The Workbook

Rolling without slipping: the contact point is at rest, which needs \(v_{\text{cm}} = R\omega\). \(\omega = \dfrac{4}{0.5} = 8\) rad/s.

1. Each point has \(v_{\text{cm}}\) plus \(\omega r\) from the spin. \(T\): \(4 + 4 = 8\) m/s forward. \(C\): \(4\) m/s. \(P\): \(4 - 4 = 0\).

2. At \(F\) the spin velocity \(\omega R = 4\) m/s points straight down. Adding \(4\) m/s forward: speed \(4\sqrt2 \approx 5.7\) m/s at \(45^\circ\) below the horizontal.

The contact point has the forward velocity \(v_{\text{cm}}\) of the centre and the backward velocity \(\omega R\) from the spin. In pure rolling these are equal, so they cancel and the point is instantaneously at rest. A moment later a different point of the rim is in contact.

Each revolution moves the centre by one circumference: \(10 \times 2\pi \times 0.5 = 10\pi \approx 31.4\) m.

\(\omega R = 4 \times 0.5 = 2\) m/s, less than \(v_{\text{cm}} = 3\) m/s, so it is not pure rolling. The contact point moves forward at \(3 - 2 = 1\) m/s. Kinetic friction opposes this slip, so it acts backward on the ball.

\(\omega = \dfrac{6}{0.3} = 20\) rad/s. Top: \(2v = 12\) m/s. Relative to the centre every rim point moves in a circle at \(\omega R\), so its acceleration is \(\omega^2R = 400 \times 0.3 = 120\) m/s\(^2\) towards the centre.

The instantaneous axis is the line perpendicular to the plane of motion through the point of zero velocity. That point may lie on the body or on its extension. The body's motion is then pure rotation about that axis. For rolling without slipping the contact point is at rest, so it is the instantaneous axis.

1. \(I\) lies on the perpendicular to each velocity through its point. \(A\) moves vertically, so \(I\) is on the horizontal through \(A\); \(B\) moves horizontally, so \(I\) is on the vertical through \(B\). \(A\) is \(1 \times 0.6 = 0.6\) m up the wall and \(B\) is \(1 \times 0.8 = 0.8\) m along the floor. So \(I\) is \(0.8\) m from the wall and \(0.6\) m above the floor.

2. \(IA = 0.8\) m. \(\omega = \dfrac{v_A}{IA} = \dfrac{2}{0.8} = 2.5\) rad/s.

3. \(IB = 0.6\) m. \(v_B = \omega \times IB = 2.5 \times 0.6 = 1.5\) m/s, along the floor away from the wall.

\(I\), \(A\), \(B\) and the corner form a rectangle, and the midpoint \(M\) of the rod is its centre. So \(M\) is \(\dfrac{L}{2} = 0.5\) m from \(I\). \(v_M = 2.5 \times 0.5 = 1.25\) m/s, perpendicular to \(IM\).

About the contact point \(P\): \(K = \dfrac{1}{2}I_P\omega^2\) with \(I_P = I_{\text{cm}} + MR^2\) (parallel axis). So \(K = \dfrac{1}{2}I_{\text{cm}}\omega^2 + \dfrac{1}{2}MR^2\omega^2 = \dfrac{1}{2}I_{\text{cm}}\omega^2 + \dfrac{1}{2}Mv^2\), using \(v = R\omega\).

\(F\) is at distance \(\sqrt{R^2 + R^2} = R\sqrt2\) from the contact point \(P\). Speed \(= \omega \times R\sqrt2 = 8 \times 0.5\sqrt2 = 4\sqrt2\) m/s, perpendicular to \(PF\), that is at \(45^\circ\) below the horizontal. This matches Q2.

Translational: \(\dfrac{1}{2}(2)(16) = 16\) J. \(I = \dfrac{1}{2}(2)(0.25) = 0.25\), \(\omega = 8\): rotational \(\dfrac{1}{2}(0.25)(64) = 8\) J. Total \(24\) J; rotational fraction \(\dfrac{8}{24} = \dfrac{1}{3}\).

\(K_{\text{rot}} = \dfrac{1}{2}Mk^2\omega^2 = \dfrac{k^2}{R^2}\cdot\dfrac{1}{2}Mv^2\), so the fraction is \(\dfrac{k^2/R^2}{1 + k^2/R^2}\). Ring (\(k^2/R^2 = 1\)): \(\dfrac{1}{2}\). Disc (\(\dfrac{1}{2}\)): \(\dfrac{1}{3}\). Sphere (\(\dfrac{2}{5}\)): \(\dfrac{2}{7}\).

