Question

A pendulum consists of a small object hanging from the ceiling at the end of a string of negligible mass. The string has a length of 0.79 m. With the string hanging vertically, the object is given an initial velocity of 1.1 m/s parallel to the ground and swings upward in a circular arc. Eventually, the object comes to a momentary halt at a point where the string makes an angle θ with its initial vertical orientation and then swings back downward. Find the angle θ.

Answer 1

Answer:

θ = 21.7 °.

Explanation:

Let the length of the pendulum be L. . Initial kinetic energy of pendulum

is 1/2 mv² . Initial  potential energy is zero .

The distance by which  the bob of pendulum is raised

= L - Lcosθ

= L (1 - cos θ )

Increase in potential energy = mgL(1- cosθ )

According to conservation of mechanical energy

1/2 mv² = mgL ( 1 - cos θ)

v² = 2gL(1-cosθ)

Putting the values of different variables given

( 1.1 )² = 2 x 9.8 x .79 (1-cosθ)

(1-cosθ) = .071

cosθ = .929

θ = 21.7 °.

Answer 2

The angle θ that the string makes with its initial vertical orientation is;

θ = 22.79°

We are given;

Length of string; L = 0.79 m

Initial velocity; v = 1.1 m/s

From conservation of energy;

Kinetic energy = Potential Energy

Thus;

½mv² = mgh

m will cancel out to give;

½v² = gh

gh = (1.1²)/2

gh = 0.605

h = 0.605/9.8

h = 0.0617 m

Now, this value of h in relation with L and the angle is given by the formula;

h = = L - Lcosθ

Where;

θ = the angle string makes with its initial vertical orientation.

Thus;

Cosθ = (L - h)/L

Cosθ = (0.79 - 0.0617)/0.79

Cosθ = 0.9219

θ = cos^(-1) 0.9219

θ = 22.79°

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