# How to solve this pendulum problem using kinematics, not the principles of conservation of energy?

I have this question, because typically problems that can be solve using conservation of energy or just energy-related principles, can usually be solved sing kinematic equations. (At least is what I saw in my book)

I am in a introductory physics course, just started. I don't exactly know how pendulum works, but do have some knowledge of tension and therefore vertical circular motion, that I think is somewhat related.

Ok, the questions goes:

A 0.02 kg pendulum bob is released from position A. As the pendulum passes position C, located 1.25 m below position A, what is its velocity?

Well, obviously the velocity varies at different position on the circular path. The cause for this seems to be the disparity in the magnitude of the centripetal force at different points. $\frac {mv^2}{r} = T - mg$ at position C. $\frac {mv^2}{r} = T - mgCos\theta$ at point A.

update Now I know that the acceleration is just that of gravity but why? The force of gravity is always vertical, but the velocity vector is not.

(Used $V_f^2 = V_i^2 + 2gd$, initial velocity, at point A, is 0 meter per second, g is positive)

• The acceleration of the system is not $g$. It changes throughout the trajectory of the pendulum. Although this problem can certainly be tackled with dynamics, it is much easier in this case to use conservation of energy. Commented Jul 23, 2015 at 2:07
• @Ultima, would you like to answer it? Commented Jul 23, 2015 at 2:23

# Energy kinematics

I have this question, because typically problems that can be solve using conservation of energy or just energy-related principles, can usually be solved sing kinematic equations.

Yep. In fact, there are two profound pieces of math, Hamiltonian and Lagrangian dynamics, which say that you can use energies to derive the actual kinematic equations for a system.

# Vectors

Well, obviously the velocity varies at different position on the circular path. The cause for this seems to be the disparity in the magnitude of the centripetal force at different points.

Kind of, but not really. What you are going to have to start appreciating is that velocity and acceleration are vectors and have both a magnitude and direction. The usual way is to write a vector in terms of its "components", so e.g. $\vec r = \left[\begin{array}{c}x \\ y\end{array}\right]$ is a vector with one component $x$ and the other component $y$. We say that it lives in $\mathbb R^2$, the vector space of pairs of real numbers, and has magnitude $|\vec r| = \sqrt{x^2 + y^2}$, and it points in a particular direction: it points from the point $(0, 0)$ to the point $(x, y)$.

There are a couple operations which "transform properly", meaning that they make sense in all coordinate systems. One of them is the vector sum $$\vec r_1 + \vec r_2 = \left[\begin{array}{c}x_1 \\ y_1\end{array}\right] + \left[\begin{array}{c}x_2 \\ y_2\end{array}\right] = \left[\begin{array}{c}x_1 + x_2 \\ y_1 + y_2\end{array}\right].$$ Another is the dot product $$\vec r_1 \cdot \vec r_2 = x_1 ~ x_2 + y_1 ~ y_2.$$ This turns out to be the number $\vec r_1 \cdot \vec r_2 = |\vec r_1| ~ |\vec r_2| \cos\theta$ where $\theta$ is the angle that the two vectors make with each other. A third one is vector scaling, $$k \vec v = \left[\begin{array}{c}k ~ x \\ k ~ y\end{array}\right].$$So that's useful.

Finally, many spaces have something called an "orientation" and thus an "antisymmetric product"; in $\mathbb R^2$ this is the number $x_1 y_2 - y_1 x_2$ which turns out to be $|\vec r_1|~|\vec r_2|~\sin\theta$; in $\mathbb R^3$ it turns out to be another vector in $\mathbb R^3$; in $\mathbb R^4$ it turns out to be two vectors in $\mathbb R^3$ or one antisymmetric $4 \times 4$ matrix, when you learn what that is.

# Vector calculus of circular motion

Many introductory physics courses do not build on a bedrock of calculus. You may need to fight against this tendency, if you don't know calculus. The basic idea is that when you zoom in on a point, a lot of curves look like straight lines, defined by that point and a slope. The slope-function of a function $f(x)$ is called the "derivative", written either as $f'(x)$ or $\frac{df}{dx}$. Mathematically we would say that if $\delta x$ is very small, then $f(x + \delta x) \approx f(x) + f'(x) ~\delta x.$ So for example $(x + \delta x)^3 = x^3 + 3 x^2 ~ \delta x + 3 x (\delta x)^2 + (\delta x)^3$, and we can say that "to first order" in $\delta x$, this is $x^3 + 3 x^2 ~\delta x,$ so the derivative of $x^3$ is $3 x^2$. Derivatives turn out to obey a couple of useful rules based on how differences work, including the sum rule $\delta(f + g) = \delta f + \delta g$, the product rule $\delta (f \cdot g) = f \cdot \delta g + \delta f \cdot g$, and the chain rule: let $f \circ g$ be the function composition $x \mapsto f(g(x))$, then $\delta (f \circ g) = (f' \circ g) \delta g$. You'll also learn things like $\sin' = \cos$ and $\cos' = -\sin$ when the angles are measured in radians.

