A small ball with a mass of 0.48kg is fastened to a string

A small ball with a mass of 0.48kg is fastened to a string .24m long and suspended from the ceiling to make a conical pendulum. The ball describes a horizontal circle and the string makes an angle of 15 degrees with vertical. (A) Find the tension of the string. (B) Find the period of the pendulum.

The correct answer and explanation is:

Let’s analyze the problem step-by-step.

Given data:

• Mass of ball, m=0.48 kgm = 0.48 \, \text{kg}
• Length of string, L=0.24 mL = 0.24 \, \text{m}
• Angle with vertical, θ=15∘\theta = 15^\circ

Objective:

(A) Find the tension TT in the string.
(B) Find the period TpT_p of the conical pendulum.

Step 1: Understanding the conical pendulum

A conical pendulum swings in a horizontal circle, with the string tracing out a cone shape. The tension TT in the string has two components:

• Vertical component balances weight: Tcos⁡θ=mgT \cos \theta = mg
• Horizontal component provides centripetal force: Tsin⁡θ=mv2rT \sin \theta = m \frac{v^2}{r}

Here,

• vv is the tangential speed of the ball in the circle,
• r=Lsin⁡θr = L \sin \theta is the radius of the horizontal circular motion,
• g=9.8 m/s2g = 9.8 \, \text{m/s}^2 is acceleration due to gravity.

(A) Find the tension TT

From the vertical balance: Tcos⁡θ=mg  ⟹  T=mgcos⁡θT \cos \theta = mg \implies T = \frac{mg}{\cos \theta}

Calculate TT: cos⁡15∘=0.9659\cos 15^\circ = 0.9659 T=0.48×9.80.9659=4.7040.9659≈4.87 NT = \frac{0.48 \times 9.8}{0.9659} = \frac{4.704}{0.9659} \approx 4.87 \, \text{N}

(B) Find the period TpT_p

We first find the velocity vv:

From horizontal force: Tsin⁡θ=mv2r  ⟹  v2=Tsin⁡θ×rmT \sin \theta = m \frac{v^2}{r} \implies v^2 = \frac{T \sin \theta \times r}{m}

But r=Lsin⁡θr = L \sin \theta, so: v2=Tsin⁡θ×Lsin⁡θm=TLsin⁡2θmv^2 = \frac{T \sin \theta \times L \sin \theta}{m} = \frac{T L \sin^2 \theta}{m}

Calculate: sin⁡15∘=0.2588\sin 15^\circ = 0.2588 v2=4.87×0.24×(0.2588)20.48=4.87×0.24×0.0670.48v^2 = \frac{4.87 \times 0.24 \times (0.2588)^2}{0.48} = \frac{4.87 \times 0.24 \times 0.067}{0.48} v2=4.87×0.01610.48=0.07840.48≈0.1633v^2 = \frac{4.87 \times 0.0161}{0.48} = \frac{0.0784}{0.48} \approx 0.1633 v=0.1633≈0.404 m/sv = \sqrt{0.1633} \approx 0.404 \, \text{m/s}

The period TpT_p is the time to complete one revolution: Tp=circumferencespeed=2πrvT_p = \frac{\text{circumference}}{\text{speed}} = \frac{2 \pi r}{v}

Calculate radius rr: r=Lsin⁡θ=0.24×0.2588=0.0621 mr = L \sin \theta = 0.24 \times 0.2588 = 0.0621 \, \text{m}

Then: Tp=2π×0.06210.404=0.3900.404≈0.965 secondsT_p = \frac{2 \pi \times 0.0621}{0.404} = \frac{0.390}{0.404} \approx 0.965 \, \text{seconds}

Final answers:

• (A) Tension TT in the string: 4.87 N\boxed{4.87 \, \text{N}}
• (B) Period of the pendulum TpT_p: 0.97 seconds\boxed{0.97 \, \text{seconds}}

Explanation (300 words):

A conical pendulum consists of a mass swinging in a horizontal circular path, with the string inclined at an angle to the vertical. The string’s tension must balance two things: the vertical component balances the gravitational force (weight), while the horizontal component provides the centripetal force required to keep the mass moving in a circle.

By analyzing the forces, the vertical component of tension is Tcos⁡θ=mgT \cos \theta = mg, which lets us solve for the tension TT. This ensures the ball doesn’t move up or down vertically. With θ=15∘\theta = 15^\circ, the tension is slightly greater than the weight because the string must also provide centripetal force.

The horizontal component Tsin⁡θT \sin \theta supplies the centripetal force mv2/rm v^2 / r, where r=Lsin⁡θr = L \sin \theta is the radius of the circular path traced by the ball. Using this, we calculate the ball’s tangential velocity vv.

Once the speed is known, the period of rotation is the circumference of the circle divided by this speed, giving the time for one full revolution.

This problem neatly illustrates how forces resolve into components in circular motion, showing the relationship between tension, angle, velocity, and period. The solution highlights the balance of forces and the geometry of the pendulum, providing insight into dynamics in rotational systems.