Curvilinear Motion of Particles

A particle moving on a plane curve is located by polar coordinates $(r,\theta)$. Using the unit vectors $\hat e_r$ (along the radius vector) and $\hat e_\theta$ (perpendicular to it, in the direction of increasing $\theta$), velocity and acceleration are found by differentiating $\vec r = r\hat e_r$.

        y
        |        P
        |       /|
        |    r / | e_theta
        |     /  |
        |    /   -> e_r along OP
        |   /\ theta
        +--------------- x
        O

Unit-vector derivatives

$$\hat e_r=\cos\theta\,\hat i+\sin\theta\,\hat j,\qquad \hat e_\theta=-\sin\theta\,\hat i+\cos\theta\,\hat j$$

Differentiating with respect to time:

$$\dot{\hat e}_r=\dot\theta\,\hat e_\theta,\qquad \dot{\hat e}_\theta=-\dot\theta\,\hat e_r$$

Velocity

$$\vec v=\frac{d}{dt}(r\hat e_r)=\dot r\,\hat e_r+r\dot\theta\,\hat e_\theta$$

So $v_r=\dot r$ and $v_\theta=r\dot\theta$.

Acceleration

$$\begin{aligned} \vec a=\frac{d\vec v}{dt}&=\ddot r\,\hat e_r+\dot r\,\dot{\hat e}_r+(\dot r\dot\theta+r\ddot\theta)\hat e_\theta+r\dot\theta\,\dot{\hat e}_\theta\\ &=\ddot r\,\hat e_r+\dot r\dot\theta\,\hat e_\theta+(\dot r\dot\theta+r\ddot\theta)\hat e_\theta-r\dot\theta^{2}\hat e_r \end{aligned}$$

Collecting terms:

$$\vec a=(\ddot r-r\dot\theta^{2})\,\hat e_r+(r\ddot\theta+2\dot r\dot\theta)\,\hat e_\theta$$

• Radial component: $a_r=\ddot r-r\dot\theta^{2}$
• Transverse component: $a_\theta=r\ddot\theta+2\dot r\dot\theta$
• Magnitude: $a=\sqrt{a_r^{2}+a_\theta^{2}}$

Here $-r\dot\theta^2$ is the centripetal term and $2\dot r\dot\theta$ is the Coriolis term. For motion on a circle ($r$ constant) $a_r=-r\dot\theta^2$ and $a_\theta=r\ddot\theta$.

Use polar coordinates with $r$ along the arm and $\theta$ the arm rotation.

Time when $\theta=35^\circ$

$$\theta=35^\circ=0.6109\ \text{rad}=0.23t^{2}\ \Rightarrow\ t^{2}=2.656,\ t=1.6297\ \text{s}$$

Kinematic quantities at this instant

$$\begin{aligned} r&=0.9-0.12t^{2}=0.9-0.12(2.656)=0.5813\ \text{m}\\ \dot r&=-0.24t=-0.3911\ \text{m/s},\qquad \ddot r=-0.24\ \text{m/s}^2\\ \dot\theta&=0.46t=0.7497\ \text{rad/s},\qquad \ddot\theta=0.46\ \text{rad/s}^2 \end{aligned}$$

Acceleration components

$$\begin{aligned} a_r&=\ddot r-r\dot\theta^{2}=-0.24-0.5813(0.7497)^{2}=-0.5667\ \text{m/s}^2\\ a_\theta&=r\ddot\theta+2\dot r\dot\theta=0.5813(0.46)+2(-0.3911)(0.7497)=-0.3190\ \text{m/s}^2\\ a&=\sqrt{a_r^{2}+a_\theta^{2}}=0.650\ \text{m/s}^2 \end{aligned}$$

Both components are negative: the acceleration points towards O and against the sense of increasing $\theta$, at $\tan^{-1}(0.319/0.567)=29.4^\circ$ from the line towards O.

Answer: $a_r=-0.567$ m/s$^2$, $a_\theta=-0.319$ m/s$^2$, total $a\approx0.650$ m/s$^2$.

Time when $\theta=45^\circ$

$$\theta=45^\circ=0.7854\ \text{rad}=0.75t^{2}\ \Rightarrow\ t^{2}=1.0472,\ t=1.0233\ \text{s}$$

Quantities at this instant

$$\begin{aligned} r&=1-0.3t^{2}=0.6858\ \text{m}\\ \dot r&=-0.6t=-0.6140\ \text{m/s},\qquad \ddot r=-0.6\ \text{m/s}^2\\ \dot\theta&=1.5t=1.5350\ \text{rad/s},\qquad \ddot\theta=1.5\ \text{rad/s}^2 \end{aligned}$$

(a) Total velocity

$$\begin{aligned} v_r&=\dot r=-0.6140\ \text{m/s}\\ v_\theta&=r\dot\theta=0.6858(1.5350)=1.0528\ \text{m/s}\\ v&=\sqrt{v_r^{2}+v_\theta^{2}}=1.219\ \text{m/s} \end{aligned}$$

The velocity points $59.8^\circ$ from the arm direction towards O, i.e. $120.3^\circ$ from $\hat e_r$.

