Worked examples · Sheet 3

Uniform circular motions

Theory used: Angular velocity Period and frequency Centripetal acceleration

Exercise 1

A toy car travels at constant speed around a circular track of radius $R = 1.5 m$ and completes $12$ laps in $30 s$ . Compute frequency, period, angular velocity, speed and centripetal acceleration.

f = \frac{N}{Δ t}, T = \frac{1}{f}, ω = 2 π f, v = ω R, a_{n} = ω^{2} R = \frac{v ^{2}}{R}

f = \frac{12}{30 s} = 0.40 Hz, T = \frac{1}{0.40 Hz} = 2.5 s

ω = 2 π \cdot 0.40 s^{- 1} = 2.51 rad/s, v = 2.51 rad/s \cdot 1.5 m = 3.77 m/s

a_{n} = (2.51 rad/s)^{2} \cdot 1.5 m = 9.47 m/ s^{2}

Track of radius

1.5 m

: the velocity is tangent and the centripetal acceleration is directed toward the center (vectors not to scale with distances)

The speed is constant, therefore the motion is uniform circular and all laps are traveled in the same time. The frequency is the number of laps per second and the period is its reciprocal. In one lap the angle described is $2 π rad$ , so the angular velocity is $2 π f$ , and the speed is obtained by multiplying by the radius. The acceleration is not zero, because the direction of the velocity changes continuously, and it is the centripetal acceleration $ω^{2} R = 9.47 m/ s^{2}$ , slightly less than $g$ . As a check, $\frac{v ^{2}}{R} = \frac{3.7 7 ^{2}}{1.5} m/ s^{2} = 9.47 m/ s^{2}$ .

Exercise 2

Treat the Earth as a sphere of radius $R_{T} = 6.37 \cdot 1 0^{6} m$ that completes one full rotation about its own axis in $T = 24 h = 8.64 \cdot 1 0^{4} s$ . Compute the angular velocity of the Earth and the speed and centripetal acceleration of a point on the equator and of a point at latitude $4 5^{\circ}$ .

ω = \frac{2 π}{T}, r = R_{T} cos λ, v = ω r, a_{n} = ω^{2} r

ω = \frac{2 π}{8.64 \cdot 1 0 ^{4} s} = 7.27 \cdot 1 0^{- 5} rad/s

λ = 0^{\circ}, r = R_{T} : v = 463 m/s, a_{n} = ω^{2} R_{T} = 0.0337 m/ s^{2}

λ = 4 5^{\circ}, r = R_{T} cos 4 5^{\circ} = 4.50 \cdot 1 0^{6} m : v = 328 m/s, a_{n} = 0.0238 m/ s^{2}

All points of the Earth rotate with the same angular velocity, because the Earth is treated as a rigid body that completes one rotation in $24 h$ . Each point, however, describes a circle around the Earth’s axis, and the radius of this circle is the distance of the point from the axis, $r = R_{T} cos λ$ at latitude $λ$ . At the equator $r = R_{T}$ , therefore the speed is $463 m/s$ and the centripetal acceleration is $0.0337 m/ s^{2}$ , about $0.34 %$ of $g$ . At latitude $4 5^{\circ}$ the radius is smaller by a factor $cos 4 5^{\circ}$ , so speed and centripetal acceleration decrease by the same factor. The centripetal acceleration due to the Earth’s rotation is therefore small compared with $g$ , and for this reason it is neglected in most problems.

Exercise 3

In a laboratory centrifuge a test tube is at $r = 10 cm$ from the axis of rotation. Compute the rotation frequency, in hertz and in revolutions per minute, for which the centripetal acceleration of the test tube is $1000$ times $g$ , and the corresponding speed.

a_{n} = ω^{2} r = 1000 g ⟹ ω = \frac{1000 g}{r}, f = \frac{ω}{2 π}, v = ω r

ω = \frac{1000 \cdot 9.81 m/ s ^{2}}{0.10 m} = 313 rad/s

f = \frac{313 rad/s}{2 π} = 49.8 Hz = 49.8 \cdot 60 rpm \approx 2990 rpm

v = 313 rad/s \cdot 0.10 m = 31.3 m/s

The test tube moves in uniform circular motion, therefore its acceleration is the centripetal acceleration $ω^{2} r$ . Setting it equal to $1000 g$ gives an equation in the single unknown $ω$ , which has the positive solution $ω = 313 rad/s$ . The frequency is the number of revolutions per second, $\frac{ω}{2 π} = 49.8 Hz$ , and multiplying by $60$ one obtains about $2990$ revolutions per minute. Since $a_{n} = ω^{2} r$ , for the same angular velocity the acceleration grows in proportion to the distance from the axis, and a test tube placed at $5 cm$ would have acceleration $500 g$ . As a check, $\frac{v ^{2}}{r} = \frac{31. 3 ^{2}}{0.10} m/ s^{2} = 9.80 \cdot 1 0^{3} m/ s^{2}$ , which is $1000 g$ within rounding.