Question

A simple pendulum consisting of a 1.0m long string and a 0.20kg bob is pulled to the side so that the string makes an angle of 20° with the vertical. If the gravitational potential energy of the bob-Earth system is zero at the lowest point of the pendulum’s arc, the angle at which the gravitational potential energy of the bob-Earth system is equal to the kinetic energy of the bob is most nearly

Answer 1

The definition of energy and simple harmonic motion we can find the result for the point at which the kinetic energy and potential energy are equal is:

         θ = 9.8º

The mechanical energy is the sum of the kinetic energy and the potential energies, in the case that there is no friction, this energy is constant and gives advantage to one of the most important principles of physics.

The kinetic energy is:        K = ½ m v²

The potential energy is:    U = m g (y-y₀)

The mechanical energy is Em = K + U

Where m is mass, v is linear velocity, g is the acceleration of gravity, y height.

The simple pendulum is a simple harmonic system that for small angles has a shape solution.

           θ = θ₀ cos (wt + Ф)

where θ is the angle of motion, w is the angular velocity t time and Ф a phase constant that depends on the initial conditions.

They indicate that the system that has a length of l = 1.0 m, is released at the angle of θ₀ = 20º (π rad / 180º)  = 0.349 rad, remember that in the rotation movement all the angles must be in radian, let's look for the initial constant.

Velocity is defined by the change in position or angles with respect to time.

         w = $\frac{d \theta}{dt}$  = -θ₀  w sin (wt + Ф)

At the initial movement for time t = 0 the velocity is zero

          0 = - θ₀  w sin Ф

For the equality to be correct, the sine function must be zero, which implies that the phase angle are zero Ф = 0, therefore the solution for the angle and the velocity are:

           θ = θ₀ cos wt

           w = - θ₀  w sin wt

The angular velocity is given by the relation

     w² = g / l

we calculate

      w = $\sqrt{\frac{9.8}{1} }$  

      w = 3.13 rad / s

let's look for the expression for the energies.

      K = ½ m v²

   

Linear and angular variables are related.

      v = w l

     

We substitute

     K = ½ m l² w²

     

We substitute the expression for the angular velocity.

     K = ½ m l² ( -θ₀ w sin  wt) ²

We look for the potential energy, where we make the initial height zero.

      U = m g θ₀ cos wt

Ask the point where kinetic energy and potential energy  are equal.

          K = U

           ½ m l² θ₀² w² sin² wt = m g θ₀ cos wt

          $sin^2 wt = \frac{2 g }{l^2 \ \theta_o \ w^2 } \ cos \ wt$

          θ = wt

 

Let's calculate.

         $sin^2 \theta = \frac{2 \ 9.8}{ 1^2 \ 0.349 3.13^2 } \ cos \ \theta$  

        sin² θ = 5.73  cos θ

The solution of

           θ' = 1.4 + 2π n        n= 0, 1, 2, ...

First occurs for n = 0

           θ = 1.40 rad = 80.2º

This solution is the angle is measured from the horizontal, therefore the angle measured from the vertical corresponds:

          θ = 90- 8.02

          θ = 9.8º

In conclusion using the definition of energy and simple harmonic motion we can find the result for the point at which the kinetic energy and potential energy are equal is:

         θ = 9.8º

Learn more here:  brainly.com/question/17315536