If that didn't make sense, go and learn some calculus. :)

Now, basic trigonometry. A point on a circle can be described by the vector position $$\vec r = \left[\begin{array}{c}r~\cos\theta \\ r~\sin\theta \end{array}\right].$$The time derivative of a position is a velocity; in this case it is (using $\dot f = \frac{df}{dt}$)$$\vec v = \dot r ~\left[\begin{array}{c}\cos\theta \\ \sin\theta \end{array}\right] + r ~\dot \theta~ \left[\begin{array}{c}-\sin\theta \\ \cos\theta \end{array}\right]$$and the time-derivative of the velocity is the acceleration:$$\vec a = \left(\ddot r - r ~ \dot\theta^2 \right)~\left[\begin{array}{c}\cos\theta \\ \sin\theta \end{array}\right] + \left(2 \dot r ~\dot \theta + r \ddot \theta \right)~ \left[\begin{array}{c}-\sin\theta \\ \cos\theta \end{array}\right]$$Only one of these terms is the $v^2/r$ term that you're using: assuming as in your case that $\dot r = 0 = \ddot r$ we have $v = |\vec v| = r~|\dot\theta|$, so the $v^2/r$ term is the $-r ~ \dot\theta^2$ term, the "centrifugal acceleration." The remaining term $r~\ddot\theta$ is the "tangential acceleration".

# Newton's laws

Newton's laws say that the sum of forces equals the mass times the acceleration; here the tension force will have magnitude $m v^2/r$ to keep it on the circle, so all we care about is the tangential acceleration, the centrifugal acceleration is "balanced out by the constraints". Defining $\phi = \theta + (\pi/2)$ so that $\phi = 0$ is $(x, y) = (0, -r)$ the general expression is:$$m ~ \vec a = m \left(r~\dot\phi^2 \left[\begin{array}{c}-\sin\phi \\ \cos\phi \end{array}\right] + r~\ddot\phi \left[\begin{array}{c}\cos\phi \\ \sin\phi \end{array}\right]\right) = T~\left[\begin{array}{c}-\sin\phi \\ \cos\phi \end{array}\right] + m ~g~\left[\begin{array}{c}0 \\ -1\end{array}\right]$$ The trick for the last term involves the Pythagorean theorem for trigonometry, $(\sin\theta)^2 + (\cos\theta)^2 = 1$. It says that $$\left[\begin{array}{c}0 \\ -1\end{array}\right] = -\cos\phi ~ \left[\begin{array}{c}-\sin\phi \\ \cos\phi \end{array}\right] - \sin\phi \left[\begin{array}{c}\cos\phi \\ \sin\phi \end{array}\right]$$So the relevant kinematic equation is simply $$m r~\ddot\phi = - m g \sin\phi.$$

Unfortunately to actually integrate this all the way you'll need elliptic integrals, which are beyond the scope of this comment. However, we can definitely integrate once with respect to time. To solve this, you can multiply both sides by $\dot \phi$ with the observation that $\frac{d}{dt} [\frac 12 ~\dot\phi^2 ] = \dot\phi~\ddot\phi$, so one antiderivative is$$\frac 12 ~\dot\phi^2 - \frac 12 ~\dot\phi_0^2 = \frac{g}{r} ~\left(\cos\phi - \cos\phi_0\right).$$

Of course, now you can multiply by r^2 to find the anticlimactic result $$\frac 12 ~v^2 - \frac 12 ~v_0^2 = -g ~\left(y - y_0\right)$$which is the exact same as you got doing it with an energy balance.

• I don't know anything about calculus. Do I need to know any calculus in order to understand what you explained in "basic trigonometry"? (Because I can't figure out what those symbols mean...) Commented Jul 24, 2015 at 1:46
• @Doeser You need calculus to understand everything that has a dot above it. The rest is linear algebra.
– Omry
Commented Jul 24, 2015 at 11:07
• @Doeser: $\theta,\phi$ are angles measured in radians; $\sin$ and $\cos$ are ratios of sides of right triangles defined in trigonometry. What you need calculus for is to understand how these "derivatives" work, what "rules" they follow. So for example, one calculation is $\frac d{dt}(r ~\dot\theta~\cos\theta)$ which is $\frac{dr}{dt}~\dot\theta~\cos\theta+r~\frac{\dot\theta}{dt}~\cos\theta+r~(\dot\theta)~\frac{d}{dt}\cos\theta$, and then the last term needs the "chain rule," to get $\dot r~\dot\theta~\cos\theta+r~\ddot\theta~\cos\theta-r~\dot\theta(\dot\theta~\sin\theta).$ Commented Jul 24, 2015 at 13:05
• Was what you did simply a proof of the formula vf^2 = vi^2 -2gd.? Commented Jul 24, 2015 at 13:09
• @Doeser Whoa, there's some bug in the LaTeX and I can't edit anymore. Anyway, sort of: it is not a proof of that formula, but you can always derive similar formulas wherever they hold. It just happens to be the case that $g ~r ~\cos \phi = -y$ in these particular coordinates so that you get the same results when you try to learn the answer based on $r$ and $\phi$ that you get when you try to learn the answer based on energy conservation with $y$. Commented Jul 24, 2015 at 13:15

Assuming that the angle theta is measured relative to the vertical (e.g., the position of the string when the pendulum is at rest), a careful free body analysis indicates that the acceleration of the pendulum is g * sin(theta). This means that the acceleration of the pendulum continuously varies as it swings. This is relevant because the kinematic equations that you are using were developed based on the assumption of constant acceleration. Thus, the kinematic equations shouldn't be used for the pendulum problem.