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Related Rates Problems · Tutorial 668 of 1495

Linear Speed Along a Circular Path

Connect angular and linear speed using arc length, and distinguish motion around a fixed circle from motion whose radius is changing.

Intermediate 9 min read

What You'll Learn

  • Derive the arc-length relation using angles measured in radians
  • Convert angular velocity into tangential speed when the radius is fixed
  • Track units and distinguish signed angular velocity from speed magnitudes
  • Explain why differentiating radius times angle does not generally give total path speed
  • Find radial, tangential, and total speed when the radius is changing

From Rotation to Distance

A point moving around a circle can be described in two related ways. Its angular position tells us how far it has turned around the centre; its linear motion tells us how far it has travelled along the circular path. The connection is simple when the radius stays fixed, but needs care when the radius changes.

Angles in the formulas below are measured in radians. This is essential: the arc-length formula uses radians, not degrees.

Definition: For motion around a circle of fixed radius \(r\), the signed arc coordinate \(s\) and angular position \(\theta\) satisfy \(s=r\theta\). Differentiating with respect to time gives the signed tangential velocity \(ds/dt=r\,d\theta/dt\). The corresponding speed magnitudes satisfy \(v=r|\omega|\), where \(\omega=d\theta/dt\).

Here \(r\) is measured in units of length, \(\theta\) in radians, and time in units such as seconds. Radians are dimensionless in dimensional analysis, so \(r\omega\) has units of length per time.

Why the Arc-Length Formula Works

A complete turn has angle \(2\pi\) radians and travels one circumference, \(2\pi r\). An angle \(\theta\) therefore covers the fraction \(\theta/(2\pi)\) of the circumference when \(\theta\) is nonnegative. Multiplying that fraction by the circumference gives the arc length:

$$ s=\frac{\theta}{2\pi}(2\pi r)=r\theta. $$

For signed angles, \(s=r\theta\) records displacement along the circle in a chosen orientation; reversing direction makes \(s\) decrease. The physical distance along the arc for an angular change \(\Delta\theta\) is \(r|\Delta\theta|\).

When the radius is constant, differentiating the relation is direct. The derivative of the constant \(r\) is zero, so the rate of change of arc position is proportional to angular velocity.

$$ \frac{ds}{dt} = \frac{d}{dt}(r\theta) = r\frac{d\theta}{dt} = r\omega. $$

The signed quantity \(ds/dt\) gives direction along the circle. Speed is nonnegative, so it is the magnitude \(v=|ds/dt|=r|\omega|\). If the point moves in the positive angular direction, the signed rate is positive; if it moves in the negative direction, it is negative.

Worked Example: Speed on a Fixed Circular Track

A point moves on a circular track of radius \(0.75\) metres. At one instant, its angular velocity is \(2.4\) radians per second in the positive direction. Find its tangential speed.

The radius is fixed, so \(v=r|\omega|\). Substituting the values gives

$$ v=(0.75\ \text{m})(2.4\ \text{rad/s}) =1.80\ \text{m/s}. $$

The direction is positive around the track, so the signed tangential velocity is also \(1.80\ \text{m/s}\). As a check, in one second the angle changes by \(2.4\) radians, covering an arc of \(0.75(2.4)=1.80\) metres. Thus the distance-per-time calculation agrees with the rate formula.

Converting Revolutions into Angular Velocity

Sometimes the given rotational rate is in revolutions per unit time rather than radians per unit time. One complete revolution is \(2\pi\) radians. Convert the rate to radians per unit time before applying \(v=r|\omega|\).

Worked Example: A Rotating Wheel

A point on a wheel is \(0.40\) metres from its axis. The wheel turns at \(90\) revolutions per minute. Find the point’s tangential speed in metres per second.

First convert revolutions per minute to radians per second:

$$ 90\ \frac{\text{revolutions}}{\text{minute}} \times \frac{2\pi\ \text{radians}}{1\ \text{revolution}} \times \frac{1\ \text{minute}}{60\ \text{seconds}} = 3\pi\ \text{rad/s}. $$

Then use the fixed-radius formula:

$$ v=(0.40\ \text{m})(3\pi\ \text{rad/s}) =1.2\pi\ \text{m/s} \approx 3.77\ \text{m/s}. $$

The approximation is rounded to two decimal places. To check, the point travels \(2\pi(0.40)=0.8\pi\) metres in each revolution. The wheel makes \(90/60=1.5\) revolutions per second, so its distance per second is \((0.8\pi)(1.5)=1.2\pi\) metres per second, as calculated.

When the Radius Changes

The formula \(s=r\theta\) describes the arc on a circle of radius \(r\). If the radius is changing, it is important not to treat \(r\theta\) as the total distance the moving point has travelled. A changing-radius point can move both around the centre and toward or away from it. Its total speed combines these two perpendicular components.

