Speed of ball on pendulum with Mechanical Energy

[VaitVhat52]

Homework Statement

A pendulum consists of a small object of mass m fastened to the end of an inextensible cord of length L. Initially, the pendulum is drawn aside through an angle of 60° with the vertical and held by a horizontal string as shown in the diagram (Attached). This string is burned so that the pendulum is released to swing to and fro.

Relevant Equations

Gravitational Potential Energy = mgh
Kinetic Energy = 1/2mv^2

Started by analyzing the change in energy from the initial position to the final position which gives us mgh=1/2mv^2
Since we are trying to find speed, we rearrange the equation to solve for v, which gives us √2gL.

My question is, do we need to take a component of L for √2gL because it is at an angle or is it just L since that is the height at the final position?

 

[PeroK]

Homework Statement: A pendulum consists of a small object of mass m fastened to the end of an inextensible cord of length L. Initially, the pendulum is drawn aside through an angle of 60° with the vertical and held by a horizontal string as shown in the diagram (Attached). This string is burned so that the pendulum is released to swing to and fro.
Relevant Equations: Gravitational Potential Energy = mgh
Kinetic Energy = 1/2mv^2

View attachment 337046

View attachment 337051

Started by analyzing the change in energy from the initial position to the final position which gives us mgh=1/2mv^2
Since we are trying to find speed, we rearrange the equation to solve for v, which gives us √2gL.

My question is, do we need to take a component of L for √2gL because it is at an angle or is it just L since that is the height at the final position?

Gravitational PE depends on the height difference. Does the object start at height $l$ above the lowest point?

 

[kuruman]

A foolproof statement of energy conservation is in terms of changes in kinetic and potential energy, $$\Delta K+\Delta U=0.$$ Here $\Delta K = (\frac{1}{2}mv^2-0)$ and $\Delta U= mg(\text{(Final height)-(Initial height)}.$ You can now answer you own question, although I see that @PeroK has already done so.

 

[VaitVhat52]

Gravitational PE depends on the height difference. Does the object start at height $l$ above the lowest point?

No, it starts at the as high as the vertical component of L.
Would that then make it √2g(L-LCosθ) or am I missing something?

 

[PeroK]

No, it starts at the as high as the vertical component of L.
Would that then make it √2g(L-LCosθ) or am I missing something?

You're still missing something. Check your trigonometry. Which I see you've just done! Well spotted.

 

[PeroK]

PS that expression $L(1 - \cos \theta)$ comes up in a lot of problems. It's worth making a mental note of that.

 

[VaitVhat52]

One last thing:
Since it asks for the answer in terms of $g$ and $L$, are we allowed to have $Cosθ$ as part of the answer?

 

[kuruman]

One last thing:
Since it asks for the answer in terms of $g$ and $L$, are we allowed to have $Cosθ$ as part of the answer?

Isn't the value of θ given in the diagram?

 

[VaitVhat52]

Isn't the value of θ given in the diagram?

Ah that's correct.
So the final answer would be $\sqrt{2gL(1-\frac{1}{2})}$

 

[haruspex]

Ah that's correct.
So the final answer would be $\sqrt{2gL(1-\frac{1}{2})}$

which simplifies to…?