# Newtonian Mechanics - Ball Falling through Syrup

• derravaragh
In summary, the equation of motion for a steel ball bearing falling through a viscous syrup is m(dv/dt) = mg - (alpha)v, with a terminal velocity of 0.2 m/s. By using differential equations, it can be determined that the value of alpha is (mg)/v_t. The speed at t = 0.04 s can be calculated using v(t) = (1-e^((-g*t)/v_t))*v_t, resulting in a velocity of 0.172 m/s. The distance that the ball bearing falls from t = 0 to t = 0.1 s can be determined by integrating the position equation x(t) = t*v_t + (v_t/g)*
derravaragh

## 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.

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.

derravaragh said:
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,

My response to your friend is: HUH? You don't know what the function v(t) is. In fact, that is precisely what you are supposed to solve for. So how exactly do you expect to solve this equation just by directly integrating with respect to time, when v is in the integrand?
derravaragh said:
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?

You seem to have obtained the correct solution for v(t) (you'll notice that your v asympotitcally approaches vt), although I'm puzzled as to how you did it. As far as I can see, the pesky g term makes this equation non-separable, which means that you can't just collect like terms and then integrate. Once I had the equation in the form ##\dot{v} + (\alpha / m)v = g##, I used the method of integrating factors to turn the left hand side into an exact differential. THEN I was able to integrate both sides with respect to time. Note: the dot denotes a derivative w.r.t. time in physics.

Ok, so using your equation (dv/dt) + (alpha/m)*v = g, I came to a new equation for v(t):
v(t) = v_t - 1/e^((gt)/v_t)) Does this check out? It has been a year since I've taken differential equations, and I sadly haven't used it since, so I'm feeling very unsure of my work.

Also, now for the final part, I integrated the equation for position, obtaining x(t) = (v_t)t + (v_t/g)*e^(-(gt)/v_t) - (v_t)/g, which gives me a negative displacement of -.00026 m. I can't help but feel that is incorrect. My work for this integration was using:
dx = (v_t - 1/e^((tg)/v_t))dt = (v_t)dt - dt/e^((tg)/v_t). Is that even remotely correct?

Last edited:
derravaragh said:
Ok, so using your equation (dv/dt) + (alpha/m)*v = g, I came to a new equation for v(t):
v(t) = v_t - 1/e^((gt)/v_t)) Does this check out? It has been a year since I've taken differential equations, and I sadly haven't used it since, so I'm feeling very unsure of my work.

Like I said before, your original solution for v(t) (that was of the form A[1-exp(-t/b)]) was correct. I just have no idea how you arrived at it. In any case, did you look up the Method of Integrating Factors?

derravaragh said:
Also, now for the final part, I integrated the equation for position, obtaining x(t) = (v_t)t + (v_t/g)*e^(-(gt)/v_t) - (v_t)/g, which gives me a negative displacement of -.00026 m. I can't help but feel that is incorrect. My work for this integration was using:
dx = (v_t - 1/e^((tg)/v_t))dt = (v_t)dt - dt/e^((tg)/v_t). Is that even remotely correct?

It's kind of hard to parse your equations, and I don't have time right now. Just integrate:

x(t) = ∫v(t)dt

and since v(t) is of the general form A[1 - exp(-t/b)], you get

= At + Abexp(-t/b) + C

where you can fill in the constants A and b with what they are actually supposed to be, and
figure out the constant of integration C yourself.

derravaragh said:
v(t) = v_t - 1/e^((gt)/v_t))
Doesn't look right. You should get something like vt(1-e-kt).

haruspex said:
Doesn't look right. You should get something like vt(1-e-kt).

Which is the solution that he/she had in the OP, like I said above (twice).

Well there we go. I am finished with this problem and confident in my answers. Thank you both for your help, and sorry for confusing the process, I misunderstood your (cepheid) comment on my initial v(t) equation, but now I can regroup with my partner and finalize this. Again, thank you.

## 1. How does viscosity affect the falling of a ball through syrup in Newtonian mechanics?

Viscosity is the measure of a fluid's resistance to flow. In Newtonian mechanics, the higher the viscosity of a fluid, the slower the ball will fall. This is because the syrup's resistance to flow will act against the force of gravity, causing the ball to fall at a slower rate.

## 2. What role does gravity play in Newtonian mechanics when a ball is falling through syrup?

Gravity is a fundamental force in Newtonian mechanics that dictates the motion of objects. In the case of a ball falling through syrup, gravity is the force that pulls the ball towards the earth. The rate at which the ball falls is directly proportional to the strength of gravity.

## 3. How does the size and weight of the ball affect its falling speed through syrup in Newtonian mechanics?

In Newtonian mechanics, the size and weight of the ball do not affect its falling speed through syrup. This is because the force of gravity acts equally on all objects regardless of their size or weight. However, the shape of the ball may affect its speed due to differences in air resistance.

## 4. What other factors can affect the falling speed of a ball through syrup in Newtonian mechanics?

Aside from viscosity, the falling speed of a ball through syrup in Newtonian mechanics can also be affected by the temperature of the syrup. Higher temperatures can decrease the viscosity of the syrup, resulting in a faster falling speed. Additionally, factors such as the shape and surface texture of the ball can also impact its speed due to differences in air resistance.

## 5. Can the principles of Newtonian mechanics be applied to the falling of objects in non-Newtonian fluids?

No, the principles of Newtonian mechanics cannot be directly applied to the falling of objects in non-Newtonian fluids. This is because non-Newtonian fluids have complex properties that cannot be accurately described by the simple equations of Newtonian mechanics. Different models and theories must be used to understand the motion of objects in non-Newtonian fluids.

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