Velocity of a Circular Motion

Section 5.2 Velocity of a Circular Motion

You know that average velocity is displacement divided by interval.

\begin{equation*} \vec v_\text{av} = \dfrac{\Delta \vec r}{\Delta t}. \end{equation*}

For infinitesimal intervals, this formula defines instantaneous velocity. We found in previous section that for small intervals in a circular motion, the displacment is

\begin{equation*} \Delta \vec r = R \Delta \theta\, \hat u_\theta, \end{equation*}

where direction is given by unit vector \(\hat u_\theta\text{,}\) the direction of tangent to the circle. Therefore, average velocity for small time intervals will be

\begin{equation*} \vec v_\text{av} = R \dfrac{\Delta \theta}{\Delta t}\; \hat u_\theta. \end{equation*}

The quantity \(\dfrac{\Delta \theta}{\Delta t}\) is called average angular “velocity” although it is not a vector. It is denoted by the Greek letter omega, \(\omega_\text{av}\text{.}\)

\begin{equation*} \omega_\text{av} = \dfrac{\Delta \theta}{\Delta t}. \end{equation*}

By convention, \(\Delta \theta \gt 0\) for motion in the counterclockwise sense. Velocity \(\vec v\) is in the tangetial direction in the unit vector \(\hat u_\theta\) direction for motion in counterclockwise sense. On the other hand, when the motion is in the clockwise sense, \(\Delta \theta \lt 0\text{,}\) and while velocity is still tangetial to circle, but it will be in opposite direction to the unit vector \(\hat u_\theta\) direction.

Figure 5.12. Velocity for counterclockwise sense is in the direction of the tangential unit vector \(\hat u_\theta\) and in the opposite direction for clockwise sense of motion.

This can be used to replace the derivative in instantaneous velocity giving us a compact formula.

\begin{equation*} \vec v = R \omega\, \hat u_\theta. \end{equation*}

Note the factor \(R \omega\) in velocity is analogous to the arc length \(R\theta\text{.}\) Thus, instantaneous velocity has magnitude \(R|\omega|\text{,}\) which is instantaneous speed \(v\text{,}\) and direction in the direction of the tangent to the circle.

\begin{equation*} v = R|\omega|, \end{equation*}

where I have placed absolute sign to denote angular speed, which should be positive, regardles sof the clockwise or coungercloskwise sense of motion.

Remark 5.13.

(Calculus) When we take infinitesimal interval limit, we get velocity at instant \(t\) as derivative

\begin{equation*} \vec v = R \frac{d \theta}{d t}\, \hat u_\theta. \end{equation*}

The instantaneous rate of change of subtended angle is called instantaneous angular velocity, which is denoted by \(\omega\text{.}\)

\begin{equation*} \omega = \frac{d \theta}{d t}. \end{equation*}

Example 5.14. Angular Speed from Speed in Circular Motion.

A car is rounding a circular bend of radius \(50\text{ m}\) with speed \(20\text{ m/s}\text{.}\) What is the angular speed of the car?

Answer.

\(0.4\text{ rad/s}\text{.}\)

Solution.

We use \(v = R|\omega|\) to find angular speed.

\begin{align*} |\omega| \amp = \dfrac{v}{R}\\ \amp = \dfrac{20\text{ m/s}}{50\text{ m}} = 0.4\text{ rad/s}. \end{align*}

Note that I put \(\text{rad}\) in the unit for \(|\omega|\) since the displacement it refers to is the angle subtended expressed in radians.

Example 5.15. Arc Length, Arc Angle, Average Speed, Average Angular Speed, and Average Velocity.

Figure 5.16 shows a body fixed at the edge of a circular rotating platform of radius \(5 \text{ m}\text{.}\) In \(2 \text{ sec}\text{,}\) the body goes from the location marked A to the location marked B without making any full turns, where the marking are in space and not on the platform.

(a) What is the value of the angle rotated in radian?

(b) How much net distance has the body moved?

(d) What is the average angular speed of the body?

(e) Draw arrows on the figure to show the direction of the velocity (i) at the instant the body was at A and (ii) at the instant the body was at B.

(f) (i) Draw an arrow on the figure to show the direction of the average velocity during the interval between A and B. (ii) What is the magnitude of the average velocity?

(g) Suppose the body ended up at location B, but during the \(2 \text{ sec}\) interval, the body made three full turns before ending up at B. What would be the average speed and average angular speed now?

(h) Suppose the body ended up at location B, but during the \(2 \text{ sec}\) interval, the body made three full turns before ending up at B. Will the average velocity now be same or different from the situation when the body did not make any full turns? Why or why not?

Answer.

