Retarding Force proportional to the velocity falling particl

In summary, the problem involves finding the displacement and velocity of a particle undergoing vertical motion in a medium with a retarding force proportional to the velocity. The equation for this force is F = m(dv/dt) = -mg-kmv. By solving for velocity using integration, we get v = (dz/dt) = (-g/k) + ((k(Vo) + g)/k)e^(-kt). The e^(-kt + kc) term can be simplified to ((k(Vo) + g)/k) by subtracting g from both sides and dividing by k. The constant of integration, c, can also be substituted for to obtain the desired result.
  • #1
Decypher
2
0
Example problem from a Classical Mechanics book.
Find the displacement and velocity of a particle undergoing vertical motion in a medium having a retarding force proportional to the velocity.

F = m(dv/dt) = -mg-kmv

dv/(kv + g) = -dt

(1/k)ln(kv + g) = -t + c

kv + g = e^(-kt+kc)

v = (dz/dt) = (-g/k) + ((k(Vo) + g)/k)e^(-kt)

I don't see how the e^(-kt + kc) is turning into ((k(Vo) + g)/k
I know a property of exponents allows you to separate the exponents powers when they are being added, then each one becomes the power of an exponent which is where e^-kt comes from, but I am not sure about where the other half of the exponent goes.
 
Physics news on Phys.org
  • #2
Decypher said:
F = m(dv/dt) = -mg-kmv
Hi Decypher:

I think y ou have a sigh error. mg is the force of gravity pulling the mass towards rthe Earth, and kmv is a retarding force, slowing down the pull. The direction of these forces are opposite.

Hope this is helpful.

Regards,
Buzz
 
  • #3
I spent a great amount of time on this part you are mentioning. The book does not provide a great description on this statement.
"Where -kmv represents a positive upward force since we take z and v = z(dot) to be positive upward, and the motion is downward-that is, v<0, so that -kmv > 0."
I think they are saying that because we are treating a positive velocity as one that is going higher on the z axis, whe have to treat -kmv as kmv because when a negative velocity is plugged in, kmv becomes an opposing force once again. It makes more sense to me to have it be g-kv just like a force opposing gravity should be.
 
  • #4
Decypher said:
Find the displacement and velocity of a particle undergoing vertical motion
Decypher said:
we are treating a positive velocity as one that is going higher on the z axis
Hi Decypher:

OK. I think the first quoted sentence would have been clearer if the word "upward" had preceded "vertical".

Decypher said:
I don't see how the e^(-kt + kc) is turning into ((k(Vo) + g)/k
The above quote is not what you want to do. I suggest you take the equation
Decypher said:
kv + g = e^(-kt+kc)
and fill in the missing step: subtract g from both sides, and divide both sides by k. Then make the substitution v = dz/dt.

Then you should be able to make a substitution for c (which is an arbitrary constant of integration) which will give you what you want.

Hope this helps.

Regards,
Buzz
 

FAQ: Retarding Force proportional to the velocity falling particl

1. What is the retarding force proportional to the velocity falling particle?

The retarding force proportional to the velocity falling particle is a force that opposes the motion of a particle as it falls due to the force of gravity. It is directly proportional to the velocity of the particle, meaning that as the particle's velocity increases, the retarding force also increases.

2. How does the retarding force affect the motion of a falling particle?

The retarding force acts in the opposite direction of the particle's motion, slowing it down as it falls. This force reduces the acceleration of the particle, causing it to eventually reach a state of constant velocity known as terminal velocity.

3. What factors affect the strength of the retarding force?

The strength of the retarding force is affected by the velocity of the falling particle and the surface area of the particle. A larger surface area will experience a greater retarding force, while a higher velocity will also result in a stronger retarding force.

4. How does the retarding force change as the particle falls?

As the particle falls, the retarding force will initially increase as the velocity increases. However, once the particle reaches terminal velocity, the retarding force will remain constant and equal to the force of gravity acting on the particle.

5. What is the significance of the retarding force proportional to the velocity falling particle in real-world applications?

The concept of the retarding force proportional to the velocity falling particle is important in understanding and predicting the motion of objects falling through a fluid, such as air or water. This concept is also used in designing parachutes and other devices for safely slowing down the descent of objects.

Back
Top