# Energy Conservation in Pendulum Swing: Finding v for Complete Circle

• mirandasatterley
In summary, the problem involves a bullet of mass m and speed v passing through a pendulum bob of mass M suspended by a stiff rod of length l. The minimum value of v is calculated such that the pendulum bob will barely swing through a complete vertical circle. This is done by using conservation of momentum and energy equations. The result is that the kinetic energy at the top of the swing is zero and the velocity of the bob is equal to the square root of Mg(2l)/2M. The velocity of the bullet and the bob are denoted by v and V, respectively.
mirandasatterley
http://www.physics.auburn.edu/~boivin/homework9.htm

"A bullet of mass m and speed v passes completely through a pendulum bob of mass M. The bullet emerges with a speed of v/2. The pendulum bob is suspended by a stiff rod of length l and negligable mass.
A) What is the minimum value of v such that the pendulum bob will barely swing through a complete verticle circle?
Energy is conserved for the bob-Earth system between bottom and top of swing. At the top, the stiff rod is in compression and the bob nearly at rest.

I'm not really sure how to do this problem, but i think that at the top, it would barely swing through because it is nearly at rest, but I am not sure what to do with that.

#### Attachments

• Picture 1.jpg
13.8 KB · Views: 378
Since it is the minimum speed at which it will make it over, you calculate v such that the kinetic energy at the top is zero (no velocity)

then:

mv = m(v/2) + MV (conservation of momentum)

1/2MV^2 = Mg(2l)

that should be enough to solve it.

I undestand the formula's but I'm really bad at physics so I'm not sure if i know what to do next. I never understood how to combine these two formulas.
Can I solve for V in the conservation of momentum equation and then sub that into the conservation of energy equation? Are V and v supposed to be different or the same?

V and v are different, one is velocity of the bullet, the big V is that of the bob. yes, you have 2 equations and 2 unknowns (v and V) so solve one and substitute as you suggested

Thank you for all your help

## 1. How does energy conservation apply to pendulum swings?

Energy conservation is a fundamental law of physics that states that energy cannot be created or destroyed, only transferred from one form to another. In the case of a pendulum swing, the potential energy at the highest point is converted into kinetic energy as the mass swings back and forth. This process repeats itself, with the total energy of the system remaining constant.

## 2. How can we measure the velocity of a pendulum for a complete circle?

The velocity of a pendulum can be measured by using the equation v = 2πr/T, where v is the velocity, r is the length of the pendulum, and T is the period of the swing. To find the velocity for a complete circle, simply multiply the result by 2π.

## 3. How does the length of a pendulum affect its energy conservation?

The length of a pendulum affects its energy conservation by changing the potential and kinetic energy at each point of the swing. As the length increases, the potential energy increases while the kinetic energy decreases, and vice versa. However, the total energy of the system remains constant.

## 4. What factors can impact the energy conservation of a pendulum swing?

There are several factors that can impact the energy conservation of a pendulum swing, such as air resistance, friction, and the angle of release. These factors can cause energy to be lost from the system, resulting in a decrease in the amplitude of the swing over time.

## 5. How can we use the principles of energy conservation to increase the efficiency of a pendulum swing?

To increase the efficiency of a pendulum swing, we can minimize the factors that can cause energy loss, such as air resistance and friction. We can also adjust the angle of release to ensure that the pendulum follows a smooth and consistent path, maximizing the transfer of energy from potential to kinetic and back again.

• Introductory Physics Homework Help
Replies
9
Views
785
• Introductory Physics Homework Help
Replies
11
Views
3K
• Introductory Physics Homework Help
Replies
16
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
225
• Introductory Physics Homework Help
Replies
21
Views
1K
• Introductory Physics Homework Help
Replies
10
Views
525
• Introductory Physics Homework Help
Replies
6
Views
1K
• Introductory Physics Homework Help
Replies
20
Views
1K
• Introductory Physics Homework Help
Replies
23
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
9K