1. Friction does no work in pure rolling: \(mgh = \dfrac{1}{2}mv^2\left(1 + \dfrac{2}{5}\right)\), so \(v^2 = \dfrac{2gh}{1.4} = \dfrac{28}{1.4} = 20\) and \(v = 2\sqrt5 \approx 4.5\) m/s.

2. The sphere. Its \(k^2/R^2 = \dfrac{2}{5}\) is the smallest, so the least share of the energy goes into spin and its acceleration is the largest. The ring, with \(k^2/R^2 = 1\), is last.

\(g\sin 37^\circ = 6\) m/s\(^2\). Sphere: \(\dfrac{6}{1.4} = \dfrac{30}{7} \approx 4.3\) m/s\(^2\). Disc: \(\dfrac{6}{1.5} = 4\) m/s\(^2\). Ring: \(\dfrac{6}{2} = 3\) m/s\(^2\).

Along the incline: \(mg\sin\theta - f = ma\) with \(a = 4\) m/s\(^2\): \(f = 3 \times 6 - 3 \times 4 = 6\) N. Normal force \(= 3 \times 10 \times 0.8 = 24\) N. \(\mu_{\min} = \dfrac{6}{24} = 0.25\). In general \(f = \dfrac{mg\sin\theta}{1 + R^2/k^2}\).

1. Friction \(\mu mg\) backward: \(a = -\mu g = -2\) m/s\(^2\). Torque about the centre \(\mu mgR = \dfrac{2}{5}mR^2\alpha\), so \(\alpha = \dfrac{5\mu g}{2R}\), spinning it up in the rolling sense.

2. \(v = 7 - 2t\) and \(\omega R = \alpha Rt = \dfrac{5\mu g}{2}t = 5t\). Pure rolling when \(v = \omega R\): \(7 - 2t = 5t\), \(t = 1\) s. Then \(v = 5\) m/s, which is \(\dfrac{5}{7}v_0\).

3. \(s = 7 \times 1 - \dfrac{1}{2}(2)(1) = 6\) m.

For a ring \(I = mR^2\): \(a = -\mu g\) and \(\alpha = \dfrac{\mu mgR}{mR^2} = \dfrac{\mu g}{R}\). So \(v = v_0 - \mu gt\) and \(\omega R = \mu gt\). Setting \(v = \omega R\): \(v_0 - \mu gt = \mu gt\), so \(\mu gt = \dfrac{v_0}{2}\) and \(v = v_0 - \dfrac{v_0}{2} = \dfrac{v_0}{2}\).

Kinetic friction acts only while the surfaces slip. Once the contact point is at rest there is no slipping and no kinetic friction. On a level floor nothing tries to change \(v\) or \(\omega\), so no static friction is needed either. The ball rolls on at constant speed.

1. The bottom moves backward, so friction \(\mu mg\) acts forward: \(v = \mu gt\). Its torque slows the spin: \(\alpha = \dfrac{\mu mgR}{\frac{1}{2}mR^2} = \dfrac{2\mu g}{R}\), so \(\omega = \omega_0 - \dfrac{2\mu g}{R}t\).

2. Pure rolling when \(v = \omega R\): \(\mu gt = \omega_0R - 2\mu gt\), so \(\mu gt = \dfrac{\omega_0R}{3}\). \(v = \dfrac{12 \times 0.5}{3} = 2\) m/s, and \(t = \dfrac{2}{\mu g} = \dfrac{2}{2} = 1\) s.

Friction and the normal force act at the contact point, so they have no torque about it. Gravity acts at the centre, directly above the contact point, so its torque about that point is zero too. With no torque about the contact point, the angular momentum about it stays constant.

\(\omega = \dfrac{15}{0.3} = 50\) rad/s. In rpm: \(\dfrac{50 \times 60}{2\pi} = \dfrac{1500}{\pi} \approx 477\).

For a solid cylinder \(k^2/R^2 = \dfrac{1}{2}\): \(K = \dfrac{1}{2}mv^2\left(1 + \dfrac{1}{2}\right) = \dfrac{3}{4}(2)(9) = 13.5\) J.

About the contact point the angular momentum is constant (Q21). At first: \(L = mv_0R\) (the ball has no spin). When rolling at speed \(v\): \(L = mvR + I\omega = mvR + \dfrac{2}{5}mR^2\cdot\dfrac{v}{R} = \dfrac{7}{5}mvR\). Equating: \(v = \dfrac{5}{7}v_0\), the same as Q17.