# Why doesn't the magnitude of velocity of an object in circular motion increase?

I've got a question about circular motion that's been bugging me for a while.

Suppose we have a ball in circular motion. At a certain point in time, the ball experiences a centripetal force that is perpendicular to its velocity.

By right, the centripetal force should cause the velocity of the ball to increase (accelerate) perpendicular to its initial velocity. Then, the two velocities combine and we get a third velocity with a larger magnitude than the initial velocity and points in a new direction. Basically, like a ball in projectile motion.

But obviously, this isn't the case because the magnitude of velocity (speed) of the ball stays constant, only the direction changes. So something is obviously wrong with my reasoning.

The issue here is that you are assuming that over a small time interval you are considering, the acceleration on the ball is constant. However, the acceleration changes no matter how small of a 'time interval' you consider. What you are essentially doing is this: $$v(t_0+\delta t)=v(t_0)+\frac{dv}{dt}(t_0)\delta t$$ You're expanding to first order in the derivative of the velocity to figure out what the velocity is at $$t_0+\delta t$$, which is fine, but there will be an error if you take finite (non-zero) interval of time. This is an identical issue of following a tangent line at, say point (1,0) to the unit circle centre origin. You can approximate a point on the unit circle that is just above this point by following the tangent line, but you will be off by a certain amount. The position vector to that point in the tangent line will have a length slightly greater than one. That error is because by using the tangent line you ignore that the true path, the circle, is continuously curving.

• Yep, that's what I was thinking! – Luo Zeyuan Aug 18 '18 at 8:07
• So to clarify, in the question, i assumed that time passed in small intervals of "dt". But in reality, dt is infinitely small. So, as dt tends toward zero, then the change in resultant velocity is essentially zero? – Luo Zeyuan Aug 18 '18 at 8:11
• Exactly! The magnitude is sqrt(1+(dt)^2). – Sean Thrasher Aug 18 '18 at 8:13
• Does that answer your question? If so could you choose it as the accepted answer? – Sean Thrasher Aug 18 '18 at 8:22

Nothing is wrong. The force causing the centripetal acceleration is perpendicular motion of the ball and hence that force does no work on the ball. Put another way $\vec F \cdot \vec v = 0$.

• Yes, at that instance, the parallel component of the velocity of the ball is unchanged. But shouldn't the overall velocity increases because of the perpendicular component of velocity introduced by the centripetal acceleration? – Luo Zeyuan Aug 18 '18 at 7:15
• @LuoZeyuan, there is one more thing to consider. Acceleration is defined as change in velocity with respect to time. Velocity is a vector quantity having both magnitude and direction. The usual way of defining acceleration involves a change in magnitude divided by a change in time. There is also another way to define acceleration, and that is a change in direction divided by a change in time. Thus, centripetal acceleration can have a constant speed but still be an acceleration, because the direction of motion is continuously changing. – David White Dec 31 '18 at 23:15

The thing to remember is that the direction of the force is always changing. Imagine that the initial velocity of the ball is along the +x axis and the force is along the +y axis--that is, the ball is initially directly below the center of the circular path. In the first instant, the ball gains a bit of +y velocity with no change in the +x velocity. However, by the time this happens, the ball has moved further passed the center in the +x direction. A centripetal force always points at the center, so now the force has a component in the -x direction to pull the ball towards the center. This begins to decrease the ball's x-velocity as the y-velocity increases, keeping the total speed constant.