Uniform Circular Motion: Speed And Acceleration

Describing Motion Around a Circle

A point moving around a circle can change its direction of travel even when its speed stays the same. This makes circular motion a useful setting for distinguishing speed, which measures how quickly distance is covered, from velocity and acceleration, which also have directions.

For uniform circular motion, the circle has a fixed centre and radius, and the angle describing the point’s position changes at a constant rate. We use radians for the angle. A positive angular rate describes counterclockwise motion; a negative angular rate describes clockwise motion.

Definition: Uniform circular motion with centre \((h,k)\), radius \(R>0\), initial angle \(\theta_0\), and constant signed angular rate \(\omega\) is described by \(x(t)=h+R\cos(\theta_0+\omega t)\) and \(y(t)=k+R\sin(\theta_0+\omega t)\). Here \(\theta(t)=\theta_0+\omega t\) is the angle at time \(t\), and \(\omega\) is measured in radians per unit of time.

The corresponding position vector is \(\mathbf r(t)=(x(t),y(t))\). The centre’s position vector is \(\mathbf c=(h,k)\), so the vector from the centre to the moving point is \(\mathbf r(t)-\mathbf c\). It always has length \(R\).

Velocity and Speed

Velocity is the derivative of position with respect to time. Differentiating both coordinate functions by the chain rule gives the velocity vector. Its magnitude is the speed.

\mathbf v(t)=\mathbf r'(t) =R\omega\bigl(-\sin\theta(t),\cos\theta(t)\bigr).

To find the speed, take the magnitude of this vector. The trigonometric identity \(\sin^2\theta+\cos^2\theta=1\) removes the dependence on the angle. Therefore the speed is constant, even though the direction of the velocity changes as the point moves around the circle.

\|\mathbf v(t)\| =\sqrt{R^2\omega^2\sin^2\theta(t)+R^2\omega^2\cos^2\theta(t)} =R|\omega|.

The absolute value matters: speed is nonnegative, while the sign of \(\omega\) records the direction of rotation. If \(\omega=0\), the point does not traverse the circle; it remains stationary at its initial position.

Acceleration and the Centre

Acceleration is the derivative of velocity with respect to time. Differentiating each component of \(\mathbf v(t)\) introduces another factor of \(\omega\), because \(\theta'(t)=\omega\). The result can be written directly in terms of the point’s displacement from the centre.

\mathbf a(t)=\mathbf v'(t) =-R\omega^2\bigl(\cos\theta(t),\sin\theta(t)\bigr) =-\omega^2\bigl(\mathbf r(t)-\mathbf c\bigr).

Since \(\omega^2>0\) when the motion is nonstationary, the acceleration vector is a negative multiple of the vector from the centre to the point. It therefore points toward the centre. Its magnitude is \(R\omega^2\), which is often called the centripetal acceleration magnitude.

Theorem: In uniform circular motion with \(R>0\) and \(\omega\ne 0\), the speed is the constant \(R|\omega|\), and the acceleration is \(-\omega^2(\mathbf r-\mathbf c)\). Thus the acceleration points toward the centre and has constant magnitude \(R\omega^2\).

Why the Acceleration Formula Works

The formula follows by differentiating the position twice. This derivation also shows why acceleration points inward, rather than merely asserting that it does.

Write the angle as a function of time.
Set \(\theta(t)=\theta_0+\omega t\), so \(\theta'(t)=\omega\).

Differentiate position.
From \(\mathbf r(t)=\mathbf c+R(\cos\theta(t),\sin\theta(t))\), the chain rule gives \(\mathbf v(t)=R\omega(-\sin\theta(t),\cos\theta(t))\).

Differentiate velocity.
Using \(\theta'(t)=\omega\) again gives \(\mathbf a(t)=-R\omega^2(\cos\theta(t),\sin\theta(t))\).

Compare with the radius vector.
Because \(\mathbf r(t)-\mathbf c=R(\cos\theta(t),\sin\theta(t))\), the acceleration equals \(-\omega^2(\mathbf r(t)-\mathbf c)\), a negative multiple of the outward radius vector.

The same formulas show that velocity and acceleration are perpendicular. Their dot product is zero because \((-\sin\theta)(\cos\theta)+(\cos\theta)(\sin\theta)=0\). Thus, in this motion, acceleration changes the direction of velocity without changing its magnitude. This agrees with the constant-speed calculation.

Worked Examples

Worked Example: Position, Velocity, and Acceleration at a Specified Time

A point moves on a circle of radius \(5\) units centred at \((2,-1)\), with angle \(\theta(t)=\frac{\pi}{6}+3t\), where \(t\) is in seconds. Find its position, velocity, speed, and acceleration at \(t=\frac{\pi}{9}\) seconds.

First evaluate the angle:

\theta\left(\frac{\pi}{9}\right) =\frac{\pi}{6}+3\left(\frac{\pi}{9}\right) =\frac{\pi}{6}+\frac{\pi}{3} =\frac{\pi}{2}.

At this angle, \(\cos(\pi/2)=0\) and \(\sin(\pi/2)=1\). Therefore the position is

\mathbf r\left(\frac{\pi}{9}\right) =(2+5\cdot 0,\,-1+5\cdot 1) =(2,4)\text{ units}.

Here \(R=5\) and \(\omega=3\) radians per second. The velocity is

\mathbf v\left(\frac{\pi}{9}\right) =5\cdot3\bigl(-1,0\bigr) =(-15,0)\text{ units/s}.

