Wednesday, 13 December 2017

newtonian mechanics - A simple pendulum


The idealized simple pendulum model (see the following figure) assumes that at every point of time the string to which the bob is attached exerts an equal and opposite force on the bob as does the component of the force of gravity acting on the bob along the line determined by the string. Is this assumption the mathematical way of modeling the string, or can it be derived from the (in my opinion more intuitive) assumption that the bob's movement is constrained to take place along the circumference of the circle centered at the string's axis whose radius is the string's length?





Here's how I have attempted to solve this problem. Unfortunately, this solution makes no sense, but I can't see where my mistake is.


Let $\mathbf{p} : \mathbb{R} \rightarrow \mathbb{R}_{2, 1}$ be a twice differentiable function from $\mathbb{R}$ to the space of real $2 \times 1$ column vectors. $\mathbf{p}$ will represent the location of the bob in a two-dimensional space. Denoting the length of the string by $L$ ($L > 0$) and assuming the string's axis coincides with the origin, the model's assumptions reduce to the following equation: $$ \|\mathbf{p}\|^2 = L^2 $$ Using this notation, we wish to show that the force acting on the point-mass is entirely perpendicular to the line determined by the string. Since this force is proportional to the mass's acceleration, this reduces to the following equation: $$ \mathbf{p}^T\ddot{\mathbf{p}} = 0 $$


By differentiating both sides of the assumption, we get $$ \mathbf{p}^T \dot{\mathbf{p}} = 0 $$


By differentiating both side of the last equation, we get $$ \|\dot{\mathbf{p}}\|^2 + \mathbf{p}^T\ddot{\mathbf{p}} = 0 $$


Therefore, if the proposition that we wish to prove is true, the last equality implies $\dot{\mathbf{p}} = \mathbf{0}$, i.e. the pendulum is motionless. However, this is evidently false in general. So what did I do wrong?



Answer



It is not true that "the string to which the bob is attached exerts an equal and opposite force on the bob as does the component of the force of gravity acting on the bob along the line determined by the string". The string's tension is not equal, in general, to the component of gravity along the line determined by the string. If you look closely at the picture you linked to, you'll see that the two arrows are not the same length. Also consult the following gif.


No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...