Chapter 3

Motion in Two Dimensions and Rotating Frames

Theory, worked examples and 17 previous-year questions with full solutions — IIT JAM, GATE, JEST and CSIR NET-JRF, 2020 onwards.

Theory

3.1 Motion in a plane: the Cartesian description

A particle moving in a plane needs two coordinates to fix its position. The simplest choice is a pair of Cartesian coordinates \((x,y)\) referred to fixed perpendicular axes, with unit vectors \(\hat{\mathbf x}\) and \(\hat{\mathbf y}\) along the axes. The position vector is

\[\mathbf r=x\,\hat{\mathbf x}+y\,\hat{\mathbf y}.\]

The essential property of the Cartesian unit vectors is that they are the same at every point of the plane (Figure 3.1). Their magnitudes are fixed by definition, and their directions do not depend on the position of the particle, so they are constant in time however the particle moves. Differentiation of \(\mathbf r\) therefore acts only on the components:

\[\mathbf v=\dot{\mathbf r}=\dot x\,\hat{\mathbf x}+\dot y\,\hat{\mathbf y},\qquad \mathbf a=\ddot{\mathbf r}=\ddot x\,\hat{\mathbf x}+\ddot y\,\hat{\mathbf y}.\]

Figure 3.1 The Cartesian unit vectors \(\hat{\mathbf x}\) and \(\hat{\mathbf y}\) have the same magnitude and direction at every point of the plane.

Newton’s second law \(\mathbf F=m\mathbf a\) then separates into two independent scalar equations,

\[\boxed{m\ddot x=F_x,\qquad m\ddot y=F_y}.\]

This separation is what makes Cartesian coordinates the natural choice when the force has a fixed direction, as in projectile motion (Section 3.4). When the force is directed towards a fixed point, or when the particle is constrained to a circle, the Cartesian equations are coupled and it is far better to use plane polar coordinates.

3.2 Plane polar coordinates

(a) Coordinates and unit vectors

In plane polar coordinates a point is specified by its distance \(r\) from the origin and the angle \(\theta\) that the line from the origin to the point makes with the \(x\)-axis, measured anticlockwise (Figure 3.2). The two descriptions are related by

\[x=r\cos\theta,\quad y=r\sin\theta;\qquad r=\sqrt{x^2+y^2},\quad \theta=\tan^{-1}\frac yx,\]

where the quadrant of \(\theta\) is fixed by the signs of \(x\) and \(y\).

Figure 3.2 Plane polar coordinates \((r,\theta)\) and the unit vectors \(\hat{\mathbf r}\) and \(\hat{\boldsymbol\theta}\) at a point, together with the Cartesian unit vectors there.

At each point we define two unit vectors. The radial unit vector \(\hat{\mathbf r}\) points in the direction of increasing \(r\) at fixed \(\theta\), that is, directly away from the origin. The transverse or azimuthal unit vector \(\hat{\boldsymbol\theta}\) points in the direction of increasing \(\theta\) at fixed \(r\), that is, along the tangent to the circle through the point, in the anticlockwise sense. The two vectors satisfy

\[\lvert\hat{\mathbf r}\rvert=\lvert\hat{\boldsymbol\theta}\rvert=1,\qquad \hat{\mathbf r}\cdot\hat{\boldsymbol\theta}=0,\]

so they form an orthonormal pair, and \(\hat{\mathbf r}\times\hat{\boldsymbol\theta}=\hat{\mathbf z}\), the unit vector normal to the plane. Unlike \(\hat{\mathbf x}\) and \(\hat{\mathbf y}\), however, their directions change from point to point (Figure 3.3). They depend on \(\theta\), though not on \(r\).

Figure 3.3 The unit vectors \(\hat{\mathbf r}\) and \(\hat{\boldsymbol\theta}\) at several points. Their directions depend on the position of the point.

Resolving \(\hat{\mathbf r}\) and \(\hat{\boldsymbol\theta}\) along the fixed axes (Figure 3.4) gives

\[\boxed{\hat{\mathbf r}=\cos\theta\,\hat{\mathbf x}+\sin\theta\,\hat{\mathbf y},\qquad \hat{\boldsymbol\theta}=-\sin\theta\,\hat{\mathbf x}+\cos\theta\,\hat{\mathbf y}}.\]

Figure 3.4 Resolution of \(\hat{\mathbf r}\) and \(\hat{\boldsymbol\theta}\) along the Cartesian axes.

