Calculus Proof that a=v^2/r

TL;DR

The magnitude of centripetal acceleration is equal to the magnitude of the velocity squared divided by the radius.

Transcript

Welcome back. Well, I'm now going to prove to you that the magnitude of centripetal acceleration is equal to the magnitude of the velocity when you're going around the circle divided-- velocity squared divided by the radius. So let's start with the drawing, just so that we know what we're doing, just as much for me as it is for you. So that's the c... Read More

Key Insights

• 😑 The position vector of an object in circular motion can be expressed in terms of its x and y components.
• 🫡 The velocity vector is tangent to the circle and can be found by taking the derivative of the position vector with respect to time.
• ❎ The acceleration vector is the derivative of the velocity vector and is equal to the negative of the angular velocity squared times the position vector.

Questions & Answers

Q: What is the position vector of an object moving in a circular path?

The position vector is defined by the radius of the circle times the cosine of the angle for the x-component and the sine of the angle for the y-component.

Q: How can the velocity vector be found?

The velocity vector is found by taking the derivative of the position vector with respect to time. It is equal to the angular velocity times the sine and cosine of the angle.

Q: What is the acceleration vector?

The acceleration vector is the derivative of the velocity vector with respect to time. It is equal to the negative of the angular velocity squared times the position vector.

Q: What is the formula for centripetal acceleration?

The magnitude of the centripetal acceleration is equal to the velocity squared divided by the radius.

Summary & Key Takeaways

• The position vector of an object moving in a circular path can be defined in terms of its x and y components, which are given by the radius times the cosine and sine of the angle, respectively.

• The velocity vector at any given point is tangent to the circle and can be found by taking the derivative of the position vector with respect to time. It is equal to the angular velocity times the sine and cosine of the angle.

• The acceleration vector is the derivative of the velocity vector with respect to time and is equal to the negative of the angular velocity squared times the position vector.

• By finding the magnitude of the acceleration vector, it is shown that the centripetal acceleration is equal to the velocity squared divided by the radius.