Theory, worked examples and 15 previous-year questions with full solutions — IIT JAM, GATE, JEST and CSIR NET-JRF, 2020 onwards.
Theory
8.1 Angular momentum
(a) Angular momentum of a particle
Let a particle of mass \(m\) have position vector \(\mathbf r\) relative to a chosen origin \(O\) and linear momentum \(\mathbf p=m\mathbf v\). Its angular momentum about \(O\) is defined as
\[\boxed{\mathbf L=\mathbf r\times\mathbf p}.\]
Unlike the linear momentum, \(\mathbf L\) depends on the choice of origin: the same particle has different angular momenta about different points. It is perpendicular to the plane containing \(\mathbf r\) and \(\mathbf p\), and its magnitude \(rp\sin\phi\), where \(\phi\) is the angle between \(\mathbf r\) and \(\mathbf p\), equals the momentum multiplied by the perpendicular distance of the line of motion from \(O\). A particle moving along a straight line through \(O\) therefore has no angular momentum about \(O\), however fast it moves.
For motion in the \(xy\)-plane the only non-zero component is \(L_z\). Consider first a particle moving parallel to the \(x\)-axis at height \(y\) with momentum \(p_x\) (Figure 8.1a). Its line of motion is a distance \(y\) from \(O\), and it turns clockwise about \(O\), so \(L_z=-yp_x\). A particle moving parallel to the \(y\)-axis at distance \(x\) with momentum \(p_y\) turns anticlockwise and has \(L_z=xp_y\) (Figure 8.1b). In the general case both motions are present at once (Figure 8.1c), and the two contributions add:
\[L_z=xp_y-yp_x.\]
Figure 8.1 The \(z\)-component of the angular momentum of a particle moving in the \(xy\)-plane: (a) motion along \(x\), (b) motion along \(y\), (c) the general case.
The same result follows directly from the cross product with \(\mathbf r=(x,y,0)\) and \(\mathbf p=(p_x,p_y,0)\):
In three dimensions the full set of components is \(L_x=yp_z-zp_y\), \(L_y=zp_x-xp_z\) and \(L_z=xp_y-yp_x\).
(b) Angular momentum of a system of particles
For a system of particles with masses \(m_i\), positions \(\mathbf r_i\) and velocities \(\mathbf v_i\), the total angular momentum about \(O\) is \(\mathbf L=\sum_i\mathbf r_i\times m_i\mathbf v_i\). Let \(\mathbf R\) be the position and \(\mathbf V\) the velocity of the centre of mass, with total mass \(M=\sum_im_i\), and write \(\mathbf r_i=\mathbf R+\mathbf r_i'\) and \(\mathbf v_i=\mathbf V+\mathbf v_i'\), where the primed quantities are measured relative to the centre of mass. Then
The angular momentum about any point is thus the angular momentum of the total mass concentrated at the centre of mass (the orbital part), plus the angular momentum of the motion relative to the centre of mass (the spin part). The second term does not depend on the choice of origin.
For a rigid body rotating with angular velocity \(\boldsymbol\omega\) about an axis through its centre of mass, every particle has \(\mathbf v_i'=\boldsymbol\omega\times\mathbf r_i'\). When the body moves in a plane and \(\boldsymbol\omega\) is perpendicular to that plane along an axis of symmetry (a principal axis, Chapter 7), the spin part is simply \(\mathbf L_{\text{cm}}=I_{\text{cm}}\boldsymbol\omega\), where \(I_{\text{cm}}\) is the moment of inertia about the axis through the centre of mass. Hence, for a rigid body in plane motion,
where \(\mathbf r\) is the position vector of the centre of mass from \(O\) and \(\mathbf v_{\text{cm}}\) is its velocity (Figure 8.2). The moments of inertia of the common bodies are calculated in Chapter 7; we use in particular \(\frac25MR^2\) for a solid sphere, \(\frac12MR^2\) for a solid cylinder or disc, \(MR^2\) for a ring, and \(\frac1{12}Ml^2\) for a thin rod about its centre.
Figure 8.2 A body of radius \(R\) whose centre of mass, at position \(\mathbf r\) from \(O\), moves with velocity \(\mathbf v_{\text{cm}}\) while the body turns with angular velocity \(\omega\).