In matrix form the change of basis is a rotation through the angle \(\theta\):

\[\begin{aligned}\begin{pmatrix}\hat{\mathbf r}\\ \hat{\boldsymbol\theta}\end{pmatrix}&=\begin{pmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{pmatrix}\begin{pmatrix}\hat{\mathbf x}\\ \hat{\mathbf y}\end{pmatrix},\\ \begin{pmatrix}\hat{\mathbf x}\\ \hat{\mathbf y}\end{pmatrix}&=\begin{pmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{pmatrix}\begin{pmatrix}\hat{\mathbf r}\\ \hat{\boldsymbol\theta}\end{pmatrix}.\end{aligned}\]

The matrix is orthogonal, so its inverse is its transpose.

(b) Derivatives of the unit vectors

Since \(\hat{\mathbf r}\) and \(\hat{\boldsymbol\theta}\) depend only on \(\theta\), differentiation of the expressions above gives

\[\frac{d\hat{\mathbf r}}{d\theta}=-\sin\theta\,\hat{\mathbf x}+\cos\theta\,\hat{\mathbf y}=\hat{\boldsymbol\theta},\qquad \frac{d\hat{\boldsymbol\theta}}{d\theta}=-\cos\theta\,\hat{\mathbf x}-\sin\theta\,\hat{\mathbf y}=-\hat{\mathbf r},\]

while \(\partial\hat{\mathbf r}/\partial r=\partial\hat{\boldsymbol\theta}/\partial r=0\). As a particle moves, \(\theta\) changes with time, and the chain rule gives

\[\boxed{\frac{d\hat{\mathbf r}}{dt}=\dot\theta\,\hat{\boldsymbol\theta},\qquad \frac{d\hat{\boldsymbol\theta}}{dt}=-\dot\theta\,\hat{\mathbf r}}.\]

These results have a simple geometrical meaning. A unit vector cannot change its length, so its derivative must be perpendicular to it. When the line from the origin turns through \(d\theta\), the tip of \(\hat{\mathbf r}\) moves through a distance \(d\theta\) in the direction of \(\hat{\boldsymbol\theta}\), and the tip of \(\hat{\boldsymbol\theta}\) moves through \(d\theta\) in the direction of \(-\hat{\mathbf r}\). Both vectors rotate rigidly with angular velocity \(\dot\theta\,\hat{\mathbf z}\), and indeed \(\dot\theta\,\hat{\mathbf z}\times\hat{\mathbf r}=\dot\theta\,\hat{\boldsymbol\theta}\) and \(\dot\theta\,\hat{\mathbf z}\times\hat{\boldsymbol\theta}=-\dot\theta\,\hat{\mathbf r}\). This is the simplest instance of the rule \(d\mathbf A/dt=\boldsymbol\omega\times\mathbf A\) for a vector of fixed length rotating with angular velocity \(\boldsymbol\omega\), which will be central in Section 3.8.

(c) The position vector

Since \(\mathbf r=x\hat{\mathbf x}+y\hat{\mathbf y}=r(\cos\theta\,\hat{\mathbf x}+\sin\theta\,\hat{\mathbf y})\), the position vector in polar form is simply

\[\boxed{\mathbf r=r\,\hat{\mathbf r}}.\]

This expression may seem puzzling at first, because it contains only one of the two coordinates. The angle has not disappeared: it is hidden in the direction of \(\hat{\mathbf r}\), which would be better written \(\hat{\mathbf r}(\theta)\) to emphasize its dependence on \(\theta\). By convention the argument is omitted, but it must never be forgotten, since it is exactly this dependence that produces the extra terms in the velocity and acceleration.

3.3 Velocity and acceleration in polar coordinates

(a) Velocity

Differentiating \(\mathbf r=r\hat{\mathbf r}\) with the product rule and using \(d\hat{\mathbf r}/dt=\dot\theta\hat{\boldsymbol\theta}\),

\[\mathbf v=\frac{d}{dt}(r\hat{\mathbf r})=\dot r\,\hat{\mathbf r}+r\frac{d\hat{\mathbf r}}{dt},\]

that is,

\[\boxed{\mathbf v=\dot r\,\hat{\mathbf r}+r\dot\theta\,\hat{\boldsymbol\theta}},\qquad v=\sqrt{\dot r^2+r^2\dot\theta^2}.\]

The radial velocity \(v_r=\dot r\) is the rate at which the distance from the origin changes, and the transverse velocity \(v_\theta=r\dot\theta\) is the speed with which the particle moves round the origin. The two contributions are seen separately in Figure 3.5. If \(\theta\) is fixed the particle moves along a radial line with \(\mathbf v=\dot r\hat{\mathbf r}\); if \(r\) is fixed it moves on a circle with \(\mathbf v=r\dot\theta\hat{\boldsymbol\theta}\). In general both occur together.

