07Classical Mechanics

Circular motion

A body that rotates about a fixed axis keeps every point at a fixed distance, the radius, and its motion is described by an angle instead of a position. The equations are the same as in one dimension, with \(\theta\), \(\omega\) and \(\alpha\) in place of \(x\), \(v\) and \(a\): the angular invariant. The radius then translates the angular quantities into linear ones: the arc travelled, the speed of the point and its two accelerations, the tangential one that changes the speed and the centripetal one that changes the direction.

Laboratory

\(\theta(t)\) and \(\omega(t)\) play the roles of \(x(t)\) and \(v(t)\). The centripetal acceleration grows with \(\omega^2\).

The invariant

Equations (1) to (4) are the invariant of constant angular acceleration, each one without one of the quantities. The rest relate the angle with the number of turns, the angular quantities with the linear ones through the radius, and the angular velocity with the period and the frequency.

Step-by-step solution

Notable events

Turns, stops and the state at the end of the interval.

Theory

Angle, arc and radians

The angle swept by the radius is measured in radians: the length of the arc divided by the radius,

\[\theta = \frac{s}{r}.\]

A full turn is an arc \(2\pi r\), that is \(2\pi\) rad, so the number of turns is \(N = \theta/2\pi\). Because it is a ratio of lengths, the radian has no dimensions, which is why \(s = r\theta\) gives metres directly.

Angular velocity, period and frequency

The angular velocity \(\omega = d\theta/dt\) is the rate at which the angle grows, in rad/s, and the angular acceleration \(\alpha = d\omega/dt\) is the rate at which \(\omega\) changes. In a turn the body sweeps \(2\pi\) rad in one period \(T\), and the frequency is the number of turns per second, \(f = 1/T\):

\[\omega = \frac{2\pi}{T} = 2\pi f.\]

In revolutions per minute: \(1\ \text{rpm} = \frac{2\pi\ \text{rad}}{60\ \text{s}} = 0.1047\ \text{rad/s}\).

The angular invariant

From \(\alpha = d\omega/dt\), \(\alpha\,dt = d\omega\); integrating with \(\alpha\) constant, \(\int_{\omega_0}^{\omega} d\omega = \int_0^t \alpha\,dt\), so \(\omega = \omega_0 + \alpha t\). Since \(\omega = d\theta/dt\), integrating again,

\[\theta = \theta_0 + \omega_0 t + \tfrac12\alpha t^2,\]

and with the average angular velocity \(\bar\omega = \tfrac12(\omega_0 + \omega)\) one gets \(\omega^2 = \omega_0^2 + 2\alpha\theta\). These are the equations of one dimension with \(x \to \theta\), \(v \to \omega\), \(a \to \alpha\), as the course notes point out.

From angular to linear

With the radius constant, differentiating \(s = r\theta\) gives the speed of the point along the circle and differentiating again its tangential acceleration:

\[v = \frac{ds}{dt} = r\,\frac{d\theta}{dt} = r\omega,\qquad a_t = \frac{dv}{dt} = r\alpha.\]

All the points of a rigid body share \(\theta\), \(\omega\) and \(\alpha\), but the farther from the axis, the larger their \(s\), \(v\) and \(a_t\).

Uniform circular motion and the centripetal acceleration

In uniform circular motion the speed is constant but the direction of the velocity changes all the time, and that change is an acceleration. Comparing the triangle of the two velocities with the triangle of the two radii (Resnick 4-4) and taking the limit \(\Delta t \to 0\):

\[a_c = \frac{v^2}{r} = \omega^2 r,\]

always pointing to the centre: it is called centripetal. As the body goes round, \(\vec a_c\) turns with it, so it is not a constant vector and the invariant of constant acceleration cannot be applied to the motion in \(x\) and \(y\).

Tangential and total acceleration

If the speed also changes, the acceleration has two perpendicular parts: the tangential one, \(a_t = dv/dt = r\alpha\), along the velocity, which changes the speed, and the centripetal one, \(a_c = v^2/r\), towards the centre, which changes the direction (Resnick 4-5). The magnitude of the total acceleration is

\[a = \sqrt{a_t^2 + a_c^2}.\]

A car that brakes in a curve feels both.

Problem bank

Problems adapted from Serway (chapters 2, 4, 5 and 6), Resnick and the course notes. Load any of them into the laboratory: the data fill the panel and the solver does the rest, so you can then change any value and see how the answer reacts.

  1. An electric motor starts up

    Class notes (McKelvey and Grotch)

    An electric motor starts from rest and reaches its normal speed of \(2000\ \text{rpm}\) in \(2\ \text{s}\), after which it runs at constant speed. Calculate (a) the angular acceleration, (b) the angular velocity 1 s after starting, (c) the number of turns of the shaft during the 2 s and (d) the angular velocity after 10 turns. (Query t = 1 s and N = 10 for b and d.)

  2. 30 turns per minute

    Course workshop, problem 5

    A point describes a circular path of radius \(1\ \text{m}\) uniformly, making 30 turns every minute. Calculate the period, the frequency, the angular velocity, the tangential velocity and the centripetal acceleration.

  3. A discus thrower

    Serway 4.25

    An athlete rotates a \(1.00\ \text{kg}\) discus along a circular path of radius \(1.06\ \text{m}\). The maximum speed of the discus is \(20.0\ \text{m/s}\). Determine the magnitude of its maximum radial acceleration.

  4. David’s sling

    Serway 4.27

    Young David, who slew Goliath, found he could revolve a sling of length \(0.600\ \text{m}\) at a rate of \(8.00\ \text{rev/s}\). What was the speed of the stone and its centripetal acceleration?

  5. The acceleration of the Moon

    Resnick, sample problem 4-5

    The Moon revolves around the Earth in \(27.3\) days. Assuming a circular orbit of radius \(3.82\times10^8\ \text{m}\), what is the magnitude of its acceleration towards the Earth?

  6. A point on the equator

    Serway 4.24

    Calculate the radial acceleration of a point on the surface of the Earth at the equator due to the rotation of the Earth about its axis (radius \(6.37\times10^6\ \text{m}\), one turn every 24 h).

  7. A Ferris wheel

    Serway 6.52

    A Ferris wheel turns four times every minute and carries each car around a circle \(18.0\ \text{m}\) in diameter. What is the centripetal acceleration of a passenger?

  8. A train braking in a curve

    Serway 4.29

    A train slows down as it rounds a sharp horizontal curve, going from \(90.0\ \text{km/h}\) to \(50.0\ \text{km/h}\) in the \(15.0\ \text{s}\) it takes to round the curve. The radius of the curve is \(150\ \text{m}\). Compute the acceleration at the moment its speed reaches \(50.0\ \text{km/h}\), assuming it keeps slowing down at the same rate.

  9. A fan that is switched off

    Own example

    A fan turning at \(1200\ \text{rpm}\) is switched off and stops uniformly in \(8.0\ \text{s}\). Find its angular acceleration and the number of turns it makes before stopping.

Theory after R. Resnick, D. Halliday and K. Krane, Física, vol. 1, chapter 4 (sections 4-4 and 4-5); problems from R. Serway and J. Jewett, Física para ciencias e ingeniería, vol. 1, chapters 4 and 6, and from the course notes on motion in the plane.