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.  This acceleration is called radial acceleration or centripetal acceleration, and it points towards the center of the circle.  The magnitude of the centripetal acceleration vector is a_c = v²/r.  (We skip the derivation of this expression.)

Link:  Uniform circular Motion

Problem:

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.

Solution:

• Reasoning:
  (a)  The driver must make the car travel in a straight line.
  (b)  The driver must make the car travel in a circle (or in a straight line with a = 0).

Problem:

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.

Solution:

• Reasoning:
  The distance the moon travels in one orbital period T is d = 2πr.
  Its speed is v = distance/time = d/T.
  The centripetal acceleration of the moon is v²/r.
• Details of the calculation:
  (a) The distance the moon travels in 27.3 days is d = 2πr = 2.41*10⁹ m.
  Its speed is v = d/(27.3 days) = (d/(2.36*10⁶ s)) = 1023 m/s.
  (b) The centripetal acceleration of the moon is v²/r = 2.725*10^-3 m/s².

Problem:

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?

Solution:

• Reasoning:
  If the centripetal acceleration was equal to the gravitational acceleration at the equator, then v²/r = g.
• Details of the calculation:
  (a)  The radius of the earth is 6368 km.  We want v²/r = 9.8 m/s², or v² = (9.8 m/s²)*6.368*10⁶ m, or v = 7900 m/s.
  (b)  The length of a day would be t = 2πr/v = 5065 s = 84.4 min.

Non-uniform circular motion

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 = a_r + a_t.  The tangential component causes a change in speed and its magnitude is given by a_t = dv/dt, and the radial component causes a change in direction and its magnitude is given by a_r = v²/r, where r is the radius of curvature at the point in question.  The components a_r and a_t are perpendicular to each other and the magnitude of a is a = (a_r² + a_t²)^½.

Problem:

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.

Solution:

• Reasoning:
  Throughout the first loop the magnitude of the tangential acceleration a_t is constant.  The magnitude of the radial acceleration a_r increases as v²/r, as the speed of the skater increases as  v = a_tt = (2a_td)^½, where d is the distance traveled.  When the skater starts out the acceleration vector is tangential to the loop.  But the angle θ it makes with a tangential line increases as tanθ = a_r/a_t = v²/(ra_t) = 2d/r.  Throughout the second loop a_t = 0 and a_r is constant, and the acceleration vector points towards the center of the loop.

Problem:

An automobile, the speed of which is increasing at a rate of 0.6 m/s², travels along a circular road of radius 20 m.  When the instantaneous speed of the automobile is 4 m/s, find
(a)  the tangential acceleration component and
(b)  the magnitude and direction of the total acceleration.

Solution:

• Reasoning:
  The tangential acceleration of the car has a given, constant magnitude.  Its direction is tangential to the circle.
  The radial acceleration has magnitude v²/r.  Its direction is perpendicular to the tangential acceleration and points towards the center of the circle.
• Details of the calculation:
  (a)  The tangential acceleration of the car has a constant magnitude of 0.6 m/s².
  (b)  When v = 4 m/s, the radial acceleration has magnitude a_r = v²/r = (16 m/s²)/(20 m) = 0.8 m/s², a = (a_r² + a_t²)^½ = 1 m/s².
  Let θ be the angle the acceleration vector makes with the line pointing from the position of the car towards the center of the circle.  Then tanθ = a_t/a_r = 0.75, θ  = 36.9^o.
  

Problem:

The figure on the right represents the total acceleration of a particle moving clockwise in a circle of radius 2.5 m 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.

Solution:

• Reasoning:
  We have to find the component of the vector a.
• Details of the calculation
  (a)  a_r = a cos(30^o) = 12.99 m/s².
  (b)  a_r = v²/r, v² = ra_r, v = 5.7 m/s.
  (c)  a_t = a sin(30^o) = 7.5 m/s².

Module 3: Question 2

Can centripetal acceleration change the speed of circular motion?

Discuss this with your fellow students in the discussion forum!

Common units for angles are degrees and radians.  We have 360^o = 2π radians.  The circumference of a circle of radius r is 2πr.  The length of a circular arc subtaining 180^o is πr.  The length of an arc subtaining 90^o is (π/2)r.  We see that the length of a circular arc subtaining an angle θ is θr, where θ is measured in radians.

For small angles θ, sinθ ~ θ, if θ is measured in radians.  You can verify this looking at the table or the graph.