# Solving a Physics Problem: Velocity at Bottom of Circular Path

• mattmns
In summary, the conversation discussed a physics question involving a string with a mass hung on the end and the initial velocity being zero when the string is horizontal. The question asked for the velocity at the bottom of its circular path. The easiest way to solve this problem is through conservation of energy, where the loss of gravitational potential energy is equal to the gain of kinetic energy. The final formula for solving for the velocity is v=sqrt(2gy).
mattmns
I had a problem on my physics test today that said something like. You have a string with radius of 25 or 50 cm (cant remember which) and you have a mass hung on the end of the string. And the intial velocity is zero when the string is horizontal. The question was what is the velocity at the bottom of its circular path. How can you solve this? I was stumped, it was a multiple choice question and the answers were approximate numbers ranging from 2 m/s to 6 m/s. I just do not see how you could solve this problem without knowing the mass, and getting an approximate number. Also it did not say that the strings mass was neglible. Any clues?

Well, this question is bordering on "homework help," but, since it's about a concept, I'll let it remain here.

The easiest way to solve this problem is via the conservation of energy. The ball loses gravitational potential energy as it descends, and gains the same amount of kinetic energy.

Gravitational potential energy lost is m g y, where m is the mass of the object and y is the (vertical) distance it fell.

Kinetic energy is expressed as 1/2 m v2.

Equating the loss of gravitational potential energy and the gain of kinetic energy look like this:

1/2 m v2 = m g y

And you can simply solve for the velocity:

v = sqrt(2 g y)

Does this make sense?

- Warren

Yeah I was wondering if I should have put it here. And yes that makes complete sense, thanks. The mass cancels out, geeze I even had that v=sqrt(2gy) forumla written down. I guess I just got to hope I guessed right

## 1. What is the equation for calculating velocity at the bottom of a circular path?

The equation for calculating velocity at the bottom of a circular path is v = √(rg), where v is the velocity, r is the radius of the circular path, and g is the acceleration due to gravity.

## 2. How do you find the radius of a circular path?

The radius of a circular path can be found by measuring the distance from the center of the circle to the edge of the circle.

## 3. What is the significance of the velocity at the bottom of a circular path?

The velocity at the bottom of a circular path is significant because it is the minimum velocity required for an object to maintain a circular motion. If the velocity is any lower, the object will fall out of the circular path.

## 4. How does the mass of an object affect its velocity at the bottom of a circular path?

The mass of an object does not affect its velocity at the bottom of a circular path. This is because the equation for velocity at the bottom of a circular path (v = √(rg)) does not contain the mass of the object.

## 5. Can the velocity at the bottom of a circular path ever be greater than the initial velocity?

No, the velocity at the bottom of a circular path can never be greater than the initial velocity. This is because energy is conserved in a circular motion, and the initial velocity contains all the energy that is required to maintain the circular motion.

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