Question

How fast do you need to swing a 250-g ball at the end of a string in a horizontal circle of 0.6-m radius so that the string makes a 39 degree angle relative to the horizontal?

Answer 1

Answer:

2.7 m/s

Explanation:

Draw a free body diagram of the ball.  There are two forces:

Weight force mg pulling down

Tension force T pulling 39° above the horizontal

Sum of the forces in the y direction:

∑F = ma

T sin θ − mg = 0

T = mg / sin θ

Sum of the forces in the radial (+x) direction:

∑F = ma

T cos θ = m v² / r

Substitute:

(mg / sin θ) cos θ = m v² / r

mg / tan θ = m v² / r

g / tan θ = v² / r

v = √(gr / tan θ)

Given that r = 0.6 m and θ = 39°:

v = √(9.8 m/s² × 0.6 m / tan 39°)

v ≈ 2.7 m/s

Answer 2

We have that the speed at which you are to  swing a 250-g ball at the end of a string in a horizontal circle of 0.6-m radius so that the string makes a 39 degree angle relative to the horizontal is

$V \approx 3m/s$

From the question we are told

How fast do you need to swing a 250-g ball at the end of a string in a horizontal circle of 0.6-m radius so that the string makes a 39 degree angle relative to the horizontal?

Generally the equation for the  Velocity is mathematically given as

$V=\sqrt{\frac{gr}{tan\theta}}\\\\Therefore\\\\V=\sqrt{\frac{9.9*0.6}{tan39}}\\\\V=2.708373887$

$V \approx 3m/s$

Therefore

The speed at which you are to  swing a 250-g ball at the end of a string in a horizontal circle of 0.6-m radius so that the string makes a 39 degree angle relative to the horizontal is

$V \approx 3m/s$

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