Rotational Dynamics

1.1 Introduction

• Revolution: every particle of the object moves in a circle about a point outside the object, or about another object.
• Rotation: the motion is about an axis of rotation that passes through the object itself.

1.2 Characteristics of Circular Motion

• It is an accelerated motion — the direction of velocity changes at every instant, even when the speed is constant.
• It is a periodic motion — the particle repeats the same path, so the motion is periodic in space.
• Uniform circular motion (U.C.M.): circular motion in which the speed of the particle remains constant. Only the direction of $\vec v$ changes, and $\vec v$ is always tangential.
• Non-uniform circular motion: the speed changes as well, so both $\vec a_r$ (magnitude no longer constant) and $\vec a_T$ are present.
• Tangential acceleration in U.C.M. is always zero.
• $\vec v = \vec\omega\times\vec r$; $\vec\omega$ and $\vec\alpha$ lie along the axis of rotation, direction given by the right-hand thumb rule (curl the fingers along the sense of rotation, the outstretched thumb gives $\vec\omega$).
• For increasing speed $\vec\alpha$ is along $\vec\omega$; for decreasing speed it is opposite to $\vec\omega$.
• If $\vec\alpha$ has a component perpendicular to the axis, it changes the direction of $\vec\omega$ and hence the plane of rotation. If $\vec\alpha$ is constant in magnitude but always perpendicular to $\vec\omega$, it changes only the direction of $\vec\omega$ and never its magnitude.
• Kinematical equations $\omega=\omega_0+\alpha t$, $\theta=\omega_0t+\dfrac{1}{2}\alpha t^2$, $\omega^2=\omega_0^2+2\alpha\theta$ hold only for constant, axial $\alpha$.

Centripetal and centrifugal force

• Centripetal force is the resultant of all the real forces acting on the body, directed towards the centre: $\vec F_{CP}=-m\omega^2\vec r$, $F_{CP}=mr\omega^2=\dfrac{mv^2}{r}=mv\omega$.
• "Centripetal" is not the name or type of a force — it is an adjective describing its direction (centre-seeking). The force itself is gravity, tension, friction or normal reaction.
• The converse is not true: a force directed towards a centre need not produce circular motion (S.H.M. is a counter-example). A matching tangential velocity is also essential.
• Centrifugal force is the net pseudo force in the rotating (non-inertial) frame: $\vec F_{CF}=+m\omega^2\vec r$. It is non-real, but not imaginary — an instrument in that frame records it.
• Only in the frame of the rotating body can we say that centrifugal force balances centripetal force; there the resultant is zero.

1.3 Applications of Uniform Circular Motion

Vehicle on a horizontal circular track

• Forces: weight $mg$ down, normal reaction $N$ up ($N=mg$), and static friction $f_s$ towards the centre. Friction is static because it prevents outward slipping (skidding).
• $\dfrac{f_s}{N}=\dfrac{v^2}{rg}$, and since $f_s \le \mu_s N$, the speed has an upper limit $v_{max}=\sqrt{\mu_s rg}$ — independent of the mass.
• A two-wheeler rider must lean at $\theta = \tan^{-1}\!\left(\dfrac{v^2}{rg}\right)$ with the vertical so that the torque of the weight balances the torque of the centrifugal and frictional forces. A four-wheeler does not lean because its four normal reactions (unequal during a turn) already supply the balancing torque.

Well (wall) of death

• Vertical cylindrical wall of radius $r$. Normal reaction $N$ is horizontal and is the entire centripetal force; static friction $f_s$ acts vertically upwards and equals $mg$ (the only upward force).
• $f_s \le \mu_s N$ gives a lower limit: $v_{min}=\sqrt{\dfrac{rg}{\mu_s}}$. The largest allowed period is $T_{max}=\dfrac{2\pi r}{v_{min}}$.
• $f_s=\mu_s N$ holds only at the minimum speed; at all higher speeds $f_s$ is still exactly $mg$.
• Banking angle here is $90^\circ$, which is why it imposes a lower limit. For $0<\theta<90^\circ$ there are both an upper and a lower limit.