(b) Total acceleration

$$\begin{aligned} a_r&=\ddot r-r\dot\theta^{2}=-0.6-0.6858(1.5350)^{2}=-2.216\ \text{m/s}^2\\ a_\theta&=r\ddot\theta+2\dot r\dot\theta=0.6858(1.5)+2(-0.6140)(1.5350)=-0.856\ \text{m/s}^2\\ a&=\sqrt{2.216^{2}+0.856^{2}}=2.376\ \text{m/s}^2 \end{aligned}$$

(c) Acceleration relative to the arm

The collar moves along the arm, so relative to the arm (a rotating frame) its motion is rectilinear along $r$:

$$a_{rel}=\ddot r=-0.6\ \text{m/s}^2$$

That is $0.6$ m/s$^2$ directed towards O along the arm.

Answer: (a) $v=1.219$ m/s; (b) $a=2.376$ m/s$^2$; (c) $a_{rel}=0.6$ m/s$^2$ towards O.

A particle moving on a plane curve is located by polar coordinates $(r,\theta)$. Using the unit vectors $\hat e_r$ (along the radius vector) and $\hat e_\theta$ (perpendicular to it, in the direction of increasing $\theta$), velocity and acceleration are found by differentiating $\vec r = r\hat e_r$.

        y
        |        P
        |       /|
        |    r / | e_theta
        |     /  |
        |    /   -> e_r along OP
        |   /\ theta
        +--------------- x
        O

Unit-vector derivatives

$$\hat e_r=\cos\theta\,\hat i+\sin\theta\,\hat j,\qquad \hat e_\theta=-\sin\theta\,\hat i+\cos\theta\,\hat j$$

Differentiating with respect to time:

$$\dot{\hat e}_r=\dot\theta\,\hat e_\theta,\qquad \dot{\hat e}_\theta=-\dot\theta\,\hat e_r$$

Velocity

$$\vec v=\frac{d}{dt}(r\hat e_r)=\dot r\,\hat e_r+r\dot\theta\,\hat e_\theta$$

So $v_r=\dot r$ and $v_\theta=r\dot\theta$.

Acceleration

$$\begin{aligned} \vec a=\frac{d\vec v}{dt}&=\ddot r\,\hat e_r+\dot r\,\dot{\hat e}_r+(\dot r\dot\theta+r\ddot\theta)\hat e_\theta+r\dot\theta\,\dot{\hat e}_\theta\\ &=\ddot r\,\hat e_r+\dot r\dot\theta\,\hat e_\theta+(\dot r\dot\theta+r\ddot\theta)\hat e_\theta-r\dot\theta^{2}\hat e_r \end{aligned}$$

Collecting terms:

$$\vec a=(\ddot r-r\dot\theta^{2})\,\hat e_r+(r\ddot\theta+2\dot r\dot\theta)\,\hat e_\theta$$

• Radial component: $a_r=\ddot r-r\dot\theta^{2}$
• Transverse component: $a_\theta=r\ddot\theta+2\dot r\dot\theta$
• Magnitude: $a=\sqrt{a_r^{2}+a_\theta^{2}}$

Here $-r\dot\theta^2$ is the centripetal term and $2\dot r\dot\theta$ is the Coriolis term. For motion on a circle ($r$ constant) $a_r=-r\dot\theta^2$ and $a_\theta=r\ddot\theta$.

Assumption (figure not shown): the slope at A is $3:4$, so $\tan\theta=3/4$, $\cos\theta=0.8$, $\sin\theta=0.6$. The 30 m/s launch is along the slope, upward. B is 25 m below A. Take $g=9.81\ \text{m/s}^2$ and neglect air resistance.

Components of launch velocity

$$v_x=30(0.8)=24\ \text{m/s},\qquad v_y=30(0.6)=18\ \text{m/s}$$

Time of flight

Vertical motion from A to B ($y=-25$ m):

$$\begin{aligned} -25&=18t-\tfrac12(9.81)t^{2}\\ 4.905t^{2}-18t-25&=0\\ t&=\frac{18+\sqrt{18^{2}+4(4.905)(25)}}{2(4.905)}=4.744\ \text{s} \end{aligned}$$

Range

$$R=v_x t=24(4.744)=113.86\ \text{m}$$

Answer: time of flight $t\approx 4.74$ s; horizontal range $R\approx 113.9$ m.

The car moves on a circle with constant speed, so the speed is entirely transverse: $v=r\dot\theta$.

