The Workbook

\(v = \sqrt{\mu rg} = \sqrt{0.5 \times 80 \times 10} = \sqrt{400} = 20\) m/s.

On a level road only friction can point towards the centre. The largest friction is \(\mu N = \mu mg\), so \(\mu mg \ge \dfrac{mv^2}{r}\), giving \(v \le \sqrt{\mu rg}\). The mass appears on both sides and cancels: friction and the needed force both grow with \(m\).

The cyclist leans so that the resultant of \(N\) and friction passes through the centre of mass: \(\tan\theta = \dfrac{v^2}{rg} = \dfrac{100}{200} = 0.5\).

1. Friction. Vertical: \(N = mg\). Horizontal: \(f = \dfrac{mv^2}{r}\).

2. \(f = \dfrac{1000 \times 100}{50} = 2000\) N.

3. Available friction \(= 0.4 \times 10000 = 4000\) N, more than \(2000\) N. Safe.

1. Vertical: \(N\cos\theta = mg\). Horizontal: \(N\sin\theta = \dfrac{mv^2}{r}\).

2. Dividing the second by the first: \(\tan\theta = \dfrac{v^2}{rg}\).

3. \(N\) is perpendicular to the road. Because the outer edge is higher, \(N\) tilts towards the inside of the curve, and its horizontal part supplies the centripetal force.

\(v^2 = rg\tan\theta = 40 \times 10 \times 1 = 400\), so \(v = 20\) m/s.

\(\tan\theta = \dfrac{v^2}{rg} = \dfrac{100}{1000} = 0.1\). For a small angle the rise is \(h = d\tan\theta = 1 \times 0.1 = 0.1\) m.

The lift plays the part of \(N\): \(\tan\theta = \dfrac{v^2}{rg}\), so \(r = \dfrac{v^2}{g\tan 45^\circ} = \dfrac{10000}{10} = 1000\) m.

1. At \(v_0\) the horizontal part of \(N\) is exactly the centripetal force; friction is zero.

2. Faster: more centripetal force is needed, so friction acts down the slope to add to it.

3. Slower: less is needed, and the car tends to slide down the bank, so friction acts up the slope.

\(v_{\max}^2 = 1000 \times \dfrac{0.75 + 0.25}{1 - 0.25 \times 0.75} = 1000 \times \dfrac{1}{0.8125} \approx 1231\), so \(v_{\max} \approx 35\) m/s.

1. Below the design speed the horizontal part of \(N\) is more than the car needs. The car tends to slip down the slope towards the centre, and friction opposes this by acting up the slope.

2. Above the design speed the car needs more centripetal force than \(N\) gives. It tends to slide up and out, so friction acts down the slope and adds an inward force.

3. At the design speed \(v_0 = \sqrt{rg\tan\theta}\).

\(v_{\min}^2 = 1000 \times \dfrac{0.75 - 0.25}{1 + 0.25 \times 0.75} = 1000 \times \dfrac{0.5}{1.1875} \approx 421\), so \(v_{\min} \approx 20.5\) m/s.

A car at rest on the bank is a block on an incline. It stays if \(mg\sin\theta \le \mu mg\cos\theta\), that is, \(\mu \ge \tan\theta\). Then the formula for \(v_{\min}\) gives zero or a negative \(v_{\min}^2\): any speed down to zero is safe.

1. Vertical: \(T\cos\theta = mg\). Horizontal, with \(r = L\sin\theta\): \(T\sin\theta = m\omega^2r = m\omega^2L\sin\theta\).

2. From the second, \(T = m\omega^2L\). Put it into the first: \(m\omega^2L\cos\theta = mg\), so \(\omega^2 = \dfrac{g}{L\cos\theta}\).

3. \(T_{\text{period}} = \dfrac{2\pi}{\omega} = 2\pi\sqrt{\dfrac{L\cos\theta}{g}} = 2\pi\sqrt{\dfrac{h}{g}}\).

1. \(\omega^2 = \dfrac{10}{1 \times 0.5} = 20\), so \(\omega = 2\sqrt{5} \approx 4.5\) rad/s.

2. \(T = \dfrac{2\pi}{2\sqrt{5}} = \dfrac{\pi}{\sqrt{5}} \approx 1.4\) s.

3. \(T = \dfrac{mg}{\cos\theta} = \dfrac{5}{0.5} = 10\) N.

1. \(\sin\theta = \dfrac{r}{L} = \dfrac{0.3}{0.5} = 0.6\), so \(\theta = 37^\circ\).

2. \(T\cos 37^\circ = mg\): \(T = \dfrac{2}{0.8} = 2.5\) N.

3. \(T\sin 37^\circ = \dfrac{mv^2}{r}\): \(1.5 = \dfrac{0.2v^2}{0.3}\), so \(v^2 = 2.25\) and \(v = 1.5\) m/s.

For small \(\theta\), \(\cos\theta \approx 1\), so the conical pendulum's period \(2\pi\sqrt{L\cos\theta/g}\) becomes \(2\pi\sqrt{L/g}\), the same as the simple pendulum's.

\(h = 0.4\) m: \(T = 2\pi\sqrt{\dfrac{0.4}{10}} = 2\pi \times 0.2 = 0.4\pi \approx 1.3\) s. The string length does not matter.

\(\omega^2 = \dfrac{g}{L\cos\theta}\): a larger \(\omega\) needs a smaller \(\cos\theta\), so \(\theta\) increases. \(T = \dfrac{mg}{\cos\theta}\) increases too. At \(\theta = 90^\circ\), \(\cos\theta = 0\), which would need infinite \(\omega\) and infinite tension; the string can never be horizontal.

1. The normal force from the wall: \(N = m\omega^2r\).

2. Static friction along the wall balances the weight: \(f = mg\), and \(f \le \mu N\), so \(\mu m\omega^2r \ge mg\).

3. \(\omega^2 \ge \dfrac{g}{\mu r} = \dfrac{10}{0.5 \times 2} = 10\), so \(\omega_{\min} = \sqrt{10} \approx 3.2\) rad/s and \(v = \omega r = 2\sqrt{10} \approx 6.3\) m/s.

\(\mu mg \ge m\omega^2r\): \(\omega^2 \le \dfrac{\mu g}{r} = \dfrac{4}{0.1} = 40\), so \(\omega_{\max} = 2\sqrt{10} \approx 6.3\) rad/s. In rpm: \(\dfrac{6.3}{2\pi} \times 60 \approx 60\).

The required friction is \(\dfrac{mv^2}{r} \propto v^2\). At \(25\) m/s it is \(\left(\dfrac{25}{20}\right)^2 = \dfrac{25}{16} \approx 1.6\) times the friction available at the limit, so the car skids.

\(\tan\theta = \dfrac{v^2}{rg} = \dfrac{100}{100} = 1\), so \(\theta = 45^\circ\).

Needed centripetal force \(= \dfrac{1000 \times 100}{40} = 2500\) N. Along the slope, taking friction up the slope as positive: \(mg\sin\theta - f = \dfrac{mv^2}{r}\cos\theta\), so \(f = 10000 \times 0.71 - 2500 \times 0.71 \approx 5.3 \times 10^{3}\) N, up the slope. The car is well below the design speed and tends to slide down.