Vehicle on a banked road

• Roads are banked so that the horizontal component of the normal reaction supplies the centripetal force and we need not depend on friction, whose value is limited and not uniform over a road surface.
• Frictionless analysis: $N\cos\theta = mg$, $N\sin\theta = \dfrac{mv^2}{r}$, hence $\tan\theta = \dfrac{v^2}{rg}$ and $v_s=\sqrt{rg\tan\theta}$ — the most safe speed, not a maximum or minimum, and independent of the mass.
• Below $v_s$, friction acts up the incline; above $v_s$, it acts down the incline.
• $v_{min}=\sqrt{rg\left(\dfrac{\tan\theta-\mu_s}{1+\mu_s\tan\theta}\right)}$ and $v_{max}=\sqrt{rg\left(\dfrac{\tan\theta+\mu_s}{1-\mu_s\tan\theta}\right)}$.
• If $\mu_s \ge \tan\theta$ then $v_{min}=0$ (true for most rough roads banked at small angles). If $\mu_s=\cot\theta$ then $v_{max}\to\infty$; since $(\mu_s)_{max}=1$, this means no upper limit for $\theta\ge 45^\circ$.
• Even a two-wheeler benefits from banking — it then needs to lean less, and the same limits apply.
• If friction were zero the vehicle could still move at exactly $v_s$, but at any other speed it would skid; so the honest board answer is "No, the vehicle will skid in the absence of friction."

Conical pendulum

• A bob revolving in a horizontal circle so that the string sweeps out a right circular cone of vertical axis.
• $T_0\cos\theta = mg$ (balances the weight), $T_0\sin\theta = \dfrac{mv^2}{r} = mr\omega^2$ (the centripetal force). Dividing, $\tan\theta = \dfrac{v^2}{rg}$.
• With $r=L\sin\theta$: $\omega=\sqrt{\dfrac{g}{L\cos\theta}}$, period $T=2\pi\sqrt{\dfrac{L\cos\theta}{g}}$, frequency $n=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{L\cos\theta}}$.
• Period and frequency depend on $L$, $g$ and $\theta$ — never on the mass of the bob.
• The string can never become horizontal: at $\theta=90^\circ$ the vertical component of tension vanishes and nothing balances $mg$; also $T_0=\dfrac{mg}{\cos\theta}$, the kinetic energy and the frequency would all become infinite and the period zero.

1.4 Vertical Circular Motion

• Two kinds: (a) controlled, such as a giant wheel, where the speed is not governed by gravity alone; (b) controlled only by gravity, where energy is supplied once at the lowest point and thereafter K.E. and gravitational P.E. interconvert.
• Only two forces act at any position: constant weight $mg$, and the force due to tension $T$ (or normal reaction $N$) along the string towards the centre.
• The resultant is purely centripetal only at the uppermost and lowermost points. Elsewhere part of it is tangential and changes the speed.

• Energy conservation over the drop $2r$ gives $v_B^2-v_A^2=4rg$; over the drop $r$, $v_B^2-v_C^2=2rg$.
• $T_B-T_A=6mg$ and $T_C-T_A=3mg$. These are independent of $v$ and $r$ — they depend only on the weight.
• $T_A$ can never be exactly zero for a string (it would slacken), so in practice $T_B>6mg$.
• At an arbitrary angle $\theta$ from the lowest point, $T=\dfrac{mv^2}{r}+mg\cos\theta$, and the rise above the lowest point is $r(1-\cos\theta)$.
• Rod instead of string: the rod can push as well as pull, so zero speed is possible at the top. Then $(v_{lowermost})_{min}=\sqrt{4rg}$, $(v_{midway})_{min}=\sqrt{2rg}$, and still $T_{lowermost}-T_{uppermost}=6mg$.
• Sphere of death: identical dynamics with $N$ in place of $T$.
• Convex over-bridge: $mg-N=\dfrac{mv^2}{r}$; $N$ falls as $v$ rises and contact is just lost at $v_{max}=\sqrt{rg}$.
• No parameter is constant in this motion, so the kinematical equations of Table 1 do not apply, and the period cannot be found theoretically (only measured).

1.5 Moment of Inertia

• Definition: the moment of inertia of a rigid body about an axis of rotation is the sum of the products of the mass of each particle and the square of its perpendicular distance from that axis.
• $I=\sum m_i r_i^2 = \int r^2\,dm$. SI unit kg m$^2$; dimensions $[M^1L^2T^0]$.
• It depends on (i) the individual masses and (ii) how those masses are distributed about the given axis — so the same body has different $I$ about different axes.
• Rotational K.E. $=\dfrac{1}{2}I\omega^2$, the exact analogue of $\dfrac{1}{2}mv^2$; this is what identifies $I$ as the rotational analogue of mass.
• Uniform ring about its own axis: the whole mass is at distance $R$, so $I=MR^2$ with no integration.
• Uniform disc about its own axis: integrating over concentric rings, $I=\dfrac{1}{2}MR^2$.

