You switch on a ceiling fan. For a few seconds the blades gather speed, and then they settle at a steady rate. At the first instant the blades are still at rest. A moment later they are turning, and a few seconds later they are a blur. The motion of a blade tip is circular throughout, but it is not uniform.
Angular acceleration
When the angular velocity changes, the turning has an angular acceleration. Its average over a time \(\Delta t\) is \(\alpha_{\text{avg}} = \Delta\omega/\Delta t\). Its instantaneous value is the limit, the rate of change of \(\omega\). This is the twin of \(a = dv/dt\) on a line.
Like \(\vec\omega\), \(\vec\alpha\) lies along the axis. If the turning speeds up, \(\vec\alpha\) points the same way as \(\vec\omega\). If the turning slows down, \(\vec\alpha\) points opposite to \(\vec\omega\). For turning in one plane, signs do the job: anticlockwise positive. A positive \(\alpha\) with a positive \(\omega\) means speeding up. A negative \(\alpha\) with a positive \(\omega\) means slowing down.
A wheel turns anticlockwise, and anticlockwise is positive. Its \(\omega\) falls from \(10\,\text{rad/s}\) to \(4\,\text{rad/s}\) in \(3\,\text{s}\). What is its average angular acceleration?
Two accelerations at once
A body on a circle whose speed changes is in non-uniform circular motion. It has both parts of the acceleration from Lesson 1. The tangential part changes the speed. Differentiate \(v = \omega r\) at a fixed \(r\) to find it: \(a_t = dv/dt = r\,d\omega/dt = \alpha r\). The normal part still turns the velocity. It points at the centre and has the size \(a_c = v^2/r = \omega^2 r\), whatever the speed is doing, using the speed at that instant.
The two parts are at right angles, so the total acceleration has the size \(\sqrt{a_t^2 + a_c^2}\). Its direction lies between the tangent and the radius. Measure it from the radius towards the centre, and \(\tan\phi = a_t/a_c\). When \(a_t\) is large compared with \(a_c\), the acceleration lies close to the tangent. When \(a_c\) is the larger, it lies close to the radius.
A wheel starts from rest with a constant angular acceleration. At the very first instant, which way does the acceleration of a point on its rim point?
The figure answers the fan's question. The ball starts from rest on a circle of radius \(2.5\,\text{m}\) with \(a_t = 1\,\text{m/s}^2\), shown at half speed. At the start the acceleration lies along the tangent. As the speed grows, \(a_c = v^2/r\) grows as the square of the speed, while \(a_t\) stays the same. The total acceleration tilts towards the centre. By \(v = 2.7\,\text{m/s}\), \(a_c = 3\,\text{m/s}^2\), and the acceleration is only \(18°\) from the radius.
A wheel of radius \(0.4\,\text{m}\) starts from rest with a constant angular acceleration of \(5\,\text{rad/s}^2\). For a point on the rim at \(t = 1\,\text{s}\), find the speed, both parts of the acceleration, the total acceleration and its angle with the radius.
\(\omega = \alpha t = 5\,\text{rad/s}\), so \(v = \omega r = 5 \times 0.4 = 2\,\text{m/s}\).
\(a_t = \alpha r = 5 \times 0.4 = 2\,\text{m/s}^2\). \(a_c = \omega^2 r = 25 \times 0.4 = 10\,\text{m/s}^2\).
\(a = \sqrt{2^2 + 10^2} = \sqrt{104} \approx 10.2\,\text{m/s}^2\).
\(\tan\phi = a_t/a_c = 0.2\), so \(\phi \approx 11.3°\) from the radius. After one second the centripetal part already dominates. The two parts were equal, at \(45°\), when \(\omega^2 = \alpha\), that is at \(\omega = \sqrt5\,\text{rad/s}\), after about \(0.45\,\text{s}\).
A car on a circular track of radius \(50\,\text{m}\) moves at \(10\,\text{m/s}\) and is speeding up at \(1.5\,\text{m/s}^2\). What is the size of its total acceleration, in m/s²?
