Math Problem Statement

A pendulum consisting of a small heavy ball of mass m at the end of a string of length ℓ is released from a horizontal position. When the ball is at point P, the string forms an angle of 30° with the horizontal. Determine the forces acting at P, the speed at P, the tension in the string, and the tangential acceleration.

Solution

This problem involves a pendulum with a ball of mass mm attached to a string of length \ell. The string makes a 30° angle with the horizontal at point PP. Let’s break down the questions:

Part (a)

You need to draw and label the forces acting on the ball when it is at point PP. There are two main forces:

  1. Gravitational force (mgmg): Acts downward (vertically).
  2. Tension in the string (TT): Acts along the string, directed towards the pivot point.

Part (b)

To determine the speed of the ball at point PP, we can use energy conservation principles, assuming the ball starts from rest at its highest point (horizontal position).

  • The potential energy at the horizontal position is entirely converted to kinetic and potential energy at point PP.

The total mechanical energy at the top is purely gravitational potential energy: Einitial=mgh=mgE_{\text{initial}} = mgh = mg\ell At point PP, the height hh above the lowest point is h=cos(30)h = \ell - \ell \cos(30^\circ), and the energy is the sum of kinetic and potential energy: Efinal=12mv2+mg(cos(30))E_{\text{final}} = \frac{1}{2}mv^2 + mg(\ell - \ell \cos(30^\circ))

Equating the initial and final energies: mg=12mv2+mg(cos(30))mg\ell = \frac{1}{2}mv^2 + mg(\ell - \ell \cos(30^\circ)) Simplifying: 12mv2=mgcos(30)\frac{1}{2}mv^2 = mg\ell \cos(30^\circ) v2=2gcos(30)v^2 = 2g\ell \cos(30^\circ) v=2gcos(30)v = \sqrt{2g\ell \cos(30^\circ)}

Part (c)

To determine the tension in the string at point PP, apply Newton’s second law in the radial direction. The forces acting along the string are the tension and the radial component of gravity.

Using: Tmgcos(30)=mv2T - mg \cos(30^\circ) = \frac{mv^2}{\ell} Substitute v2v^2 from part (b): Tmgcos(30)=m(2gcos(30))T - mg \cos(30^\circ) = \frac{m(2g\ell \cos(30^\circ))}{\ell} T=3mgcos(30)T = 3mg \cos(30^\circ)

Part (d)

The tangential acceleration atangentiala_{\text{tangential}} is due to the tangential component of the gravitational force: atangential=gsin(30)a_{\text{tangential}} = g \sin(30^\circ)

Let me know if you'd like further clarifications or details! Here are some related questions you could explore:

  1. How would the motion change if the mass of the ball doubled?
  2. What happens to the speed at point PP if the angle were increased to 45°?
  3. How does the tension vary if the string is shortened?
  4. What would happen if frictional forces were considered?
  5. How is the period of the pendulum affected by the string length?

Tip: Always start energy conservation problems by identifying the reference points for potential energy.

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Math Problem Analysis

Mathematical Concepts

Energy conservation
Forces
Circular motion

Formulas

E_initial = mgh = mgℓ
E_final = (1/2)mv^2 + mg(h)
v = sqrt(2gℓcos(30°))
T - mg cos(30°) = mv^2 / ℓ
T = 3mg cos(30°)
a_tangential = g sin(30°)

Theorems

Conservation of Mechanical Energy
Newton's Second Law

Suitable Grade Level

Grades 11-12