The Workbook

1. \(v = \omega r = 3 \times 2 = 6\) m/s.

2. \(T = \dfrac{2\pi}{\omega} = \dfrac{2\pi}{3} \approx 2.1\) s.

3. \(f = \dfrac{1}{T} = \dfrac{3}{2\pi} \approx 0.48\) Hz.

\(120\) rpm \(= 2\) rev/s, so \(\omega = 2 \times 2\pi = 4\pi \approx 12.6\) rad/s. \(v = \omega r = 4\pi \times 0.5 = 2\pi \approx 6.3\) m/s.

\(\theta = \omega t = 6\) rad. Revolutions \(= \dfrac{6}{2\pi} = \dfrac{3}{\pi} \approx 0.95\), just under one full turn.

Minute hand: one turn in \(3600\) s, \(\omega = \dfrac{2\pi}{3600} = \dfrac{\pi}{1800}\) rad/s. Hour hand: one turn in \(43200\) s, \(\omega = \dfrac{\pi}{21600}\) rad/s. Ratio \(= 12\).

1. Arc \(= r\theta = 3 \times \dfrac{\pi}{3} = \pi \approx 3.1\) m.

2. \(OPQ\) is equilateral (two radii and \(60^\circ\)), so the chord is \(3\) m.

3. Average speed \(= \dfrac{\pi}{\pi} = 1\) m/s. \(\omega = \dfrac{\theta}{t} = \dfrac{\pi/3}{\pi} = \dfrac{1}{3}\) rad/s.

\(a_c = \dfrac{v^2}{r} = \dfrac{36}{2} = 18\) m/s\(^2\). \(\omega = \dfrac{v}{r} = 3\) rad/s, and \(\omega^2r = 9 \times 2 = 18\) m/s\(^2\).

1. In uniform circular motion the acceleration always points to the centre: downward at \(P\), to the right at \(Q\), upward at \(R\).

2. \(a = \dfrac{v^2}{r} = \dfrac{16}{2} = 8\) m/s\(^2\).

3. No. Its magnitude is fixed but its direction keeps turning, so the vector changes.

\(a = \dfrac{100}{50} = 2\) m/s\(^2\). \(F = ma = 2000\) N, supplied by the friction between the tyres and the road.

\(a = \dfrac{4 \times 9.87 \times 3.84 \times 10^{8}}{(2.36 \times 10^{6})^2} = \dfrac{1.516 \times 10^{10}}{5.57 \times 10^{12}} \approx 2.7 \times 10^{-3}\) m/s\(^2\), about \(\dfrac{g}{3600}\).

\(\omega = 2 \times 2\pi = 4\pi\) rad/s. \(v = \omega r = 4\pi\) m/s. \(a = \omega^2r = 16\pi^2 \approx 158\) m/s\(^2\).

1. \(a_t = \dfrac{dv}{dt} = 2\) m/s\(^2\).

2. At \(t = 1\) s, \(v = 2\) m/s: \(a_c = \dfrac{4}{2} = 2\) m/s\(^2\).

3. \(a = \sqrt{2^2 + 2^2} = 2\sqrt{2} \approx 2.8\) m/s\(^2\), at \(45^\circ\) to the radius (and to the tangent).

1. Speeding up: \(a_t\) points along the velocity (anticlockwise tangent), so the speed grows.

2. \(a = \sqrt{9 + 16} = 5\) m/s\(^2\). \(\tan\phi = \dfrac{a_c}{a_t} = \dfrac{4}{3}\), so \(\phi = 53^\circ\) from the velocity, leaning towards the centre.

3. \(a_t\) changes the speed; \(a_c\) changes only the direction.

\(v = a_tt = 10\) m/s. \(a_c = \dfrac{100}{100} = 1\) m/s\(^2\). \(a = \sqrt{2^2 + 1^2} = \sqrt{5} \approx 2.2\) m/s\(^2\).

1. \(\alpha = \dfrac{8 - 2}{3} = 2\) rad/s\(^2\).

2. \(\theta = \omega_0t + \dfrac{1}{2}\alpha t^2 = 6 + 9 = 15\) rad.

3. \(a_t = \alpha r = 2 \times 0.5 = 1\) m/s\(^2\).

Perpendicular when \(a_t = 0\), that is, when the speed is constant: only the centripetal part remains. Along the velocity when \(a_c = 0\), that is, at the instant the particle starts from rest, since \(v = 0\) gives \(\dfrac{v^2}{r} = 0\).

At the top \(v = u\cos 60^\circ = 10\) m/s, horizontal, and the whole of \(g\) is perpendicular to it. \(R = \dfrac{100}{10} = 10\) m.

1. \(a_\perp = g\cos 60^\circ = 5\) m/s\(^2\); the other part, \(g\sin 60^\circ\), only slows the ball.

2. \(R = \dfrac{v^2}{a_\perp} = \dfrac{400}{5} = 80\) m.

3. Near launch the speed is greater and the perpendicular acceleration smaller, so the path bends less: a larger \(R\).

\(a_t = 2\cos 30^\circ = \sqrt{3} \approx 1.7\) m/s\(^2\); \(a_\perp = 2\sin 30^\circ = 1\) m/s\(^2\). \(R = \dfrac{25}{1} = 25\) m.

\(y' = 2x = 0\) at the origin and \(y'' = 2\). \(R = \dfrac{(1 + 0)^{3/2}}{2} = 0.5\).

1. \(a = \dfrac{100}{20} = 5\) m/s\(^2\), towards the centre of the hill, that is, downward.

2. \(mg - N = ma\): \(N = 60(10 - 5) = 300\) N, half the weight.

3. \(N = 0\) when \(\dfrac{v^2}{R} = g\): \(v = \sqrt{gR} = \sqrt{200} = 10\sqrt{2} \approx 14\) m/s.

\(\omega = \dfrac{v}{r} = \dfrac{2}{0.1} = 20\) rad/s. All points share \(\omega\), so at \(0.2\) m: \(v = 20 \times 0.2 = 4\) m/s.

\(\omega = \dfrac{v}{r} = \dfrac{10}{0.5} = 20\) rad/s. Revolutions per second \(= \dfrac{\omega}{2\pi} = \dfrac{10}{\pi} \approx 3.2\).

Speed is constant by definition, and so is \(|a| = \dfrac{v^2}{r}\). The velocity is not constant, because its direction turns. The acceleration is not constant either, because it always points to the centre and so keeps turning too.

The velocity vectors at the two instants have the same length \(v\) and differ in direction by \(\Delta\theta\). Placed tail to tail they form an isosceles triangle whose third side is \(\Delta\vec v\). For a small angle its length is the arc, \(v\,\Delta\theta\).

\(a = \dfrac{|\Delta\vec v|}{\Delta t} = v\dfrac{\Delta\theta}{\Delta t} = v\omega = v \times \dfrac{v}{r} = \dfrac{v^2}{r}\). As \(\Delta\theta \to 0\), \(\Delta\vec v\) becomes perpendicular to \(\vec v\), pointing to the centre.