0.28 Phy1250: circular motion -- acceleration and centripetal force  (Page 2/6)

The velocity vector for an object that is traveling on a circular path at a constant speed is constantly changing direction. Otherwise, the object would notstay on the circular path. As you are aware, velocity is a vector. That is, it has both magnitude and direction.

As you are also aware, acceleration (the time rate of change of velocity) is also a vector. That is to say, it also has magnitude and direction. Thedirection of the acceleration vector is the same as the direction of the force that caused the acceleration.

The direction of the acceleration vector

The direction of the acceleration vector for an object under uniform circular motion always points to the center of the circle.

That may be difficult for you to believe. I will try to convince you using several different approaches, some graphically satisfying, and some moreanecdotal.

Subtraction of vectors

For this approach, I would like for you to use your graph board and follow along by creating and subtracting vectors. Use your graph board, somestring, and a few pushpins to create an arc of a circle centered on the origin of a Cartesian coordinate system. All you will need isabout one-half of the circle, on and to the right of the vertical axis.

Uniform circular motion

We will assume that on object is moving counter-clockwise around that circle with a uniform speed.

The initial velocity vector

Draw a vector with a length of ten units with it's tail at the intersection of the circle and the positive horizontal axis. This vector should pointstraight up in the direction of the positive y-axis. This vector represents the velocity of the object as it crosses the horizontal axis on its trip around thecircle.

The final velocity vector

Now go up to a point on the circle about 30 degrees relative to the positive horizontal axis (the exact angle isn't critical). Draw a vector with a length of10 units that is tangent to the circle at that point.

It isn't easy to draw a vector that is tangent to a circle. However, the direction of that vector should be perpendicular to a line that extends from the point on the circle to the centerof the circle. That will cause the vector to be tangent to the circle at that point.

The initial and final velocity vectors

We now have two vectors that represent the velocity vectors for the object at two different points along its circular path separated by a timeinterval. (The exact time interval doesn't matter at this point in the discussion.) Label the vector at the horizontal axis as Vi (for initial vector).Label the vector at the 30-degree point as Vf (for final vector).

Although there is no way for you to label it on the graph, we will represent the time interval required for the object to move from the first point to thesecond point as dt (which is an abbreviation for delta-time, or change in time).

Average acceleration

We know that the average acceleration is the time rate of change of velocity. Therefore, we can write

Aavg = (Vf - Vi)/dt

where

• Aavg represents the average acceleration
• Vf represents the velocity at the second point
• Vi represents the velocity at the first point
• (Vf - Vi) represents the change in velocity
• dt represents the time interval over which that change took place