- #1

derravaragh

- 24

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## Homework Statement

A steel ball bearing falls through a viscous syrup. The equation of motion is

m(dv/dt) = mg - (alpha)v where g = 9.8 m/s^2. The terminal velocity is 0.2 m/s. (a) Determine the value of alpha. (b) Determine the speed at t = 0.04 s. [initial velocity = 0; mass m = 12.6E-3 kg] (c) Determine the distance that the ball bearing will fall from t=0 to t=0.1s.

## Homework Equations

ma = mg-(alpha)v

## The Attempt at a Solution

OK, so for this problem, I would like confirmation on my work, not necessarily the solution to the question. For part (a) I have alpha = (mg)/v_t where v_t is the terminal velocity. Part (b) is where I am unsure of my work. Using differential equations, I obtain the equation for velocity to be v(t) = (1-e^((-g*t)/v_t))*v_t, which at t=0.04s gives me a velocity of 0.172 m/s. However, when I met with a group member to work on this, his velocity equation was v(t) = (gt)/(1+(gt)/v_t) which gave v(0.04s) = .132 m/s. The difference in our integration was at the step (dv/dt) = g - (v*alpha)/m. My partner merely switched the dt over and integrated dv = gdt - ((v*alpha)/m)dt, whereas I integrated dv/(1-((alpha*v)/(mg))) = gdt because I collected the velocity terms on the left had side. I wanted to know which was the correct way, if either?

For part c, using my velocity equation, the position equation I obtained was

x(t) = t*v_t + (v_t/g)*e^((-gt)/v_t) + (v_t)/g which from t = 0 to t = 0.1 s gave me a displacement of -0.0002m, which doesn't seem right. Again, I don't really want solutions so much as someone to check my work. Thanks in advance.