Figure 3.5 The two contributions to the velocity: a change of \(r\) at fixed \(\theta\) (case 1) and a change of \(\theta\) at fixed \(r\) (case 2).

(b) Acceleration

Differentiating the velocity term by term,

\[\begin{aligned}\frac{d\mathbf v}{dt}&=\ddot r\,\hat{\mathbf r}+\dot r\frac{d\hat{\mathbf r}}{dt}+\dot r\dot\theta\,\hat{\boldsymbol\theta}+r\ddot\theta\,\hat{\boldsymbol\theta}+r\dot\theta\frac{d\hat{\boldsymbol\theta}}{dt}\\&=\ddot r\,\hat{\mathbf r}+\dot r\dot\theta\,\hat{\boldsymbol\theta}+\dot r\dot\theta\,\hat{\boldsymbol\theta}+r\ddot\theta\,\hat{\boldsymbol\theta}-r\dot\theta^2\,\hat{\mathbf r}.\end{aligned}\]

Collecting terms,

\[\boxed{\mathbf a=\left(\ddot r-r\dot\theta^2\right)\hat{\mathbf r}+\left(r\ddot\theta+2\dot r\dot\theta\right)\hat{\boldsymbol\theta}}.\]

The radial acceleration \(a_r=\ddot r-r\dot\theta^2\) contains, besides \(\ddot r\), the centripetal term \(-r\dot\theta^2\), which is present even when \(r\) is constant: a particle moving on a circle accelerates towards the centre. The transverse acceleration \(a_\theta=r\ddot\theta+2\dot r\dot\theta\) contains, besides the angular acceleration term \(r\ddot\theta\), the term \(2\dot r\dot\theta\), which is the inertial-frame counterpart of the Coriolis acceleration discussed in Section 3.8. Half of it comes from the turning of \(\dot r\hat{\mathbf r}\), and half from the change of the transverse speed \(r\dot\theta\) caused by the change of \(r\). A useful identity is

\[a_\theta=r\ddot\theta+2\dot r\dot\theta=\frac1r\frac{d}{dt}\left(r^2\dot\theta\right).\]

Exam tip. The radial acceleration is \(\ddot r-r\dot\theta^2\), not \(\ddot r\). Even if \(\dot r\) and \(\dot\theta\) are both constant, the acceleration is \(-r\dot\theta^2\hat{\mathbf r}+2\dot r\dot\theta\hat{\boldsymbol\theta}\), which is not zero.

(c) Newton’s second law in polar form

Resolving the force into components \(F_r\) along \(\hat{\mathbf r}\) and \(F_\theta\) along \(\hat{\boldsymbol\theta}\), Newton’s law \(\mathbf F=m\mathbf a\) gives

\[\boxed{F_r=m\left(\ddot r-r\dot\theta^2\right),\qquad F_\theta=m\left(r\ddot\theta+2\dot r\dot\theta\right)}.\]

These are the working equations for all problems on circular motion and central forces (Chapter 6). When the particle is constrained to move on a curve, the constraint relation, such as \(r=R\) for a circle, is added to these equations and determines the unknown constraint force.

(d) Cylindrical coordinates

In three dimensions the polar coordinates in the \(xy\)-plane are supplemented by the height \(z\); these are cylindrical coordinates \((\rho,\varphi,z)\), with \(\rho\) and \(\varphi\) playing the parts of \(r\) and \(\theta\). The unit vector \(\hat{\mathbf z}\) is fixed, so the position vector and its derivatives are

\[\mathbf r=\rho\,\hat{\boldsymbol\rho}+z\,\hat{\mathbf z},\qquad \mathbf v=\dot\rho\,\hat{\boldsymbol\rho}+\rho\dot\varphi\,\hat{\boldsymbol\varphi}+\dot z\,\hat{\mathbf z},\]

\[\mathbf a=\left(\ddot\rho-\rho\dot\varphi^2\right)\hat{\boldsymbol\rho}+\left(\rho\ddot\varphi+2\dot\rho\dot\varphi\right)\hat{\boldsymbol\varphi}+\ddot z\,\hat{\mathbf z}.\]

Note that the position vector has no \(\hat{\boldsymbol\varphi}\) component: the azimuth enters only through the direction of \(\hat{\boldsymbol\rho}\). A particle with \(\rho\) and \(\dot\varphi\) constant and \(\dot z\) constant moves on a helix of radius \(\rho\) wound on a circular cylinder.

