PhysicsScience

The Physics Philes, lesson 11: You Say You Want a Revolution

In which circles are drawn, hammers are thrown, and pi is served.

Until this point, the only motion we’ve been dealing with has been motion in a straight line. But guess what! Objects don’t always move in a straight line. Sometimes they move in a circle. But don’t worry. There are some equations for that.

Think about swinging an object around, like a hammer throw. The speed of the hammer may be constant, but there is an acceleration vector. There has to be. Remember that an object is accelerating whenever it changes direction, regardless of any change in speed. Think about swinging something around, like a hammer throw. That hammer is always changing direction, therefore is always accelerating.

What keeps the particle moving in a circle is an acceleration component that is perpendicular to the the particle’s path. Think about that hammer throw again. When you let go, the hammer will fly off in a straight line. Newton’s first law of motion tells us this. It needs that force pulling toward the center of the circle to keep it along the circular path. Otherwise that particle will fly off in a straight line.

There are two types of circular motion: uniform and nonuniform circular motion. I’ll start first with uniform circular motion.

Uniform circular motion is when a particle moves in a circle at a constant speed. There is no component of acceleration that is tangent – or parallel – to the particle’s path. The acceleration vector is normal – or perpendicular – to the particle’s path and directed inward. Like this:

OK, it’s sloppy. But it’s more or less how it goes down.

This inward force is often called centripetal acceleration. The magnitude of the acceleration vector is related in simple way to the speed of the radius of the circle. To explain this, I’m afraid I’m going to need to draw another picture. Sorry.

Here is a the diagram of a particle moving along a circular path.

Ooooo a triangle!

P sub 1 and P sub 2 is the particle as it moves along its circular path. Delta s (Δs) is the distance the particle has moved. R represents the radius of the circle. Those green lines are the velocity vectors of the particle at those particular instances.

Now, here’s the triangle made of v sub 1 and v sub 2:

Hmmm…looks kind of familiar.

These two triangles are (supposed to be) similar. The angle ΔΦ are the same in each diagram because v sub 1 is perpendicular to line OP sub 1 and v sub 2 is perpendicular to line OP sub 2. The ratios of the corresponding sides of similar triangles are equal. That means we get our first formula of the day!

| v |/v sub 1 = Δs/R

or

| v | = (v sub 1/R)(Δs)

The magnitude of the average acceleration during Δt is:

a sub av = | v |/Δt = (v sub 1/R)(Δs/Δt)

The magnitude of instantaneous acceleration at P sub 1 is the limit of the expression as we move P sub 2 closer to P sub 1.

a = lim Δt→0 (v sub 1/R)(Δs/Δt) = v sub 1 lim Δt→0 Δs/Δt

The limit of Δs/Δt is v sub 1 at P sub 1. This point 1 can be anywhere along the path and v can represent the speed at any point. This means that

a sub rad = v squared/R

Do you understand all of that? Really? Because I’m not sure I do. But the end result is, in uniform circular motion, the magnitude of instantaneous acceleration is equal to to the square of the speed divided by the radius of the circle and its direction is perpendicular to the velocity vector and inward along the radius of the circle.

But this is just one way of expressing the magnitude of the acceleration vector. We can also do it in terms of the period.

The period is the time it takes for one revolution of the circle. In time T the particle travels a distance equal to the circumference of the circle, which just so happens to be 2πR. So we can find the speed by using this equation:

v = 2πR/T

We can also substitute a = v squared/R and get an alternate expression:

a sub rad = 4π squared R/T squared

These equations are sufficient only when we are working with uniform circular motion. For nonuniform circular motion, we need more. Nonuniform motion is when a particle does not move at a constant speed. This means that the particle has an acceleration component that is tangent – or parallel – to the instantaneous velocity. This component is equal to the rate of change of speed. The vector acceleration of a particle moving in a circle with varying speed is the vector sum of the radial and tangential components of acceleration. The tangent acceleration is in the same direction as the velocity when a particle is speeding up and in the opposite direction when it’s slowing down.

a sub rad = v squared/R

a sub tan = d| v |/dt

Oh no! A derivative equation! I can’t explain it to you. Sorry. I’m not there yet. The minute I figure it out, you’ll know.

In the meantime, this is where I leave you. Next week we’ll work through some problems involving circular motion. As ever, let me know if I have misunderstood something or if something needs further explanation. I reserve the right to be outrageously wrong, but only if you take the time to correct me.

Featured image credit: foxypar4

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Mindy

Mindy

Mindy is an attorney and Managing Editor of Teen Skepchick. She hates the law and loves stars. You can follow her on Twitter and on Google+.

2 Comments

  1. August 28, 2012 at 9:51 pm —

    As usual, this was great. These are really fun, and make me nostalgic 🙂

    • August 28, 2012 at 10:03 pm —

      Thanks! I’m having fun writing this series. I’m glad you’re enjoying it 🙂

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