When \(\boldsymbol\omega\) is not along a principal axis, \(\mathbf L_{\text{cm}}\) is not parallel to \(\boldsymbol\omega\); Section 8.5(c) gives the essentials and Chapter 13 the full treatment.
8.2 Translation, rotation and rolling
(a) Combination of translation and rotation
Consider a sphere (or a cylinder, or a wheel) of radius \(R\) moving on a horizontal surface. In pure translation every point of the body has the same velocity, equal to the velocity \(v_{\text{cm}}\) of the centre of mass. In pure rotation about the centre of mass with angular velocity \(\boldsymbol\omega\), the centre is at rest and a point at position \(\mathbf r'\) relative to it has velocity \(\boldsymbol\omega\times\mathbf r'\); the highest and lowest points both move with speed \(\omega R\), but in opposite directions. The general motion of a rigid body in a plane is a superposition of the two (Figure 8.3), so that the velocity of any point of the body is
For a body moving to the right and turning clockwise, the highest point therefore moves with speed \(v_{\text{cm}}+\omega R\) and the lowest point, the point of contact \(P\), with velocity \(v_{\text{cm}}-\omega R\) to the right. If \(\omega R>v_{\text{cm}}\) this velocity is to the left.
Figure 8.3 Rolling as the superposition of a pure translation and a pure rotation about the centre of mass; on the right, rolling without slipping.
(b) Rolling without slipping
If the surface is rough enough, the point of contact does not slide on it. This requires its velocity to vanish, \(v_{\text{cm}}-\omega R=0\), and so
the second relation following by differentiation, with \(\alpha=\dot\omega\) the angular acceleration. These are the conditions for rolling without slipping. The point of contact is then instantaneously at rest, the centre moves with \(v_{\text{cm}}\) and the highest point with \(2v_{\text{cm}}\). At each instant the body is turning about the line of contact, which is the instantaneous axis of rotation: every point moves with a speed proportional to its distance from the point of contact. Because the point of contact is at rest, the friction that acts there is static friction and does no work, so the mechanical energy of a body rolling without slipping is conserved when the other forces are conservative.
When the conditions do not hold, the body slips. Kinetic friction then acts at the point of contact, opposite to the velocity of slipping \(v_{\text{cm}}-\omega R\), and changes \(v_{\text{cm}}\) and \(\omega\) until the rolling condition is reached (Example 8.11).
8.3 Energy of a rigid body in plane motion
(a) Kinetic energy
The kinetic energy of a system of particles can be divided in the same way as the angular momentum. With \(\mathbf v_i=\mathbf V+\mathbf v_i'\),
and the middle term vanishes. For a rigid body the motion relative to the centre of mass is a rotation, \(v_i'=\omega r_{i\perp}'\), where \(r_{i\perp}'\) is the distance from the axis through the centre of mass, so the last term is \(\frac12I_{\text{cm}}\omega^2\). Hence
The first term is the translational kinetic energy of the centre of mass and the second is the rotational kinetic energy about the centre of mass. For a body rolling without slipping, \(v_{\text{cm}}=\omega R\) and
where \(I_P=I_{\text{cm}}+MR^2\) is the moment of inertia about the line of contact, by the parallel-axis theorem (Chapter 7). This is the kinetic energy of pure rotation about the instantaneous axis.
(b) The compound pendulum
A rigid body free to oscillate in a vertical plane about a fixed horizontal axis is called a compound pendulum or physical pendulum. Let the axis pass through the point \(O\), let \(C\) be the centre of mass, at distance \(OC=l\) from the axis, and let \(\theta\) be the angle between \(OC\) and the downward vertical (Figure 8.4). The mass is \(M\) and the moment of inertia about the parallel axis through \(C\) is \(I_{\text{cm}}\).
Figure 8.4 A compound pendulum: a rigid body turning about a horizontal axis through \(O\), with its centre of mass \(C\) at distance \(l\) from the axis.
