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} = v^{2}/r. (We skip the
derivation of this expression.)

Link: Uniform circular Motion

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:

(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).

The orbit of the moon around
the earth is approximately circular, with a mean radius of 3.85*10^{8}
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:

(a) The distance the moon travels in 27.3 days is d = 2πr = 2.41*10^{9
}m.

Its speed is v = d/(27.3 days) = d/(2.36*10^{6 }s) = 1023 m/s.

(b) The centripetal acceleration of the moon is v^{2}/r =
2.725*10^{-3 }m/s^{2}.

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:

(a) The radius of the earth is 6368 km. We want v^{2}/r
= 9.8 m/s^{2}, or v^{2 }= (9.8 m/s^{2})*6.368*10^{6
}m, or v = 7900 m/s.

(b) The length of a day would be t = 2πr/v = 5065 s =
84.4 min.

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 }= v^{2}/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_{r}^{2} + a_{t}^{2})^{½}.

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:

Throughout the first loop the magnitude of the tangential acceleration a_{t}
is constant. The magnitude of the radial acceleration a_{r}
increases as v^{2}/r, as the speed of the skater increases as
v = a_{t}t = (2a_{t}d)^{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 }= v^{2}/(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.

An automobile, the speed of which is increasing at a rate of 0.6 m/s^{2},
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:

(a) The tangential acceleration of the car has a
constant magnitude of 0.6 m/s^{2}.

(b) When
v = 4 m/s, the radial acceleration has magnitude v^{2}/r = (16 m/s^{2})/(20
m) = 0.8 m/s^{2}, a = (a_{r}^{2} + a_{t}^{2})^{½}
= 1 m/s^{2}.

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}.

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:

(a) a_{r }= a cos(30^{o}) = 12.99 m/s^{2}.

(b) a_{r }= v^{2}/r, v^{2 }=
ra_{r}, v = 5.7 m/s.

(c) a_{t }= a sin(30^{o}) = 7.5 m/s^{2}.

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.

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 |