TITLE: Calculating the acceleration of an object on the end of a lever.
QUESTION [1 upvotes]: I have a lever that is 16 feat long and pivots on a fixed point 4 ft from the left/heavy end, and 12ft from the right/light end.  I wanted to calculate the acceleration of the right end due to gravity.  The right end ways 10 lbs.  The left end weighs 1000 lbs.
I calculated the acceleration, but I am skeptical of my accuracy, and was wondering if I did it correctly.  What I did was this:
I calculated the force on each side(unit: lb*ft/s^2):
 - F = 1000*32ft/s^2 = 32000
 - (force) = (mass, 1000lbs) * acceleration due to gravity)
 - F = 70*32ft/s^2 = 2240
 - (70 because I am accounting for the weight of the beam, which overall is about 90lbs)
Because of the mechanical advantage of the right side, I multiplied the force of the left side by .3 (3.5ft/11.5ft) (the weight would really be attached approximately 6 inches from the end of the board), which gave me 9600lb*ft/s^2.
Then, since these forces are on opposite ends of the pivot point, they counter each other, so, I subtracted the answers, and got 9600-2240 = 7360lb*ft/s^2. So since I want acceleration, which is measured in ft/s^2, I divided by the mass of the light end. 7360/70 = 105.14286 ft/s^2.  Did I reach the answer correctly?  It's been a while since I took physics :).
Note: I am aware I did have a few inaccuracies in rounding, and not accounting for the mass of the beam on the left side, I am more concerned that in principal, I used valid math.
Also, once I have the acceleration, if I wanted to figure out how fast it would be going in say... 3 seconds, I would just do: 105*3^2 = 105*9 = 945?(Which obvoiusly doesn't acc

REPLY [2 votes]: First of all, you were given the weights, which already are forces; you do not multiply by acc due to gravity.  Anyway, the following is what I think is a standard method of attack in physics:
The net torque $\tau$ on the lever is $(1000) (4) - (10) (12) = 3880 \,\text{ft} \,\text{lb}$.  We find the angular acceleration by dividing the torque by the moment of inertia $I$, which is split into two components: $I_w$ (due to the weights on the ends), and $I_b$ (the uniform load)
$$I_w = [(1000) (4^2) + (10) (12^2)]/32.2 \approx 541.6 \, \text{ft}\, \text{lb} \, \text{sec}^2 $$
$$I_b = \left[\frac{1}{12} (97) (16^2) + (97) (4^2)\right]/32.2 \approx 112.5 \, \text{ft}\, \text{lb} \, \text{sec}^2$$
(The second term is an addition due to the parallel axis theorem; the lever has an off-center fulcrum.  Also, the factor of $32.2$ is acc due to gravity.)  The angular acceleration of the lever is then
$$\alpha = \frac{\tau}{I} \approx 5.93/\text{sec}^2 $$
The linear acceleration of the right end is then the length of the right arm times the angular acceleration, or about $(5.93)(12) \approx 71.2 \,\text{ft}/\text{sec}^2 $.


TITLE: Recurrence Relations, understanding definition of formula
QUESTION [0 upvotes]: I have been trying to practice specific discrete math questions in my textbook and am having issues understanding how the terms are defined for a formula. Two examples that I do not understand.

(1) Find a recurrence relation for $C_n$, the number of ways to parenthesize the product of $n + 1$ numbers,
  $x_1, x_2, \ldots, x_n$, to specify the order of multiplication. e.g. $C_3 = 5$ there are five ways to parenthesize $x_0 \cdot x_1 \cdot x_2 \cdot x_3$ to determine the order of multiplication:

$((x_0 · x_1) · x_2) · x_3$
$(x_0 · (x_1 · x_2)) · x_3$
$(x_0 · x_1) · (x_2 · x_3)$
$x_0 · ((x_1 · x_2) · x_3)$
$x_0 · (x_1 · (x_2 · x_3))$.

(2) Find a recurrence relation and give initial conditions for the number of bit strings of length n
  that do not have two consecutive 0s. How many such bit strings are there of length five?

For the first one, I don't understand how to approach this kind of question other than to set one pair of brackets and set the other to be completely within it or completely outside of it. But I am stuck beyond that.
For the second problem, I know that you are supposed to find the number of bitstrings that end in $1$ or $10$ and then add them. However, I do not really understand why this is the case and how this formula is achieved and further, why the number of bitstrings with two consecutive zeros would be this number plus $2^{n-2}$? It doesn't make a lot of sense to me.
Any help understanding or pointing me toward resources would be great I would really like to understand this. Thank you.

REPLY [1 votes]: We are trying to count the number of ways to parenthesize $x_1 \cdot \cdot \cdot x_n $.  This number is $ C_n $.  But we can rewrite this as $ x_1 \cdot (x_2 \cdot \cdot \cdot x_n )$. There are then $ C_{n-1} $ ways to parenthesize the stuff in the parentheses. But we can also rewrite the original product as $ (x_1 \cdot x_2) \cdot ( x_3 \cdot \cdot \cdot x_n ) $. Now there are $ C_2 $ ways to parenthesize the left and $ C_{n-2} $ ways to parenthesize the right.  Keep going.  Do you see how we might continue?  We can also start with $ (x_1 \cdot x_2 \cdot x_3) \cdot (x_4 \cdot \cdot \cdot x_n)$ which admits $ C_3 \cdot C_{n-3} $ sub-parenthesizations.  Continuing in this fashion, we can count all $ C_n $ ways of parenthesizing $ x_1 \cdot \cdot \cdot x_n $ in terms of the quantities $ C_1, \ldots, C_{n-1} $.


TITLE: How to find the center of a particle moving in uniform circular motion
QUESTION [0 upvotes]: At time $t=0$ a particle in uniform circular motion in the plane has velocity $\vec{v}=6\vec{i}-4\vec{j}$ and acceleration $\vec{a}=2\vec{i}+3\vec{j}$. Find the radius and center of its orbit if at time $t=0$ it is at the point $P=(0,0)$.
The radius of its orbit is found using the formula, $R=\frac{\left\|\vec{v}\right\| ^{2}}{\left\|\vec{a}\right\|}$.
Thus, $R=\frac{52}{\sqrt{13}}$.
How to find the center of its orbit?

REPLY [0 votes]: Hint:
The centre will lie along the acceleration vector.Thus,centre(C) will be $(R\cos A,R\sin A )$,where A is the angle which PC makes with positive $x$-axis.