1.6 Radius of Gyration

• Definition: the distance between the axis of rotation and the point at which the whole mass of the body may be supposed to be concentrated so as to give the same moment of inertia about that axis. $I=MK^2$.
• Its use: $I$ can be calculated only for mathematically integrable shapes, but it can be measured for any object. Writing $I=MK^2$ separates out the mass and leaves $K$ carrying the information about how the mass is distributed.
• $K$ depends on the distribution of mass about the given axis (and on the shape and the choice of axis); it does not depend on the mass of the body.
• Larger $K$ means the mass sits farther from the axis. For the same $M$ and $R$: $K_{ring}=R$, $K_{disc}=\dfrac{R}{\sqrt2}$, so $K_d<K_r$ — among objects of the same mass and radius, the ring has the largest $K$ and hence the largest $I$.
• Like the centre of mass, $K$ is a single equivalent location that stands in for the whole distribution — but it is defined with respect to an axis, not a point.

1.7 Theorems of Parallel and Perpendicular Axes

• In terms of radius of gyration, the parallel axes theorem reads $K_O^2=K_C^2+h^2$.
• Useful transforms: disc about a diameter $\dfrac{1}{4}MR^2$; disc about a tangent in its plane $\dfrac{5}{4}MR^2$; disc about a tangent perpendicular to its plane $\dfrac{3}{2}MR^2$; ring about a diameter $\dfrac{1}{2}MR^2$; sphere about a tangent $\dfrac{7}{5}MR^2$; rod about a transverse axis through one end $\dfrac{1}{3}ML^2$.

1.8 Angular Momentum

• $\vec L=\vec r\times\vec p$; in magnitude $L=pr\sin\theta$, the product of linear momentum and its perpendicular distance from the axis.
• For a rigid body rotating about a fixed axis, $L=I\omega$ — analogous to $p=mv$, which is the physical significance of $I$.
• K.E. $=\dfrac{1}{2}I\omega^2=\dfrac{1}{2}L\omega=\dfrac{L^2}{2I}$.
• SI unit kg m$^2$/s; dimensions $[M^1L^2T^{-1}]$.

1.9 Torque

• $\tau = I\alpha$, obtained by summing $m_i r_i^2\alpha$ over all particles — analogous to $f=ma$.
• $\vec\tau=\vec r\times\vec F=\dfrac{d\vec L}{dt}$. SI unit N m; dimensions $[M^1L^2T^{-2}]$.
• Work done by a torque $W=\tau\theta$; power $P=\tau\omega=2\pi n\tau$.
• A "couple of 20 N m" in a numerical simply means a torque of 20 N m.

1.10 Conservation of Angular Momentum

• Statement: the angular momentum of an isolated system is conserved in the absence of an external unbalanced torque.
• Proof in one line: $\dfrac{d\vec L}{dt}=\vec r\times\vec F+m(\vec v\times\vec v)=\vec\tau$, and $\vec v\times\vec v=0$; so $\vec\tau=0$ gives $\vec L$ = constant.
• Working form: $I_1\omega_1=I_2\omega_2$, i.e. $I_1n_1=I_2n_2$. Decreasing $I$ raises $n$ and vice versa.
• Standard examples: ballet dancers and ice skaters pulling limbs in for fast small-radius rounds and stretching out for slow large-radius ones; a diver folding up after leaving the board to spin faster and stretching out for a clean entry; a planet speeding up near the sun in an elliptical orbit; a person on a rotating platform changing speed by moving weights in and out.
• For a planet in a circular orbit the angular momentum is trivially constant too — conservation applies to both circular and elliptical orbits, because the gravitational force is central and exerts no torque about the sun.

1.11 Rolling Motion

• Pure rolling is treated as rotation of the whole body about its own symmetry axis plus translation of the whole mass concentrated at the centre of mass; the centre of mass performs purely translational motion, and $v=R\omega$.
• Static friction is essential for pure rolling — it prevents sliding. It is why a pushed sphere first slips and only then starts rolling.
• $E=\dfrac{1}{2}Mv^2+\dfrac{1}{2}I\omega^2=\dfrac{1}{2}Mv^2\left[1+\dfrac{K^2}{R^2}\right]$.
• Down an incline: $v=\sqrt{\dfrac{2gh}{1+\dfrac{K^2}{R^2}}}$ and $a=\dfrac{g\sin\theta}{1+\dfrac{K^2}{R^2}}$. For pure sliding these become $\sqrt{2gh}$ and $g\sin\theta$ — the factor $\left(1+\dfrac{K^2}{R^2}\right)$ is the only difference.
• Because $\dfrac{K^2}{R^2}$ contains no mass and no radius, all bodies of the same shape roll down an incline with the same acceleration, whatever their size or mass.
• When a rod rolls, it is really a cylinder that is rolling.

Analogy Between Translational and Rotational Motion