(a) \(\dfrac{\pi}{3}\text{ rad}\text{,}\) (b) \(5.24\text{ m} \text{,}\) (c) \(2.62 \text{ m/s}\text{,}\) (d) \(0.524 \text{ rad/s}\text{,}\) (e) and (f) see the solution for figures; \(v_{\text{av}} = 2.5\text{ m/s}. \text{,}\) (g) \(49.74\text{ m/s}, 9.95\text{ rad/s} \text{,}\) (h) Same as (f), magnitude of average velocity \(2.5\text{ m/s} \text{.}\)

Solution 1. a,b,c,d

(a) \(\theta = 60^{\circ} = 60\times\dfrac{\pi}{180} = \dfrac{\pi}{3}\text{ rad}\text{.}\)

(b) \(s = R\,\theta = 5\times \dfrac{\pi}{3} = 5.24\text{ m}\text{.}\)

(d) \(\omega_{\text{av}} = \dfrac{\theta}{t} = \dfrac{\pi}{3\times2} = 0.524 \text{ rad/s}\text{.}\)

Solution 2. e,f

(e) and (f). The figure shows the velocity directions at A (\(\vec v_A\)) and at B (\(\vec v_B\)), and the average velocity direction(\(\vec v_{\text{av}}\)).

To figure out the magnitude of the average velocity, we need the direct distance between points A and B, and not the distance on the arc, which was used for the average speed.

We can compute the direct distance from the \((x,y) \) of A and B. The formula for direct distance between two points with coordinates \((x_A, y_A) \) \((x_B, y_B) \) is

\begin{equation*} d = \sqrt{ (\Delta x)^2 + (\Delta y)^2 }, \end{equation*}

where \(\Delta x = x_B - x_A \) and \(\Delta y = y_B - y_A \text{.}\)

In the figure, we have

\begin{align*} \amp x_A = R,\ \ y_A = 0 \\ \amp x_B = 5\,\cos\, 60^{\circ} = 2.5\text{ m},\ \ y_B = 5\,\cos\, 60^{\circ} = 4.33\text{ m}. \end{align*}

Therefore, the direct distance

\begin{equation*} d = \sqrt{2.5^2 + 4.33^2} = 5\text{ m}. \end{equation*}

Therefore, the magnitude of the average velocity is

\begin{equation*} v_{\text{av}} = \dfrac{5}{2} = 2.5\text{ m/s}. \end{equation*}

Solution 3. g,h

(g) If the body made three additional full turns, then the distance traveled for average speed will be not just the arc length from A to B but that plus the distance of the three full turns. For each turn, the distance is the circumference, \(2\pi R \text{.}\) The time still is \(2\text{ sec}\text{.}\) Therefore, the average speed now will be

\begin{equation*} v_{\text{s,av}} = \dfrac{5.24 + 3\times 2\pi\times 5}{2} = 49.74\text{ m/s}. \end{equation*}

Similarly, the average angular speed will include all the angles went over.

\begin{equation*} \omega_{\text{av}} = \dfrac{\pi/3 + 3\times 2\pi}{2} = 9.95\text{ rad/s}. \end{equation*}

(h) For average velocity, whatever you did between A and B does not matter - only the end points matter. Therefore, the direct distance will not change, and since the time is the same, the magnitude of the averafge velocity will not change. The direction is also the same, from point A towards point B.

Exercises Exercises

1. Angular Velocity of a Pebble on a Rotating Tire.

A pebble is stuck in the grooves of a rotating tire of radius \(25\) cm (Figure 5.17). If the car moves at a constant velocity of \(20\) m/s what is the angular velocity of the pebble?

Solution.

Notice that when the center of the tire moves a distance \(2\pi R\text{,}\) i.e. a full circle of the tire, the pebble rotates by an angle of \(2\pi\) radians in a circle about the center of the tire whose radius is equal to the radius of the tire \(R\) (units meter). Since the center of the tire moves \(20\) m in \(1\) sec, the pebble rotates an angle of \(20/R\) radians in that time. Therefore, the angular speed of the pebble is \(80\ \text{rad/s}\text{,}\) since \(R=0.25\ \text{m}\text{.}\)

2. Direction of a Particle Moving on a Circle.

A particle moves in a circle of radius \(a\) with a uniform circular motion of constant angular velocity \(\omega = \omega_0\) (radians per second). The particle moves in the \(xy\)-plane in a counter-clockwise manner when observed from the positive \(z\)-axis with the center of the circle of motion at the origin. At \(t=0\) the particle is at the point whose Cartesian coordinates are \((a,0,0)\) with respect to origin at the center of the circle of motion. Where will the particle be at an arbitrary time \(T\text{?}\)

Answer.

\(\omega_0 T \text{ mod } 2\pi\text{.}\)

Solution.