Take axes with \(x\) vertically downwards and \(y\) horizontal, so that the centre of mass is at \(x=l\cos\theta\), \(y=l\sin\theta\). Then \(\dot x=-l\sin\theta\,\dot\theta\) and \(\dot y=l\cos\theta\,\dot\theta\), so \(v_{\text{cm}}^2=\dot x^2+\dot y^2=l^2\dot\theta^2\), and the body turns with angular velocity \(\omega=\dot\theta\). The kinetic energy is
where \(I_O=I_{\text{cm}}+Ml^2\) is the moment of inertia about the axis of rotation. This is just the kinetic energy of rotation about a fixed axis. Measuring the potential energy from the horizontal plane through \(O\), \(V=-Mgl\cos\theta\), and the total energy is
where \(k\) is the radius of gyration about the centre of mass, \(I_{\text{cm}}=Mk^2\). The compound pendulum therefore has the same period as a simple pendulum of length \(L_e=(k^2+l^2)/l\), the equivalent length. As a function of \(l\), the quantity \(k^2/l+l\) is very large both for small \(l\) and for large \(l\); it has its minimum value \(2k\) at \(l=k\). The period is therefore least,
\[T_{\min}=2\pi\sqrt{\frac{2k}{g}},\]
when the axis is at a distance equal to the radius of gyration from the centre of mass.
(c) Equations of motion from the energy
The method used for the compound pendulum is general. For a system with a single degree of freedom \(q\) and only conservative forces doing work, we write \(T\) and \(V\) in terms of \(q\) and \(\dot q\), using the constraints (a fixed axis, rolling without slipping, a rigid connection) to eliminate the other coordinates, and then set \(dE/dt=0\); after cancelling the common factor \(\dot q\) we obtain the equation of motion. Examples 8.3 to 8.8 apply the method. For systems with several degrees of freedom, the same expressions for \(T\) and \(V\) are the starting point of the Lagrangian formulation of Chapter 9.
8.4 Torque and angular momentum
(a) Torque
The torque of a force \(\mathbf F\) acting at the point \(\mathbf r\), about the origin \(O\), is
It is perpendicular to both \(\mathbf r\) and \(\mathbf F\), and its magnitude is the force multiplied by the perpendicular distance of its line of action from \(O\). For a system of particles with forces \(\mathbf F_1,\mathbf F_2,\dots\) acting at \(\mathbf r_1,\mathbf r_2,\dots\) (Figure 8.5), the total torque is \(\sum_i\mathbf r_i\times\mathbf F_i\).
Figure 8.5 Forces \(\mathbf F_1\) and \(\mathbf F_2\) acting on particles at \(\mathbf r_1\) and \(\mathbf r_2\).
Torque and force are independent quantities. There can be a torque on a body with zero net force, a net force with zero torque, or, in general, both. The three cases are shown for a disc of radius \(R\) in Figure 8.6, with torques taken about the centre: two equal and opposite tangential forces \(f\) at opposite ends of a diameter form a couple, with \(\tau=2Rf\) and \(F=0\); two equal parallel forces at opposite ends of a diameter give \(F=2f\) and \(\tau=0\); and a single tangential force gives \(F=f\) and \(\tau=Rf\).
Figure 8.6 Force and torque on a disc of radius \(R\), with torques about the centre.
(b) Torque about different points
Let \(\mathbf R\) be the position of the centre of mass relative to \(O\) and let the forces \(\mathbf F_i\) act at \(\mathbf r_i=\mathbf R+\mathbf r_i'\). Then
where \(\boldsymbol\tau_0=\sum_i\mathbf r_i'\times\mathbf F_i\) is the torque about the centre of mass and \(\mathbf F=\sum_i\mathbf F_i\) is the total force. The torque about any point is therefore the torque about the centre of mass plus the torque of the total force imagined to act at the centre of mass. In particular, the torque of a couple (\(\mathbf F=0\)) is the same about every point.
