## Section5.6Circular Motion Bootcamp

### Subsection5.6.6Miscellaneous

Figure 5.6.30 shows a pebble P stuck at the outside of a tire. As the tire rolls, the pebble's position in space changes. Suppose tire does not slide but rolls smoothly. The tire has radius $R$ and is rolling at a constant speed $v_0 \text{.}$ At $t=0\text{,}$ the pebble was at the origin.

Find the expressions for (a) position, (b) velocity, and (c) acceleration vectors (i.e., magnitudes and directions) at an arbitraru instant $t$ with respect to the coordinate system given in the figure.

Hint

(a) Introduce another coordinate system with origin at the center of the tire at all times and $x$ and $y$ axes parallel to the one given. Find position in that coordinate system first. Then translate that to the given coordinate system. (b) and (c): use derivatives.

(a) $\vec r = (-R\sin\theta + v_0 t)\; \hat i + (-R\cos\theta)\; \hat j\text{,}$ (b) $((1 -\cos\theta)\;v_0 , (-\sin\theta)v_0)\text{,}$ (c) $\text{.}$

Solution 1 (a)

(a) Let us introduce a coordinate system as shown in Figure 5.6.31. For convenience, I will define $\theta$ that measures angle from negative $y$ axis as shown.

Let $(x,y)$ and $(x',y')$ be the coordinates of P at this arbitrary instant $t$ with respect to the two coordinate systems. It is immediately clear from the figure that

\begin{align*} \amp y = y',\ \ x = x' + v_0 t.\\ \amp x' = -R\sin\theta,\ \ y'=-R\cos\theta. \end{align*}

Therefore, position $\vec r$ in original coordinate system will be

\begin{equation*} \vec r = (-R\sin\theta + v_0 t)\; \hat i + (-R\cos\theta)\; \hat j. \end{equation*}

It magnitude is

\begin{align*} r \amp = \sqrt{ x^2 + y^2 }\\ \amp = \sqrt{(-R\sin\theta + v_0 t)^2 + (-R\cos\theta)^2 }\\ \amp = \sqrt{R^2 - (2Rv_0\sin\theta) t + v_0^2t^2 }. \end{align*}

We can indicate direction in space by angle with respect to $x$ axis by using arctangent.

\begin{equation*} \theta_x = \tan^{-1}(y/x) =\tan^{-1}\left( \frac{-R\cos\theta}{-R\sin\theta + v_0 t} \right). \end{equation*}
Solution 2 (b)

(b) Now, that we know the components of position, we can obtain components of velocity by taking derivatives. Thus,

\begin{align*} \amp v_x = \frac{dx}{dt} = (-R\cos\theta)\frac{d\theta}{dt} + v_0 \\ \amp v_y = \frac{dy}{dt} = (-R\sin\theta)\frac{d\theta}{dt} \end{align*}

Since the tire is not slipping, the rate at which angle $\theta$ changes is related to the speed with which center of the tire is moving. The tire has rotated by $R\Delta\theta\text{,}$ its center has moved $v_0\Delta t\text{.}$ Equating them we get

\begin{equation*} R\Delta\theta = v_0\Delta t. \end{equation*}

This gives

\begin{equation*} \frac{d\theta}{dt} = \frac{v_0}{R}. \end{equation*}

Using this, we rewrite the components of velocity as

\begin{align*} \amp v_x = (1 -\cos\theta)\;v_0 \\ \amp v_y = (-\sin\theta)v_0. \end{align*}

Therefore, magnitude of velocity, i.e., speed is

\begin{equation*} v = \sqrt{v_x^2 + v_y^2} = v_0\sqrt{ 2 - 2 \cos\theta} = 2 v_0 \left| \sin(\theta/2) \right|. \end{equation*}

Note: this tells us that the speed of pebble when touching the ground is zero (since $\theta=0$) and when at the very top, it is $(2\;v_0)$ (since $\theta=\pi$). That is, speed of the pebble in $Oxy$ coordinate system changes as it goes around.

The direction of velocity can again be given with respect to the $x$ axis by using arctangent.

\begin{equation*} \theta_x = \tan^{-1}(v_y/v_x) = \tan^{-1}\left( \frac{-\sin\theta}{1 -\cos\theta} \right). \end{equation*}

You can further simplify this expression by using double angle formulas. Try that. You should be able to express this without arctangent operation.

Solution 3 (b)

(c) Now, that we know the components of velocity, we can obtain components of acceleration by taking derivatives. Thus,

\begin{align*} \amp a_x = \frac{dv_x}{dt} = (-v_0 \sin\theta)\frac{d\theta}{dt} \\ \amp a_y = \frac{dv_y}{dt} = (-v_0\cos\theta)\frac{d\theta}{dt} \end{align*}

Replacing $d\theta/dt$ by $v_0/R$ as found in (b), we get

\begin{equation*} a_x = -\frac{v_0^2}{R}\;\sin\theta;\ \ a_y = -\frac{v_0^2}{R}\;\cos\theta. \end{equation*}

Therefore, magnitude of acceleration is

\begin{equation*} a = \sqrt{a_x^2 + a_y^2} = \frac{v_0^2}{R}. \end{equation*}

The direction with respect to $x$ axis will again be obtained by using arctangent.

\begin{equation*} \theta_x = \tan^{-1}(a_y/a_x) = \tan^{-1}(\cot(\theta)). \end{equation*}

Of course, we can write $cot(\theta)$ as $\tan(\pi/2 - \theta)\text{.}$ Hence

\begin{equation*} \theta_x = \frac{\pi}{2} -\theta. \end{equation*}

This is actually direction towards the moving center of the tire. Take values of $\theta=0, \pi/2, \pi, 3\pi/2, 2\pi$ and draw them on the corresponding $Oxy$ plane. You will see that the direction is always towards the center of the tire as shown in Figure 5.6.32.