The Workbook

\(\vec L = \vec r \times \vec p = \vec r \times m\vec v = 3\hat i \times (2 \times 4)\hat j = 24\hat k\) kg m\(^2\)/s.

1. \(L = mv \times d = 1 \times 5 \times 2 = 10\) kg m\(^2\)/s, where \(d\) is the perpendicular distance from \(O\) to the line.

2. No. \(L = mvd\) and \(m\), \(v\) and \(d\) all stay the same as the particle moves along the line. No force acts, so no torque acts and \(L\) is constant.

(i) \(L = I\omega = 0.5 \times 20 = 10\) kg m\(^2\)/s. (ii) \(I = MR^2 = 2 \times 0.25 = 0.5\), so \(L = 0.5 \times 4 = 2\) kg m\(^2\)/s.

Each particle moves in a circle of radius \(r_i\) about the axis with speed \(v_i = r_i\omega\), and \(\vec r_i \perp \vec v_i\). So \(L = \sum m_ir_iv_i = \sum m_ir_i^2\omega = \omega\sum m_ir_i^2 = I\omega\).

\(L = mvr = 0.5 \times 3 \times 2 = 3\) kg m\(^2\)/s. \(\vec L\) is perpendicular to the plane of the circle, in the sense given by the right-hand rule.

1. \(I_1\omega_1 = I_2\omega_2\): \(4 \times 2 = 1 \times \omega_2\), so \(\omega_2 = 8\) rad/s.

2. Before: \(\dfrac{1}{2}(4)(4) = 8\) J. After: \(\dfrac{1}{2}(1)(64) = 32\) J.

3. The skater's muscles do work pulling the arms inward against the outward pull needed to keep them moving in smaller circles. That work becomes the extra kinetic energy.

\(I_2 = 2 + 0.5 \times 4 = 4\) kg m\(^2\). \(2 \times 10 = 4\omega\), so \(\omega = 5\) rad/s. Energy: \(\dfrac{1}{2}(2)(100) = 100\) J before, \(\dfrac{1}{2}(4)(25) = 50\) J after: \(50\) J lost as heat when the clay is dragged up to speed.

At the rim: \(I = 300 + 60 \times 4 = 540\) kg m\(^2\). At the centre the man adds nothing: \(I = 300\). \(540 \times 1 = 300\,\omega\), so \(\omega = 1.8\) rad/s.

\(I \propto R^2\), so \(I\) falls to a quarter and \(\omega\) rises four times. The period falls to a quarter: \(5\) days.

\(2 \times 6 = (2 + 4)\omega\), so \(\omega = 2\) rad/s. Energy: \(\dfrac{1}{2}(2)(36) = 36\) J before, \(\dfrac{1}{2}(6)(4) = 12\) J after: \(24\) J lost.

The friction between the discs is an internal force. The torque it exerts on \(A\) is equal and opposite to that on \(B\), so the total angular momentum cannot change. But friction does negative net work as the discs slip on each other, turning kinetic energy into heat.

The angular impulse of the net torque equals the change of angular momentum: \(\int\tau\,dt = L_2 - L_1\). Here \(5 \times 4 = 2\omega\), so \(\omega = 10\) rad/s.

1. Angular impulse \(= J \times L = 6 \times 1 = 6\) N m s.

2. \(I = \dfrac{ML^2}{3} = \dfrac{3 \times 1}{3} = 1\) kg m\(^2\). \(I\omega = 6\), so \(\omega = 6\) rad/s.

Angular impulse \(= 4 \times 0.5 = 2\) N m s. \(\omega = \dfrac{2}{0.5} = 4\) rad/s.

The hinge force acts at the hinge, so its lever arm about the hinge is zero. Its torque, and therefore its angular impulse about the hinge, is zero however large it is. That is why torques are taken about the hinge.

1. \(L = mvL_{\text{rod}} = 1 \times 4 \times 1 = 4\) kg m\(^2\)/s.

2. \(I = \dfrac{ML^2}{3} + mL^2 = 1 + 1 = 2\) kg m\(^2\). \(4 = 2\omega\), so \(\omega = 2\) rad/s.

3. Kinetic energy after impact: \(\dfrac{1}{2}(2)(4) = 4\) J. Rise in potential energy at angle \(\theta\): the rod's centre rises \(\dfrac{L}{2}(1 - \cos\theta)\) and the putty rises \(L(1 - \cos\theta)\). \(30 \times 0.5(1 - \cos\theta) + 10 \times 1(1 - \cos\theta) = 25(1 - \cos\theta) = 4\), so \(\cos\theta = 0.84\).

During the impact the hinge exerts a large impulsive force on the rod, an external force, so the linear momentum of rod plus putty changes. That force acts at the hinge and has no torque about it, and gravity's impulse is negligible in the short time. So angular momentum about the hinge is conserved across the impact.

1. No external horizontal force: \(1 \times 3 = 3v_{\text{cm}}\), so \(v_{\text{cm}} = 1\) m/s.

2. Take \(C\) as origin, the struck end at \(0.5\) m: \(x_{\text{cm}} = \dfrac{2 \times 0 + 1 \times 0.5}{3} = \dfrac{1}{6}\) m from \(C\) towards the particle.

3. The particle is \(\dfrac{1}{3}\) m from the centre of mass: \(L = 1 \times 3 \times \dfrac{1}{3} = 1\) kg m\(^2\)/s. \(I\) about the centre of mass: rod \(\dfrac{2}{12} + 2\left(\dfrac{1}{6}\right)^2 = \dfrac{2}{9}\), particle \(1 \times \dfrac{1}{9}\); total \(\dfrac{1}{3}\). \(\omega = \dfrac{1}{1/3} = 3\) rad/s.

The blow now passes through the centre of mass of rod plus particle, which is at \(C\). So there is no angular momentum about it. The system slides at \(v_{\text{cm}} = 1\) m/s without rotating. The linear motion is the same as in Q18; only the rotation is missing.

Linear: \(v_{\text{cm}} = \dfrac{4}{2} = 2\) m/s. Angular, about the centre: \(J\dfrac{L}{2} = I\omega\), \(4 \times 0.5 = \dfrac{2 \times 1}{12}\omega\), so \(\omega = 12\) rad/s. Struck end: \(2 + 12 \times 0.5 = 8\) m/s. Far end: \(2 - 6 = -4\) m/s, backward.

Velocity of a point at distance \(x\) from the centre, on the far side: \(2 - 12x\). It vanishes at \(x = \dfrac{1}{6}\) m from the centre, away from the blow. That point is the instantaneous centre of rotation.

Gravity is central, so \(L = mvr\) is conserved: \(v_{\text{far}}r_{\text{far}} = v_{\text{near}}r_{\text{near}}\). \(v_{\text{far}} = \dfrac{30}{3} = 10\) km/s.

Angular momentum of the child about the axis: \(mvR = 30 \times 3 \times 2 = 180\) kg m\(^2\)/s. After: \(I = 60 + 30 \times 4 = 180\) kg m\(^2\). \(\omega = \dfrac{180}{180} = 1\) rad/s.

\(\vec L = \vec r \times m\vec v\). \(\dfrac{d\vec L}{dt} = \dfrac{d\vec r}{dt} \times m\vec v + \vec r \times m\dfrac{d\vec v}{dt} = \vec v \times m\vec v + \vec r \times 0 = 0\), since \(\vec v \times \vec v = 0\) and \(\vec v\) is constant. So \(\vec L\) is constant. Equivalently, \(L = mvd\) with \(d\) the fixed perpendicular distance.