The amount of angle covered duration time \(T\) in radians will be

\begin{equation*} \Delta \theta = \omega_0 T. \end{equation*}

But, on a circle, the direction will be indicated by an angle between \(0\) and \(2\pi\text{.}\) Therefore, the angle \(\theta\) of the direction from the center will be

\begin{equation*} \theta = \omega_0 T \text{ mod } 2\pi. \end{equation*}

For example: if \(\omega_0 = 20\,\text{rad/sec}\) and \(T=5.0\,\text{sec}\text{,}\) then \(\Delta \theta = 100\,\text{rad}\text{.}\) Now, which way from the center of the circle is this direction? You will need to subtract some whole number times \(2\pi\) to get this to come between \(0\) and \(2\pi\text{.}\) One way to do this to divide \(100\,\text{rad}\) by \(2\pi\) and look at the remainder.

\begin{equation*} 100 / (2\pi) \approx 15.9155. \end{equation*}

Now, we drop the whole number and look at only the fraction part.

\begin{equation*} 0.9155\,\text{rad} = 0.9155 \times 180/\pi \,\tex{deg} = 52.5^\circ. \end{equation*}

3. Particle in a Non-uniform Circular Motion.

A particle moves in a circle of radius \(a\) with a varying angular velocity \(\omega = b t\) counterclockwise, where \(b\) is constant. At \(t=0\) the particle has Cartesian coordinates are \((a,0,0)\) with respect to the origin at the center of the circle of motion which is at the origin. Where will the particle be at an arbitrary time \(T\text{?}\)

Answer.

\(\frac{b}{2}T^2 \text{ mod } 2\pi\text{.}\)

Solution.

Let’s mae use of the basic definition of angular velocity as derivative of the angle coordinate of the particle on the circle.

\begin{equation*} frac{d\theta}{dt} = bt. \end{equation*}

Now, rearrange it and then integrate to get \(\Delta \theta\text{.}\)

\begin{equation*} \Delta\theta = \int_0^T b t dt = \frac{b}{2}T^2. \end{equation*}

Therefore, the particle will be

\begin{equation*} \theta = \frac{b}{2}T^2 \text{ mod } 2\pi. \end{equation*}

4. (Calculus) Angular Velocity from Angle as a Function of Time.

A car is moving in a circular path of radius \(40\text{ m}\) with increasing speed. Its angular position is given by \(\theta = 0.05\, t^2\) in radians when \(t\) is in seconds. (a) What is angular speed at \(t=3\text{ s}\text{?}\) (b) (a) What is speed at \(t=3\text{ s}\text{?}\)

Hint.

(a) Use \(\omega = d\theta/dt\text{,}\) (b) Use \(v = R|\omega|\text{.}\)

Answer.

(a) \(0.3\text{ rad/s}\text{,}\) (b) \(12\text{ m/s}\text{.}\)

Solution 1. a

Angular speed will be absolute value of

\begin{equation*} \omega = \dfrac{d\theta}{dt} = 0.1\, t. \end{equation*}

Evaluating this at \(t=3\text{ s}\text{,}\) we get \(\omega = 0.3\text{ rad/s}\text{.}\)

Solution 2. b

We use \(v = R|\omega|\) to find speed.

\begin{equation*} v = R|\omega| = 40\text{ m} \times 0.3\text{ rad/s} = 12\text{ m/s}. \end{equation*}

You may have noticed that when multiplying \(\text{radian}\) by another unit, \(\text{radian}\) disappears since it is itself dimensionless, being ratio of two distances.

5. Circular Motion of a Cell in a Centrifuge.

In a centrifuge a biological soup are sometimes placed in a test tube and then test tube is spun in a centrifuge. Suppose a cell is moving in a circular path of radius \(5\text{ cm}\text{.}\) If the centrifuge is spinning at angular speed of \(100,000\text{ revolutions per minute}\text{.}\) What is the speed of the cell in \(\text{m/s}\text{?}\)

Hint.

Convert rev/min to rad/sec.

Answer.

\(524\text{ m/s}\text{.}\)

Solution.

The angular speed is given in revolutions per minute. We need to convert this into rad/sec. In each revolution, the motion covers \(2\pi\text{ rad}\text{.}\) Therefore,

\begin{equation*} \omega = 100,000 \times \frac{2\pi}{60} = 10,472\text{ rad/sec}. \end{equation*}

Therefore, speed of the cell is

\begin{equation*} v = \omega R = 10,472 \times 0.05 = 524\text{ m/s}. \end{equation*}

6. Practice with a Friend: Regular and Angular Velocity of Earth.

Earth goes around Sun in almost a circular orbit of radius \(1.5\times 10^{11}\text{ m}\) once around in approximately \(365.2425\text{ days}\text{.}\) (a) What is Earth’s angular velocity for its oribital motion about the Sun? (b) How fast is the Earth moving through space?

Answer.

(a) \(2.0\times 10^{-7}\text{ rad/sec}\text{,}\) (b) \(29.9\text{ km/s}.\text{.}\)

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