Pendulum Problem with unknown angle

[jg727]

A pendulum (with string length "L") and aball of mass "m" is pulled back to a horizontal position and then released. Assuming that θ is the angle between the string and the vertical, find the speed of the ball (v) at an angle of θ as a function of m,g,L, and/or θ.


I just can't get my mind around this problem...

 

[Delphi51]

Welcome to PF, jg!
I could be horrendous using motion formulas.
Have you considered an energy approach? The initial Ek is easy. Can you find the height when at angle θ and thus the potential energy? From that you can get the Ek at θ and then the speed.

 

[jg727]

Thank You!
Ok, so I guess I am actually having problems with finding the potential energy. I'm completely blanking.

 

[Delphi51]

PE = mgh

 

[asteriatic]

I understand that this pendulum problem can be challenging to conceptualize. However, it is important to approach it systematically and break it down into smaller components. Let's start by defining the variables given in the problem: m represents the mass of the ball, g is the acceleration due to gravity, L is the length of the string, and θ is the angle between the string and the vertical.

To solve for the speed of the ball at an angle of θ, we can use the conservation of energy principle. At the initial horizontal position, the ball has only potential energy, which is equal to mgh, where h is the height of the ball above the ground. As the ball swings down, it gains kinetic energy and loses potential energy. At any point along the swing, the total energy (E) of the ball can be expressed as:

E = mgh + 1/2 * mv^2

At the bottom of the swing, the ball has reached its maximum speed (vmax) and all of its potential energy has been converted to kinetic energy. Therefore, we can set the total energy at the bottom of the swing (Emax) equal to the total energy at any point along the swing (E):

Emax = mgh + 1/2 * mvmax^2

Now, we can rearrange this equation to solve for vmax:

vmax = √(2gh + v^2)

To find the speed of the ball at an angle of θ, we need to know the height (h) of the ball at that angle. This can be found using trigonometry, where h = L - Lcosθ. Therefore, the speed of the ball at an angle of θ can be expressed as:

v = √(2gL(1-cosθ))

This equation shows that the speed of the ball at any angle is dependent on the length of the string, the acceleration due to gravity, and the cosine of the angle. This means that the speed will be highest at the bottom of the swing (θ = 0) and will decrease as the angle increases towards the vertical (θ = 90°). It is also important to note that the mass of the ball does not affect the speed, as it cancels out in the equation.

In conclusion, the speed of the ball at an angle of θ can be determined using the equation v = √(2gL