3.4 Projectile motion

A projectile is a particle moving near the Earth’s surface under gravity alone, air resistance being neglected. The force \(-mg\hat{\mathbf y}\) is constant, with \(\hat{\mathbf y}\) vertically upwards, so the Cartesian equations \(m\ddot x=0\) and \(m\ddot y=-mg\) are uncoupled. For a projectile launched from the origin with speed \(u\) at an angle \(\theta_0\) above the horizontal,

\[x=ut\cos\theta_0,\qquad y=ut\sin\theta_0-\frac12gt^2.\]

Eliminating \(t\) gives the parabolic trajectory

\[\boxed{y=x\tan\theta_0-\frac{gx^2}{2u^2\cos^2\theta_0}}.\]

The vertical velocity \(u\sin\theta_0-gt\) vanishes at \(t=u\sin\theta_0/g\), and the projectile returns to the ground at twice this time. Hence the time of flight, the greatest height and the horizontal range are

\[\boxed{T=\frac{2u\sin\theta_0}{g},\qquad H=\frac{u^2\sin^2\theta_0}{2g},\qquad R=\frac{u^2\sin2\theta_0}{g}}.\]

The range is greatest, \(R_{\max}=u^2/g\), for \(\theta_0=45^\circ\), and complementary angles \(\theta_0\) and \(90^\circ-\theta_0\) give the same range. The highest point lies at horizontal distance \(R/2\) from the origin, so seen from the point of launch it has elevation \(\alpha\) given by

\[\tan\alpha=\frac{H}{R/2}=\frac{u^2\sin^2\theta_0/2g}{u^2\sin\theta_0\cos\theta_0/g}=\frac12\tan\theta_0.\]

Although Cartesian coordinates are natural here, a projectile may also be described in polar coordinates centred at the point of launch, and examination questions often ask for this. With \(x=r\cos\theta\) and \(y=r\sin\theta\), the kinetic energy is \(\frac12m(\dot x^2+\dot y^2)=\frac12m(\dot r^2+r^2\dot\theta^2)\) and the potential energy, zero on the ground, is \(mgy=mgr\sin\theta\). The constant acceleration \(-g\hat{\mathbf y}\) has polar components

\[a_r=-g\,\hat{\mathbf y}\cdot\hat{\mathbf r}=-g\sin\theta,\qquad a_\theta=-g\,\hat{\mathbf y}\cdot\hat{\boldsymbol\theta}=-g\cos\theta,\]

which change along the path even though \(\mathbf a\) itself is constant.

3.5 Circular motion

For motion on a circle of radius \(r_0\) centred at the origin, \(r=r_0\) and \(\dot r=\ddot r=0\). The polar equations of Section 3.3(c) reduce to

\[F_r=-mr_0\dot\theta^2,\qquad F_\theta=mr_0\ddot\theta.\]

A net inward force \(mr_0\dot\theta^2\) is always required to keep the particle on the circle; a tangential force is needed only if the angular speed changes. The speed is \(v=r_0\lvert\dot\theta\rvert\).

(a) Uniform circular motion

If there is no tangential force, \(F_\theta=0\), then \(\ddot\theta=0\) and the angular speed \(\dot\theta=\omega\) is constant. The speed \(v=r_0\omega\) is constant, but the velocity is not: its direction turns continuously, and the acceleration is directed towards the centre,

\[\boxed{\mathbf a=-r_0\omega^2\,\hat{\mathbf r}=-\frac{v^2}{r_0}\hat{\mathbf r}}.\]

At every point the velocity is tangential and the acceleration is radial, as in Figure 3.6(a). The period of the motion is \(2\pi/\omega=2\pi r_0/v\).