A car goes round a bend and brakes at the same time. Where does its total acceleration point?
The vector forms
The relations \(v = \omega r\), \(a_t = \alpha r\) and \(a_c = \omega^2 r\) give sizes. The cross product gives the directions as well. Place the origin at the centre of the circle, so the position vector \(\vec r\) runs from the centre to the body. The angular velocity \(\vec\omega\) points along the axis, at right angles to \(\vec r\).
Then \(\vec v = \vec\omega \times \vec r\). Check the size: the angle between \(\vec\omega\) and \(\vec r\) is \(90°\), so \(|\vec v| = \omega r\). Check the direction: the cross product is at right angles to both \(\vec\omega\) and \(\vec r\). So it lies in the plane of the circle, along the tangent. The right-hand rule picks the correct sense of the tangent.
\[ \vec v = \vec\omega \times \vec r \]
Why. the velocity of a point on the circle, from the paragraph above.
?
Your turn. which line is it? Differentiate a product, one factor at a time, keeping the order of the cross product.
\[ \vec a = \vec\alpha \times \vec r + \vec\omega \times \vec v \]
Why. \(d\vec\omega/dt\) is the angular acceleration \(\vec\alpha\), and \(d\vec r/dt\) is the velocity \(\vec v\).
\[ |\vec\alpha \times \vec r| = \alpha r = a_t, \qquad |\vec\omega \times \vec v| = \omega v = \omega^2 r = a_c \]
Read it. each pair of vectors is at right angles. The first term lies along the tangent, the second along the radius towards the centre.
The two terms are the two parts found before. The term \(\vec\alpha \times \vec r\) is the tangential acceleration, and it vanishes when the turning is steady. The term \(\vec\omega \times \vec v\) is the centripetal acceleration. Use the right-hand rule on it: with \(\vec\omega\) out of the page and \(\vec v\) along the tangent, \(\vec\omega \times \vec v\) points at the centre. The vector form gives the size and the direction of both parts in one line.
A particle moves on a circle about the origin in the \(xy\) plane. At one instant it is at \(\vec r = 3\hat i\,\text{m}\), with \(\vec\omega = 2\hat k\,\text{rad/s}\) and \(\vec\alpha = 0.5\hat k\,\text{rad/s}^2\). Find \(\vec v\), \(\vec a_t\) and \(\vec a_c\).
\(\vec v = \vec\omega \times \vec r = 2\hat k \times 3\hat i = 6\,(\hat k \times \hat i) = 6\hat j\,\text{m/s}\). The particle at \(+x\) moves towards \(+y\): anticlockwise, as \(\vec\omega\) along \(+\hat k\) says.
\(\vec a_t = \vec\alpha \times \vec r = 0.5\hat k \times 3\hat i = 1.5\hat j\,\text{m/s}^2\), along the velocity, so the particle is speeding up.
\(\vec a_c = \vec\omega \times \vec v = 2\hat k \times 6\hat j = 12\,(\hat k \times \hat j) = -12\hat i\,\text{m/s}^2\), from the particle towards the origin. Check: \(\omega^2 r = 4 \times 3 = 12\).
A particle on a circle about the origin is at \(\vec r = 2\hat j\,\text{m}\), and \(\vec\omega = 3\hat k\,\text{rad/s}\). What is its velocity?
- Forgetting the centripetal part when the speed changes, and writing \(a = \alpha r\).\(\alpha r\) is only the tangential part. Add \(a_c = \omega^2 r\) at right angles: \(a = \sqrt{a_t^2 + a_c^2}\).
- Adding \(a_t\) and \(a_c\) as plain numbers.They are at right angles. Combine them by Pythagoras.
- Saying that a wheel starting from rest has no acceleration at the first instant.\(\omega = 0\) makes \(a_c = 0\), but \(a_t = \alpha r\) is not zero. The acceleration lies along the tangent.
- Writing \(\vec r \times \vec\omega\) for the velocity.The order is \(\vec\omega \times \vec r\). Swapping it reverses the direction.