Angular rate

$$\dot\theta=\frac{v}{r}=\frac{30}{400}=0.075\ \text{rad/s}$$

Acceleration

With $r$ constant and $v$ constant: $\dot r=\ddot r=0$ and $\ddot\theta=0$.

$$\begin{aligned} a_r&=\ddot r-r\dot\theta^{2}=-400(0.075)^{2}=-2.25\ \text{m/s}^2\\ a_\theta&=r\ddot\theta+2\dot r\dot\theta=0 \end{aligned}$$

So the acceleration is purely centripetal, $a=\dfrac{v^2}{r}=\dfrac{30^2}{400}=2.25\ \text{m/s}^2$, directed towards the centre.

Answer: $\dot\theta=0.075$ rad/s; $a=2.25$ m/s$^2$.

Once released, the water moves as a projectile with the plane's horizontal velocity and zero initial vertical velocity.

Data

$$v_x=315\ \text{km/h}=\frac{315}{3.6}=87.5\ \text{m/s},\qquad h=80\ \text{m},\qquad g=9.81\ \text{m/s}^2$$

Time to fall 80 m

$$h=\tfrac12 g t^{2}\ \Rightarrow\ t=\sqrt{\frac{2(80)}{9.81}}=4.0386\ \text{s}$$

Horizontal distance

$$d=v_x t=87.5(4.0386)=353.4\ \text{m}$$

Answer: the water must be released $d\approx 353$ m before the fire (horizontal distance).

On a circular path the acceleration has a tangential part $a_t=\dot v$ and a normal part $a_n=v^2/\rho$.

Speed at $t=4$ s

$$v=v_0+\int_0^{4}0.4t\,dt=4+0.2t^{2}=4+0.2(4)^{2}=7.2\ \text{m/s}$$

Acceleration at $t=4$ s

$$\begin{aligned} a_t&=0.4(4)=1.6\ \text{m/s}^2\\ a_n&=\frac{v^{2}}{\rho}=\frac{7.2^{2}}{50}=1.0368\ \text{m/s}^2\\ a&=\sqrt{a_t^{2}+a_n^{2}}=\sqrt{1.6^{2}+1.0368^{2}}=1.907\ \text{m/s}^2 \end{aligned}$$

Answer: $v=7.2$ m/s; $a=1.91$ m/s$^2$.

Assumption: the radar gun at O is at the centre of the circular road, so the radius to the car is constant, $r=250$ m. Then $\dot r=\ddot r=0$ (the angle $40^\circ$ is not needed for the magnitudes).

Velocity

$$v_r=\dot r=0,\qquad v_\theta=r\dot\theta=250(0.15)=37.5\ \text{m/s}$$ $$v=37.5\ \text{m/s}$$

Acceleration

$$\begin{aligned} a_r&=\ddot r-r\dot\theta^{2}=0-250(0.15)^{2}=-5.625\ \text{m/s}^2\\ a_\theta&=r\ddot\theta+2\dot r\dot\theta=250(0.025)=6.25\ \text{m/s}^2\\ a&=\sqrt{5.625^{2}+6.25^{2}}=8.409\ \text{m/s}^2 \end{aligned}$$

Answer: $v=37.5$ m/s; $a=8.41$ m/s$^2$.

The acceleration depends on velocity, so use $a=\dfrac{dv}{dt}$ and then $v=\dfrac{ds}{dt}$.

Velocity after 3 s

$$\begin{aligned} \frac{dv}{dt}&=-0.5v^{3}\\ \int_{80}^{v}\frac{dv}{v^{3}}&=-0.5\int_0^{t}dt\\ \left[-\frac{1}{2v^{2}}\right]_{80}^{v}&=-0.5t\\ \frac{1}{v^{2}}&=\frac{1}{80^{2}}+t \end{aligned}$$

At $t=3$ s:

$$\frac1{v^2}=\frac1{6400}+3=3.000156\ \Rightarrow\ v=0.5773\ \text{m/s}$$

Position after 3 s

$$v=\frac{ds}{dt}=\left(\frac1{6400}+t\right)^{-1/2}$$ $$s=\int_0^{3}\left(\frac1{6400}+t\right)^{-1/2}dt=2\left[\sqrt{\tfrac1{6400}+t}\right]_0^{3}=2(1.7321-0.0125)=3.439\ \text{m}$$

Answer: $v\approx0.577$ m/s and $s\approx3.44$ m at $t=3$ s.

Treat the ball as a projectile. Use ft units with $g=32.2\ \text{ft/s}^2$.