(b) Non-uniform circular motion

If the speed \(v(t)=r_0\dot\theta(t)\) varies, the acceleration acquires a tangential component equal to the rate of change of the speed. For a circle of radius \(r\),

\[\boxed{\mathbf a=-\frac{v^2}{r}\hat{\mathbf r}+\frac{dv}{dt}\hat{\boldsymbol\theta}},\qquad \lvert\mathbf a\rvert=\sqrt{\left(\frac{v^2}{r}\right)^2+\left(\frac{dv}{dt}\right)^2}.\]

The acceleration now points inside the circle but not at the centre, making an angle \(\alpha\) with the radius such that \(\tan\alpha=(dv/dt)/(v^2/r)\), as in Figure 3.6(b). The same decomposition holds for motion along any smooth curve, with \(r\) replaced by the local radius of curvature: the tangential acceleration \(dv/dt\) changes the speed and the normal acceleration \(v^2/r\) changes the direction.

Figure 3.6 (a) Uniform circular motion: the velocity is tangential and the acceleration points to the centre. (b) Non-uniform circular motion: the acceleration has a centripetal component \(v^2/r\) and a tangential component \(dv/dt\).

3.6 Torque and angular momentum in polar coordinates

With the force written as \(\mathbf F=F_r\hat{\mathbf r}+F_\theta\hat{\boldsymbol\theta}\), the torque about the origin is

\[\boldsymbol\tau=\mathbf r\times\mathbf F=r\hat{\mathbf r}\times\left(F_r\hat{\mathbf r}+F_\theta\hat{\boldsymbol\theta}\right)=rF_\theta\,\hat{\mathbf z}.\]

Only the transverse force produces a torque. Using \(F_\theta=m(r\ddot\theta+2\dot r\dot\theta)\),

\[\tau=m\left(r^2\ddot\theta+2r\dot r\dot\theta\right)=\frac{d}{dt}\left(mr^2\dot\theta\right).\]

The angular momentum about the origin follows directly from the velocity: \(\mathbf L=\mathbf r\times m\mathbf v=r\hat{\mathbf r}\times m(\dot r\hat{\mathbf r}+r\dot\theta\hat{\boldsymbol\theta})=mr^2\dot\theta\,\hat{\mathbf z}\). Hence

\[\boxed{L=mr^2\dot\theta,\qquad \tau=\frac{dL}{dt}=rF_\theta}.\]

If the force has no transverse component, \(F_\theta=0\), then \(d(mr^2\dot\theta)/dt=0\) and

\[mr^2\dot\theta=\text{constant}.\]

The angular momentum of a particle moving under a purely radial force is conserved. This is the starting point of the theory of central-force motion in Chapter 6, where it leads to Kepler’s second law.

3.7 Worked examples: motion in a plane

Example 3.1 A bead moves along the spoke of a wheel at a constant speed \(u\). The wheel rotates with uniform angular velocity \(\dot\theta=\omega\) about an axis fixed in space. At \(t=0\) the spoke lies along the \(x\)-axis and the bead is at the origin. Find (a) the velocity and (b) the acceleration of the bead.

Solution. The bead moves outwards along the spoke at speed \(u\) while the spoke turns, so

\[r=ut,\quad \dot r=u,\quad \ddot r=0;\qquad \theta=\omega t,\quad \dot\theta=\omega,\quad \ddot\theta=0.\]

The velocity is

\[\mathbf v=\dot r\,\hat{\mathbf r}+r\dot\theta\,\hat{\boldsymbol\theta}=u\,\hat{\mathbf r}+ut\omega\,\hat{\boldsymbol\theta},\]

with magnitude \(u\sqrt{1+\omega^2t^2}\).

The acceleration is

\[\mathbf a=\left(\ddot r-r\dot\theta^2\right)\hat{\mathbf r}+\left(r\ddot\theta+2\dot r\dot\theta\right)\hat{\boldsymbol\theta}=-ut\omega^2\,\hat{\mathbf r}+2u\omega\,\hat{\boldsymbol\theta}.\]

The radial part is the centripetal acceleration of a point at distance \(ut\) on the rotating spoke. The transverse part \(2u\omega\) is constant; it must be supplied by the sideways push of the spoke on the bead.

Example 3.2 Find the magnitude of the linear acceleration of a particle moving in a circle of radius

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Chapter 3 Motion in Two Dimensions and Rotating Frames