Its speed is \(\sqrt{(-15)^2+0^2}=15\) units/s, which agrees with \(R|\omega|=5\cdot3=15\). The acceleration is

\mathbf a\left(\frac{\pi}{9}\right) =-5\cdot3^2(0,1) =(0,-45)\text{ units/s}^2.

Check the direction: from the centre \((2,-1)\) to the point \((2,4)\) the radius vector is \((0,5)\), so the acceleration \((0,-45)\) points toward the centre. Its magnitude is \(45=R\omega^2\) units/s², as required.

Worked Example: Clockwise Motion

A point moves on a circle of radius \(0.4\) m, centred at the origin, with \(\theta(t)=\frac{\pi}{3}-5t\), where \(t\) is in seconds. Find its position, velocity, speed, and acceleration at \(t=\frac{\pi}{15}\) seconds.

The angular rate is \(\omega=-5\) rad/s, so the motion is clockwise. At the specified time,

\theta\left(\frac{\pi}{15}\right) =\frac{\pi}{3}-5\left(\frac{\pi}{15}\right) =\frac{\pi}{3}-\frac{\pi}{3} =0.

Thus the position is \((0.4,0)\) m. The velocity is

\mathbf v\left(\frac{\pi}{15}\right) =0.4(-5)(-\sin 0,\cos 0) =-2(0,1) =(0,-2)\text{ m/s}.

The speed is \(\sqrt{0^2+(-2)^2}=2\) m/s, also given by \(R|\omega|=0.4\cdot5=2\) m/s. The acceleration is

\mathbf a\left(\frac{\pi}{15}\right) =-0.4(-5)^2(\cos 0,\sin 0) =-10(1,0) =(-10,0)\text{ m/s}^2.

The point is at the rightmost point of the circle, and the acceleration points left, toward the origin. Its magnitude \(10\) m/s² equals \(R\omega^2=0.4\cdot25=10\) m/s². The negative angular rate changes the velocity’s direction of travel, but its square leaves the inward acceleration formula unchanged.

Worked Example: Using a Period to Find the Motion

A point starts at the rightmost point of a circle of radius \(2\) units centred at \((-1,3)\), and completes one counterclockwise revolution in \(4\) seconds. Find its position, velocity, speed, and acceleration after \(1\) second.

One revolution is \(2\pi\) radians, so the constant angular rate is \(\omega=\frac{2\pi}{4}=\frac{\pi}{2}\) rad/s. Starting at the rightmost point means \(\theta_0=0\), hence \(\theta(t)=\frac{\pi}{2}t\). At \(t=1\), the angle is \(\pi/2\), and the position is

\mathbf r(1) =\left(-1+2\cos\frac{\pi}{2},\,3+2\sin\frac{\pi}{2}\right) =(-1,5)\text{ units}.

The velocity at that time is

\mathbf v(1) =2\left(\frac{\pi}{2}\right)(-1,0) =(-\pi,0)\text{ units/s}.

Its speed is \(\sqrt{(-\pi)^2+0^2}=\pi\) units/s, which matches \(R|\omega|=2\cdot\frac{\pi}{2}=\pi\). The acceleration is

\mathbf a(1) =-2\left(\frac{\pi}{2}\right)^2(0,1) =\left(0,-\frac{\pi^2}{2}\right)\text{ units/s}^2.

From the centre \((-1,3)\) to the position \((-1,5)\), the radius vector is \((0,2)\), so the acceleration points in the opposite direction, toward the centre. Its magnitude is \(\frac{\pi^2}{2}=R\omega^2\) units/s².

A Common Sign Error

A frequent mistake is to differentiate the velocity but lose the negative sign in the acceleration. For example, when \(\theta=0\), an incorrect calculation might claim

\mathbf a=R\omega^2(1,0).

This points outward from the centre at the rightmost point, so it cannot be the correct centripetal acceleration. Differentiating \(v_x=-R\omega\sin\theta\) gives \(a_x=-R\omega^2\cos\theta\), not \(+R\omega^2\cos\theta\). The correct vector at \(\theta=0\) is

\mathbf a=-R\omega^2(1,0),

which points left, toward the centre. Checking the direction against the circle’s centre is a quick way to catch this sign error.

What the Formulas Tell Us

Uniform circular motion has constant speed, but it does not have zero acceleration. Acceleration measures a change in velocity, and velocity includes direction as well as magnitude. As the point moves around the circle, its velocity direction continuously turns; the inward acceleration accounts for that change.

When solving a problem, keep the signed angular rate in the velocity formula, since its sign affects the direction of travel. For speed, use \(R|\omega|\). For acceleration, use \(-\omega^2(\mathbf r-\mathbf c)\): the square makes the magnitude independent of rotation direction, while the minus sign ensures that the vector points toward the centre.

Check Your Understanding

Use the definitions and results in this tutorial to answer the following questions.

A point moves on a circle of radius \(3\) m with angular rate \(-2\) rad/s. What is its speed?
For a nonstationary uniform circular motion, write the acceleration in terms of the position vector and the centre vector.
At the top of a circle centred at the origin, which direction does the acceleration point?
Why does a negative angular rate change the velocity direction but not the acceleration magnitude?
What sign error would make the acceleration point away from the centre at the rightmost point?