To see this precisely, describe the point’s Cartesian coordinates in terms of its radius and angle: \(x=r\cos\theta\) and \(y=r\sin\theta\). Differentiating both coordinates with respect to time gives

$$ \frac{dx}{dt} = \frac{dr}{dt}\cos\theta-r\sin\theta\frac{d\theta}{dt}, \qquad \frac{dy}{dt} = \frac{dr}{dt}\sin\theta+r\cos\theta\frac{d\theta}{dt}. $$

The speed is the magnitude of the velocity, \(\sqrt{(dx/dt)^2+(dy/dt)^2}\). Squaring the two coordinate rates and adding them cancels the mixed terms. Using \(\sin^2\theta+\cos^2\theta=1\) leaves

$$ \begin{aligned} \left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2 &= \left(\frac{dr}{dt}\right)^2(\cos^2\theta+\sin^2\theta) +r^2\left(\frac{d\theta}{dt}\right)^2(\sin^2\theta+\cos^2\theta)\\ &= \left(\frac{dr}{dt}\right)^2 +r^2\left(\frac{d\theta}{dt}\right)^2. \end{aligned} $$

Taking the nonnegative square root gives the total speed. The radial component has signed rate \(dr/dt\); the tangential component has signed rate \(r\,d\theta/dt\). Their magnitudes are perpendicular components, not quantities to add directly.

Theorem: For a point whose distance \(r\) from a fixed centre and angle \(\theta\) both vary differentiably with time, its total speed is $$ v=\sqrt{\left(\frac{dr}{dt}\right)^2+r^2\left(\frac{d\theta}{dt}\right)^2}. $$ The tangential speed is \(r|d\theta/dt|\), and the radial speed magnitude is \(|dr/dt|\). If \(dr/dt=0\), the total speed reduces to the fixed-radius formula \(v=r|d\theta/dt|\).

Worked Example: A Point Moving Outward While Rotating

At an instant, a point is \(2\) metres from the centre and has angular position \(\pi/3\). Its radius is increasing at \(1/2\) metre per second, and its angular velocity is \(2\) radians per second. Find its radial speed, tangential speed, and total speed.

The radial speed magnitude is \(|dr/dt|=1/2\ \text{m/s}\). The tangential speed is

$$ r\left|\frac{d\theta}{dt}\right| = (2\ \text{m})(2\ \text{rad/s}) = 4\ \text{m/s}. $$

Since these components are perpendicular, the total speed is

$$ v = \sqrt{\left(\frac{1}{2}\right)^2+4^2}\ \text{m/s} = \sqrt{\frac{1}{4}+16}\ \text{m/s} = \frac{\sqrt{65}}{2}\ \text{m/s} \approx 4.03\ \text{m/s}. $$

The decimal is rounded to two places. For a direct check, differentiate \(x=r\cos\theta\) and \(y=r\sin\theta\) at the given instant. Since \(\cos(\pi/3)=1/2\) and \(\sin(\pi/3)=\sqrt{3}/2\),

$$ \frac{dx}{dt} = \frac{1}{2}\cdot\frac{1}{2} - 2\cdot\frac{\sqrt{3}}{2}\cdot 2 = \frac{1}{4}-2\sqrt{3}, \qquad \frac{dy}{dt} = \frac{1}{2}\cdot\frac{\sqrt{3}}{2} + 2\cdot\frac{1}{2}\cdot 2 = \frac{\sqrt{3}}{4}+2. $$

Squaring and adding, the cross terms cancel:

$$ \left(\frac{1}{4}-2\sqrt{3}\right)^2 + \left(\frac{\sqrt{3}}{4}+2\right)^2 = \frac{65}{4}. $$

The speed is therefore \(\sqrt{65/4}=\sqrt{65}/2\ \text{m/s}\), agreeing with the component calculation.

A Common Mistake: Differentiating \(r\theta\) as Total Distance

If both \(r\) and \(\theta\) vary, the product rule gives \(d(r\theta)/dt=r(d\theta/dt)+\theta(dr/dt)\). That is the derivative of the quantity \(r\theta\), but it is not generally the total speed of the moving point. The product \(r\theta\) is an arc coordinate for the circle at the current radius; when that radius changes, it does not record the complete path travelled.

For the values in the previous example, this product-rule calculation would give

$$ \frac{d}{dt}(r\theta) = 2(2)+\frac{\pi}{3}\left(\frac{1}{2}\right) = 4+\frac{\pi}{6}\ \text{m/s}. $$

This is not the point’s total speed, which is \(\sqrt{65}/2\ \text{m/s}\). Nor is \(d(r\theta)/dt\) its tangential speed, which is \(4\ \text{m/s}\). The changing-radius motion must be handled using the radial and tangential components.

Takeaway: For motion on a fixed circle, use \(v=r|\omega|\). For motion whose radius changes, \(r|\omega|\) gives only the tangential component; combine it with the radial component using the square-root formula for total speed.

Check Your Understanding

Use radians throughout, and include units where appropriate.

  1. A point moves on a fixed circle of radius \(3\) metres with angular velocity \(-2\) radians per second. What are its signed tangential velocity and speed?
  2. A point \(0.25\) metres from an axis rotates at \(120\) revolutions per minute. Convert its angular velocity to radians per second and find its tangential speed.
  3. Why must angles be measured in radians when using \(s=r\theta\)?
  4. A point has \(dr/dt=3\) metres per second and \(d\theta/dt=4\) radians per second at radius \(2\) metres. Find its radial speed, tangential speed, and total speed.
  5. When the radius changes, why is \(d(r\theta)/dt\) not generally the total speed?

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