[Physics] the Benefit of Plotting a Log of Net Force vs. Log of Acceleration Graph

accelerationdata analysisforcesnewtonian-mechanics

In preparation for a quiz, I've been going over a set of sample questions.

One question in particular asks that if I am to graph the logarithm of the horizontal net force of a system vs the logarithm of the horizontal acceleration, what does the slope and y-intercept of the graph mean?

This one has me quite puzzled. I can understand why you'd want to plot just net force against acceleration. I'm assuming the slope would be the mass and then your y-intercept is the amount of total force when the acceleration is equal to zero. However, I don't get why taking the log of each and graphing them against each other would really give you anything substantial. Any help would be greatly appreciated.

Best Answer

Suppose given an equation of the form $$y = Cx^n.$$ If you plot $y$ vs $x$ then it would be hard to find the value of $C$ or $n$ by looking at it. But if we take log on both sides we get

$$ log(y) = n ~log(x) + log(C).$$

Now plotting $log(y)$ vs $log(x)$ would give us a straight line with slope $n$ and intercept $log(C)$. In your question the relation is $F= ma$. So plotting $log(F)$ vs $log(a)$ would give a line of slope 1 and intercept of $log(m)$.

It may seem like a pointless thing to do for this case since we already know the relation. But plotting logarithms are useful when you know that two quantities are related by some power law and you want to find the exact equation from experimental data you have.

Related Solutions

[Physics] How to graph acceleration

I am more familiar with graphing in Python than in the language you are using. The following snippet of code produces the graph you are asking for:

import numpy as np
import matplotlib.pyplot as plt

# sci fi velocity graph
D = 160934400000     # m
v_i = 3070           # m/s
total_time = 40*3600 # seconds: 40 hours

# to cover half the distance in half the time, we solve
# x(t) = x(0) + v_0*t + 0.5*a*t*t
# x(0) = 0
# x(t) = D/2
# t = total_time / 2 = th
th = total_time / 2
xt=D/2
a = 2.0*(xt/th-v_i)/th

print 'acceleration needed is %.2f m/s^2'%a
# this prints out 30.96 m/s^2 - encouraging

# calculate position during acceleration phase - every second:
tm = 20*3600 # time at mid point
x=[]
for t in range(0,tm+1):
    x.append(v_i*t + 0.5 * a * t * t)

xm = x[-1] # position at mid point

for t in range(tm+1, 2*tm):
    x.append((v_i + tm*a)*(t-tm) - 0.5 * a * (t - tm) * (t - tm) + xm)

time = np.arange(0,2*tm)/3600.0

plt.close('all')
plt.figure()
plt.plot(time, np.array(x)/1000)
plt.xlabel('time (hr)')
plt.ylabel('position (km)')
plt.show()

v_t = np.diff(x)
plt.figure()
plt.plot(np.array(x[:-1])/1000,v_t/1000)
plt.xlabel('position (km)')
plt.ylabel('velocity (km/s)')
plt.show()

plt.figure()
plt.plot(time[:-1], v_t/1000)
plt.xlabel('time (hr)')
plt.ylabel('velocity (km/s)')
plt.show()

This produces the following graph:

enter image description here

You can also plot velocity as a function of time:

enter image description here

Or as a function of position:

enter image description here

[Physics] Horizontal and vertical acceleration of a ball on a slope

If velocity at bottom of slope is all you want then energy method in Gert's and Dr Xorille's answers is an elegant and easy way to get at it. However you seem to be tangled up with resolving vectors.

In working with vectors you should set up a coordinate system and stick to it until the end. If you have to set up more than one, and worse, have to switch between them often (as it is in your case), then you make sure that every component of a vector (in your case gravity vector $g$) is expressed in the same coordinate system.

In your example there are two coordinate systems: X-Y system in which X direction is parallel to ground and Y direction is perpendicular to ground and downward; P-Q system in which P direction is parallel to slope and Q direction is perpendicular to slope.

In X-Y system gravity vector $\textbf{g}\equiv(0,g)$ while in P-Q system $\textbf{g}\equiv(g\sin \theta,g\cos\theta)$. You may now solve the problem in any of the two coordinate systems of your choice.

What you have done is, you have calculated $\textbf{g}$'s components parallel and perpendicular to slope (i.e. in P-Q system) but for some reason you want to go back to X-Y system. That is fine but then you $\textit{must}$ take both components of $\textbf{g}$ in P-Q system and re-express them in X-Y system. Therefore you should not leave out the component normal to the slope, because both components, parallel and perpendicular to slope, together make the gravity vector $\textbf{g}$. Although doing this will simply return you to components of $\textbf{g}$ in X-Y system viz. $(0,g)$ and I don't see the point of it all; if you wanted to stay in X-Y system you could have done so right from the beginning.

Once you know components of $\textbf{g}$ in X-Y system you may now calculate velocity at any other instant by making use of the usual laws of motion.

Best Answer

Related Solutions

[Physics] How to graph acceleration

[Physics] Horizontal and vertical acceleration of a ball on a slope

Related Question