Question

A 5.47-kg ball hangs from the top of a vertical pole by a 2.13-m-long string. The ball is struck, causing it to revolve around the pole at a speed of 4.75 m/s in a horizontal circle with the string remaining taut. Calculate the angle, between 0° and 90°, that the string makes with the pole. Take g = 9.81 m/s2. Answer in degrees.

What is the tension of the string? Answer in N

Answer 1

The object is moving circular at constant speed, so the acceleration would be towards pole.

T_c = T sinθ (towards pole)

T_y = T cosθ (perpendicular)

Now, for y direction ,

Σ F_y = T_y - mg = m a_y = 0

where the acceleration is zero. So

T cosθ = mg

T = mg / cosθ

Now, towards the center -

Σ F_c = T_c = m a_c = m v²/r

where a_c = centripetal acceleration.

Since r = L sinθ

where L is the length of the string, so

T sinθ = m v² / (L sinθ)

=> T = m v² / (L sin²θ)

=> T = m v² / (L sin²θ) = mg / cosθ

=> mv² cosθ = mgL sin²θ

Since sin²θ = 1 - cos²θ, so

=> mv² cosθ = mgL (1 - cos²θ) = mgL - mgL cos²θ

=> gL cos²θ + v² cosθ - gL = 0

This is a quadratic in cosθ.

Formula for cosθ = - (B /2A) +/- ( √[ B² - 4AC ] / 2A)

So, A = gL , B = v² , C = -gL

cosθ = -(4.75^2 / (2*9.81*2.13)) + (sqrt[4.75^4 + 4*9.81^2*2.13^2]/(2*9.81*2.13))

Hence on solving we get,

cosθ = 0.596

θ = arccos(0.596) = 53.4 degrees

b)
Tension T = mg / cosθ
= (5.47 kg)(9.81 N/kg) / cos(53.4°)
= 90 N