An object moving in a circle of radius r with constant speed v is accelerating. The direction of its velocity vector is changing all the time, but the magnitude of the velocity vector stays constant. The acceleration vector cannot have a component in the direction of the velocity vector, since such a component would cause a change in speed. The acceleration vector must therefore be perpendicular to the velocity vector at any point on the circle. But what is its magnitude?
Let us compute the average acceleration of the object as it moves from position 1 to position 2, and its velocity changes from v1 to v2. The travel time Dt is the distance traveled divided by the speed. The distance traveled is 2qr, with the angle q measured in radians, therefore Dt = 2qr/v. The change in velocity Dv is 2vsinq pointing towards the center of the circle. The average acceleration therefore is a = Dv/Dt = v2sinq/(rq) towards the center of the circle. For small angles q, sinq @ q, if q is measured in radians.
The instantaneous acceleration a is therefore given by 
, where
is a unit vector pointing towards the center of
the circle. This acceleration is called radial
acceleration or centripetal acceleration.
Link:
 | Describe how a driver can steer a car traveling at constant speed so that (a) the acceleration is zero, or (b) the magnitude of the acceleration remains constant.
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| The orbit of the moon around the earth is approximately circular, with a
mean radius of 3.85×108 m. It takes 27.3 days for the
moon to complete one revolution around the earth. Find (a) the mean orbital speed of the moon and (b) its centripetal acceleration.
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| If the rotation of the earth increased so that the centripetal
acceleration was equal to the gravitational acceleration at the equator, (a) what would be the tangential speed of a person standing at the equator, and (b) how long would a day be?
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A rollercoaster cart moving through a loop is changing speed as well as direction of
travel. Its acceleration has a radial component and a tangential component, a
= ar + at.
The tangential component causes a change in speed and its magnitude is given by at
= dv/dt,
and the radial component causes a change in direction and its magnitude is given by ar
= v2/r,
where r is the radius of curvature at the point in question. The components ar
and at are perpendicular to each other and the magnitude of a is
.
| An ice skater is executing a figure eight, consisting of two equal tangent
circular paths. Throughout the first loop she increases her speed
uniformly, and during the second loop she moves at a constant speed.
Make a sketch of her acceleration vector at several points along
the path of motion.
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| An automobile, the speed of which is increasing at a rate of 0.6m/s2,
travels along a circular road of radius 20m. When the instantaneous
speed of the automobile is 4m/s, find (a) the tangential acceleration component and (b) the magnitude and direction of the total acceleration.
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| The figure below represents the total
acceleration of a particle moving clockwise in a circle of radius 2.5m at a
given instant of time. At this instant, find (a) the centripetal acceleration, (b) the speed of the particle, and (c) its tangential acceleration.
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Common units for angles are degrees and radians. We have 360o = 2p radians. The circumference of a circle of radius r is 2pr. The length of a circular arc subtaining 180o is pr. The length of an arc subtaining 90o is (p/2)r. We see that the length of a circular arc subtaining an angle q is qr, where q is measured in radians.
For small angles q, sinq @ q, if q is measured in radians. You can verify this looking at the table or the graph.
| angle (deg) | angle (rad) | sin(angle) |
0 |
0 |
0 |
0.1 |
0.001745 |
0.001745 |
0.2 |
0.003491 |
0.003491 |
0.3 |
0.005236 |
0.005236 |
0.4 |
0.006981 |
0.006981 |
0.5 |
0.008727 |
0.008727 |
0.6 |
0.010472 |
0.010472 |
0.7 |
0.012217 |
0.012217 |
0.8 |
0.013963 |
0.013962 |
0.9 |
0.015708 |
0.015707 |
1 |
0.017453 |
0.017452 |
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,
where d is the distance traveled. When the skater starts out the acceleration vector is
tangential to the loop. But the angle q it makes with a
tangential line increases as tanq = ar/at
=
v2/(rat) = 2d/r. Throughout the second loop at
= 0 and ar
is constant, and the acceleration vector points towards the center of the loop.
