2011-02-16, 02:31 PM
Oh my god. I just had an epiphany. If F= -bv, then it makes sense to set this as a differential equation.
-b*r'(t) = m*r''(t)
0= r''(t) +b/m *r'(t)
0= r''(t)*e^(b/m *t) +b/m *r'(t)*e^(b/m *t)
0= ∫ r''(t)*e^(b/m *t) +b/m *r'(t)*e^(b/m *t) dt
<C1,C2> = r'(t)*e^(b/m *t)
r'(t)= e^(-b/m *t) < C1, C2>
r''(t)= < C1*(-b/m)*e^(-b/m *t), C2*(-b/m)*e^(-b/m *t) >
F= ma
F(t) = < -C1*b*e^(-b/m *t), -C2*b*e^(-b/m *t) >
From here though, I run into three problems. How do I go about finding C1 and C2, which arose from my integration? Also, I understand that I ended up with the same friction function for both the x and the y directions, because I used integrating factors, but what if the particle isn't moving from rest? Finally, this forcing function implies that the force due to friction actually decreases as time goes on, or am I interpreting this wrong?
-b*r'(t) = m*r''(t)
0= r''(t) +b/m *r'(t)
0= r''(t)*e^(b/m *t) +b/m *r'(t)*e^(b/m *t)
0= ∫ r''(t)*e^(b/m *t) +b/m *r'(t)*e^(b/m *t) dt
<C1,C2> = r'(t)*e^(b/m *t)
r'(t)= e^(-b/m *t) < C1, C2>
r''(t)= < C1*(-b/m)*e^(-b/m *t), C2*(-b/m)*e^(-b/m *t) >
F= ma
F(t) = < -C1*b*e^(-b/m *t), -C2*b*e^(-b/m *t) >
From here though, I run into three problems. How do I go about finding C1 and C2, which arose from my integration? Also, I understand that I ended up with the same friction function for both the x and the y directions, because I used integrating factors, but what if the particle isn't moving from rest? Finally, this forcing function implies that the force due to friction actually decreases as time goes on, or am I interpreting this wrong?