Equations of motion (origin at A)

$$x=(v_A\cos30^\circ)t,\qquad y=(v_A\sin30^\circ)t-\tfrac12 g t^{2}$$

At B: $x=15$ ft, $y=-9$ ft. Eliminating $t$:

$$y=x\tan\theta-\frac{g x^{2}}{2v_A^{2}\cos^{2}\theta}$$ $$-9=15\tan30^\circ-\frac{32.2(15)^{2}}{2v_A^{2}\cos^{2}30^\circ}$$ $$-9=8.660-\frac{4830}{1.5\,v_A^{2}}\ \Rightarrow\ \frac{3220}{v_A^{2}}=17.660$$ $$v_A^{2}=273.5\ \Rightarrow\ v_A=16.54\ \text{ft/s}$$

Check: $t=\dfrac{15}{16.54\cos30^\circ}=1.047$ s.

Answer: the ball is kicked at $v_A\approx16.5$ ft/s (time of flight $\approx1.05$ s).

Relative velocity

If two particles A and B have velocities $\vec v_A$ and $\vec v_B$ (both measured from a fixed frame), the relative velocity of A with respect to B is the velocity of A as seen by an observer moving with B:

$$\vec v_{A/B}=\vec v_A-\vec v_B$$

Equivalently, $\vec v_A=\vec v_B+\vec v_{A/B}$. It follows from the relative position $\vec r_{A/B}=\vec r_A-\vec r_B$ by differentiating.

Relative acceleration

In the same way the relative acceleration of A with respect to B is

$$\vec a_{A/B}=\vec a_A-\vec a_B$$

(valid when the observer on B translates without rotating). Note $\vec v_{B/A}=-\vec v_{A/B}$.

Example

A car A moves east at 60 km/h and a car B moves north at 80 km/h. The velocity of A relative to B is

$$\vec v_{A/B}=60\hat i-80\hat j\ \text{km/h},\qquad |v_{A/B}|=\sqrt{60^2+80^2}=100\ \text{km/h}$$

at $\tan^{-1}(80/60)=53.1^\circ$ south of east. A passenger in B sees A moving away at 100 km/h in that direction. If A accelerates at $2\hat i$ m/s$^2$ and B at $1\hat j$ m/s$^2$, then $\vec a_{A/B}=2\hat i-1\hat j$ m/s$^2$.

Assumption (figure not shown): the cord makes $\theta=40^\circ$ with the vertical, and the tension is $T=2.5W$. The bob moves on a circle of radius $L=2$ m, so use normal and tangential components.

          O
          |\
          | \  L = 2 m
          |40\
          |   \
                O  bob (W)

Normal direction (towards O)

$$T-W\cos\theta=\frac{W}{g}\frac{v^{2}}{L}$$ $$v^{2}=gL\,(2.5-\cos40^\circ)=9.81(2)(2.5-0.7660)=34.02\ \text{m}^2/\text{s}^2$$ $$v=5.833\ \text{m/s}$$

Accelerations

$$a_n=\frac{v^{2}}{L}=\frac{34.02}{2}=17.01\ \text{m/s}^2$$

Tangential direction (only the weight component acts):

$$W\sin\theta=\frac{W}{g}a_t\ \Rightarrow\ a_t=g\sin40^\circ=6.306\ \text{m/s}^2$$ $$a=\sqrt{a_n^{2}+a_t^{2}}=\sqrt{17.01^{2}+6.306^{2}}=18.14\ \text{m/s}^2$$

The acceleration makes $\tan^{-1}(6.306/17.01)=20.3^\circ$ with the cord.

Answer: $v=5.83$ m/s; $a=18.14$ m/s$^2$ ($a_n=17.01$, $a_t=6.31$ m/s$^2$).

Take $g=9.81\ \text{m/s}^2$ and the origin at P, with $y$ upward.

Launch components

$$v_x=120\cos30^\circ=103.92\ \text{m/s},\qquad v_{y0}=120\sin30^\circ=60\ \text{m/s}$$

i) Maximum height above the level of projection

At the top $v_y=0$:

$$H=\frac{v_{y0}^{2}}{2g}=\frac{60^{2}}{2(9.81)}=183.49\ \text{m}$$

ii) Velocity on striking the target ($y=-100$ m)

$$v_y^{2}=v_{y0}^{2}+2g(100)=3600+1962=5562\ \Rightarrow\ v_y=74.58\ \text{m/s (downward)}$$ $$v=\sqrt{103.92^{2}+74.58^{2}}=127.9\ \text{m/s}$$

Direction: $\tan^{-1}(74.58/103.92)=35.7^\circ$ below the horizontal.

iii) Total time of flight

$$t=\frac{v_{y0}+|v_y|}{g}=\frac{60+74.58}{9.81}=13.72\ \text{s}$$

Answer: (i) $H=183.5$ m; (ii) $v=127.9$ m/s at $35.7^\circ$ below horizontal; (iii) $t=13.72$ s.