(c) Torque and the rate of change of angular momentum
The first term is \(\mathbf v\times m\mathbf v=0\), since the cross product of parallel vectors vanishes, and by Newton’s second law \(d\mathbf p/dt=\mathbf F\). Hence
with both sides taken about the same fixed point. For a system of particles the same equation holds with \(\mathbf L\) the total angular momentum and \(\boldsymbol\tau\) the total external torque, provided the internal forces between each pair of particles are equal, opposite and directed along the line joining them; the torques of such a pair then cancel. The equation also holds about the centre of mass, even when the centre of mass accelerates, with \(\mathbf L_{\text{cm}}\) and \(\boldsymbol\tau_0\) in place of \(\mathbf L\) and \(\boldsymbol\tau\): differentiating \(\mathbf L=M\mathbf R\times\mathbf V+\mathbf L_{\text{cm}}\) and using \(M\dot{\mathbf V}=\mathbf F\) gives \(\dot{\mathbf L}=\mathbf R\times\mathbf F+\dot{\mathbf L}_{\text{cm}}\), and comparison with \(\boldsymbol\tau=\mathbf R\times\mathbf F+\boldsymbol\tau_0\) shows that \(\dot{\mathbf L}_{\text{cm}}=\boldsymbol\tau_0\).
If the external torque about a point vanishes, the angular momentum about that point is constant:
This is the conservation of angular momentum. As with linear momentum, it applies component by component: if the torque has no component along a fixed axis, the angular momentum about that axis is conserved. The choice of point is often the key to a problem. For a body that slides and then rolls on a horizontal floor, for instance, the friction force passes through every point of the floor and the weight and normal reaction cancel, so the angular momentum about any fixed point on the floor is conserved (Example 8.11).
Exam tip. In collision and rolling problems choose the reference point so that the unknown forces (friction, impulsive reactions at a pivot) pass through it; their torques then vanish and angular momentum about that point is conserved.
8.5 Newton's laws for a rigid body
(a) The equations of motion in plane motion
The motion of a rigid body in a plane is governed by two vector equations. The centre of mass moves as if the total mass were concentrated there and acted on by the total external force,
\[\boxed{\mathbf F=M\mathbf a_{\text{cm}}},\]
and the rotation about the centre of mass obeys \(\boldsymbol\tau_0=d\mathbf L_{\text{cm}}/dt\), which for rotation about an axis perpendicular to the plane of motion becomes
where \(\boldsymbol\alpha=\dot{\boldsymbol\omega}\) is the angular acceleration. For rolling without slipping the two are connected by \(\alpha=a_{\text{cm}}/R\). Torque equations may also be written about a fixed axis, \(\tau_O=I_O\alpha\), or, for a round body rolling without slipping, about the instantaneous point of contact, \(\tau_P=I_P\alpha\); this last form holds because the centre of mass of such a body is always at the same distance \(R\) above the point of contact, and it removes the friction force from the torque equation.
(b) The direction of friction
Friction at the point of contact always opposes the relative sliding, or, in the case of static friction, the tendency to slide. Its direction is therefore not known in advance and must be found from the motion (Figure 8.7). Consider a sphere moving to the right with acceleration \(a\) of its centre and angular acceleration \(\alpha\) in the rolling sense. If the forces acting would make \(a>R\alpha\), the point of contact tends to slide forwards and friction acts backwards, opposite to \(a\) (case i). If they would make \(a<R\alpha\), the point of contact tends to slide backwards and friction acts forwards, along \(a\) (case ii). When the body rolls without slipping (case iii), \(a=R\alpha\) exactly and the point of contact \(A\) is at rest; friction then takes whatever value and direction are needed to maintain this, and these depend on how the other forces act.
Figure 8.7 The direction of friction on a sphere: (i) \(a>R\alpha\); (ii) \(a<R\alpha\); (iii) rolling without slipping; (iv) a horizontal force at the top; (v) a sphere pushed up an incline by a force through its centre.
Two cases are worth remembering. If a horizontal force \(F\) is applied at the top of a sphere (case iv), it produces a large torque about the centre, so that without friction the sphere would turn faster than the rolling condition allows; the lowest point would slide backwards, and friction therefore acts forwards, in the direction of \(F\) (Example 8.15). If instead a sphere is pushed up an incline by a force through its centre (case v), the force produces no torque about the centre; without friction the sphere would slide up without turning, so the point of contact \(A\) tends to move up the slope, and friction acts down the slope. The friction is then what makes the sphere turn. A sphere rolling freely down an incline is the reverse case: gravity acts at the centre, the point of contact tends to slide down, and friction acts up the slope (Example 8.12).
Continue reading — it's free
Register with your mobile number to unlock the full chapter and all 100+ chapters of the Physics Library.
