) and solve for T₂, we get

localid="1649070587054" $$\begin{gathered} T_{2}sin\theta =T_{1}sin\theta -mg \\ {T}_{2}sin{30}^o=14.17Nsin{30}^o-(0.3kg)(9.8m/s^2) \\ \mathrm{Solve}\mathrm{for}\mathrm{T}2,\mathrm{we}\mathrm{get}. \\ T2=8.30N \end{gathered}$$