Motion In the Presence of Resistive Forces

[webren]

Hello,
I am having a difficult time getting far into solving this problem:

"A small piece of Styrofoam packing material is dropped from a height of 2.00 m above the ground. Until it reaches terminal speed, the magnitude of its acceleration is given by a = g - bv. After falling 0.500 m, the Styrofoam effectively reaches terminal speed, and then takes 5.00 s more to reach the ground. (a) What is the value of the constant b? (b) What is the acceleration at t = 0? (c) What is the accleration when the speed is 0.150 m/s?"

The first thing I did is to draw a free-body diagram of the Styrofoam falling, with the resisitive forces pointing upward, and the weight pointing downward. Making downward being positive, the resistive forces are negative (hence the given accleration being a = g - bv).

To find b, I first need to find velocity and I don't see how I could find that. I tried setting mg - bv = m(g -bv) = m(v^2/r). This seems kind of messy, but if it's necessary, I would think velocity and mass would eventually cancel out. Is this the right way of solving this problem?

I also thought about using a free-fall kinematic equation and simply solving for velocity, but I think those are appropriate for problems that neglect air resistance. Am I correct?

Thank you.

 

[nrqed]

Hello,
I am having a difficult time getting far into solving this problem:

"A small piece of Styrofoam packing material is dropped from a height of 2.00 m above the ground. Until it reaches terminal speed, the magnitude of its acceleration is given by a = g - bv. After falling 0.500 m, the Styrofoam effectively reaches terminal speed, and then takes 5.00 s more to reach the ground. (a) What is the value of the constant b? (b) What is the acceleration at t = 0? (c) What is the accleration when the speed is 0.150 m/s?"

The first thing I did is to draw a free-body diagram of the Styrofoam falling, with the resisitive forces pointing upward, and the weight pointing downward. Making downward being positive, the resistive forces are negative (hence the given accleration being a = g - bv).

To find b, I first need to find velocity and I don't see how I could find that. I tried setting mg - bv = m(g -bv) = m(v^2/r). This seems kind of messy, but if it's necessary, I would think velocity and mass would eventually cancel out. Is this the right way of solving this problem?

I also thought about using a free-fall kinematic equation and simply solving for velocity, but I think those are appropriate for problems that neglect air resistance. Am I correct?

Thank you.

There is no circular motion here so there is reason to use mv^2/r!

The problem is actually simple because of the information given. You know that after it has reached terminal speed, it takes 5.00 seconds to fall another 1.50 meter, right?

Now, after it has reached terminal speed, the acceleration is zero, right? So g - b v_{terminal} = 0. So you can relate the terminal speed to b and g. Then, using the 5 seconds for 1.5 m, you can get the terminal speed, right?

It's that easy!

Patrick

 

[webren]

There is no circular motion here so there is reason to use mv^2/r!

I don't know why I was thinking of m(v^2/r). It's probably from doing circular motion problems all day.

I understand your statements regarding the problem and I have solved it.

There is another problem that is in the same category that I don't quite understand as well.

"Calculate the force required to pull a copper ball of radius 2.00 cm upward through a fluid at the constant speed 9.00 cm/s. Take the drag force to be proportional to the speed, with proportionality constant 0.950 kg/s. Ignore the buoyant force."

To solve the problem, I immediately drew a free-body diagram of the copper ball being pulled upward in the water, with the resistive forces pointing downward, and motion being pointing upward.

After that, I asked myself why the problem declared a radius and how this radius could be used in solving the problem. Because the copper ball is a sphere, I thought possibly the area or volume might be involved.

One thing I also thought about was what kind of object this is. My professor stated that smaller objects with low speeds use the R = -bv formula and larger objects (airplane, car, etc.) with high speeds (such as through air) use the R = 1/2DpAv^2, where D is the drag coefficient, A is the cross sectional area, p = density, and v being velocity.

I think this problem relates to the R = -bv equation, because it's a smaller object and obviously going through the low speed of water. The problem states: "Take the drag force to be proportional to the speed, with proportionality constant 0.950 kg/s.," so I made b = 0.950 kg/s.

Finally, I converted the velocity and the drag coefficient (b) to m/s instead of cm/s. After that, I said that the sum of forces equals F - R = 0 (constant speed, so there is no acceleration). I also do not think weight (mg) is involved, because the ball is getting pulled upward. I made F = R, and because R = -bv, the equation became F = -bv. I plugged in the b and v values, and got the force, which came out to be 0.855 N. The book disagrees and claims it is 3.01 N.

Are there any steps I am skipping or doing wrong? Any suggestions would be appreciated. Thanks for taking the time to read my post.

 

[Mute]

You need to include the gravitational force. Does the question not give you the density of copper? According to wikipedia, the density of copper near room temperature is 8.96 g/cm³.

With that you should arrive at the correct result.

 

[nrqed]

I don't know why I was thinking of m(v^2/r). It's probably from doing circular motion problems all day.

I understand your statements regarding the problem and I have solved it.

There is another problem that is in the same category that I don't quite understand as well.

"Calculate the force required to pull a copper ball of radius 2.00 cm upward through a fluid at the constant speed 9.00 cm/s. Take the drag force to be proportional to the speed, with proportionality constant 0.950 kg/s. Ignore the buoyant force."

To solve the problem, I immediately drew a free-body diagram of the copper ball being pulled upward in the water, with the resistive forces pointing downward, and motion being pointing upward.

After that, I asked myself why the problem declared a radius and how this radius could be used in solving the problem. Because the copper ball is a sphere, I thought possibly the area or volume might be involved.

One thing I also thought about was what kind of object this is. My professor stated that smaller objects with low speeds use the R = -bv formula and larger objects (airplane, car, etc.) with high speeds (such as through air) use the R = 1/2DpAv^2, where D is the drag coefficient, A is the cross sectional area, p = density, and v being velocity.

I think this problem relates to the R = -bv equation, because it's a smaller object and obviously going through the low speed of water. The problem states: "Take the drag force to be proportional to the speed, with proportionality constant 0.950 kg/s.," so I made b = 0.950 kg/s.

Finally, I converted the velocity and the drag coefficient (b) to m/s instead of cm/s. After that, I said that the sum of forces equals F - R = 0 (constant speed, so there is no acceleration). I also do not think weight (mg) is involved, because the ball is getting pulled upward. I made F = R, and because R = -bv, the equation became F = -bv. I plugged in the b and v values, and got the force, which came out to be 0.855 N. The book disagrees and claims it is 3.01 N.

Are there any steps I am skipping or doing wrong? Any suggestions would be appreciated. Thanks for taking the time to read my post.

I was going to say exactly the same thing as Mute. I don't know why you concluded that the weight is not involved because the motion is upward (think of throwing a baseball upward, even it it is moving upward, the force of gravity is clearly relevant!). So the force you apply on the object must overcome both the drag force and the force of gravity (the weight). (In real life, the buoyant force would "help" you, i.e. it would be acting upward also but they said to neglect it).

and this is why they gave you the *composition* and radius of the ball, so that you may calculate its mass and therefore its weight (mg).