# Finding Force of Air resistance

• anonymouz
In summary: So in summary, the conversation discusses a lab where a cylinder was rolled down a ramp and various data was gathered, such as the mass and diameter of the cylinder, initial and final velocity, and the angle of the ramp. The goal is to find the average force of air resistance, but the method for doing so is not yet known. One suggestion is to find the acceleration components in the x and y directions, but the specific equations and calculations are still being discussed and refined.
anonymouz
My class did a lab where we roll a cylinder down a ramp (the end of the ramp was at the edge of the table). We placed a photogate timer right at the position where the cylinder fell off the ramp, and a photogate timer right at the position where the cylinder hit the floor.
The information gathered was mass of cylinder, diameter of cylinder, initial and final velocity, the angle above the horizontal that the cylinder fell, the height of the table, and the distance on the floor away from the table that the cylinder hit.
I'm supposed to find the average force of air resistance.
We have not learned that the force of air drag is proportional to velocity and we are not using calculus.

I'm thinking about finding the acceleration in x and y but I need to figure out either the time, or the x and y components of the final velocity. Then i can graph the position of the cylinder with respect to time and figure out the distance of the path that the cylinder fell. Then i could find force.

So basically my question is: how do i find time or the x/y components of the final velocity? thanks all.

Surely you can find acceleration if you know the initial and final velocities and the distance between them?

so it doesn't matter that the path that the cylinder travels in isn't straight?
because the air makes the cylinder travel in a curve

does anyone know how to do this??

i have an idea..can someone check this?

vf = vi + at
d = vit + .5at^2

so you can solve for t in terms of a from the last equation
then you can square the first equation and set that equal to the second equation. then you substitute in t and you can solve for a.

i end up with this:

vi^2 - vi - 2ad + 8vi(-vi +/- sqrt(vi^2 + 2ad))/a + (8/a^2)(-vi +/- sqrt(vi^2 + 2ad))^2

is this correct?

Hi anonymouz.

First consider that the direction of the velocity vector of the cylinder changes in its trajectory downwards. The drag, R, on the cylinder will be in the opposite direction of this vector. This means that the cylinder will have two acceleration components, one in the x (decelerating the cylinder in the x-direction) and one in the y-direction, which will decrease its normal, g, acceleration. In order to determine the components of the drag one need to determine both these acceleration components.

I assume that trhe photogates were orientated vertically. This means that the speeds that the gates measured were the x components - in which case the measured speed at the bottom will be less than that at the top.

So the x-component of the acceleration is

$$a_x = \frac{v_x ^2 - u_x ^2}{2d}$$

Once you've got this average acceleration in the x-direction you can calculate the flight time, $$t_f$$, of the cylinder using the known distance, d, traveled in the x-direction.

Assuming you've got the inclination of the ramp, $$\theta$$, you can then determine the intial y-speed of the cylinder when it leaves the ramp from

$$u_y = u_x \tan(\theta)$$

this plus the flight time and the known height through which it fell enables you to calculate the y acceleration component.

Last edited:
what if the photogates weren't oriented vertically?
the top photogate was oriented perpendicular to the ramp and the second photogate was oriented approx. perpendicular to the direction of the cylinder's fall

also: are my calculations for accleration correct? (3 posts up)

I am not sure how you got to this equation. The second equation already does not involve t

$$v^2 = u^2 +2ad$$

so why develop another such relation?

My biggest concern with this approach is that as I tried to explain is that the cylinder will accelerate in both the x and y direction.

The measured velocities will be the real velocities of the cylinder at the beginning, u, and end, v, of its motion, that is its velocities tangential to its path. Which means that one need to use the components of u and v in the kinematic equations. For this one need the angles at the beginning and the end of the motion.

It seems that you measured the angle above the horizontal that the cylinder came onto the floor, $$\beta$$. Did you also measure the angle of the ramp, $$\alpha$$? If not can you still get it?

I am not sure how you got to this equation. The second equation already does not involve t

$$v^2 = u^2 +2ad$$

My biggest concern with this approach is that as I tried to explain is that the cylinder will accelerate in both the x and y direction.

The measured velocities will be the real velocities of the cylinder at the beginning, u, and end, v, of its motion, that is its velocities tangential to its path. Which means that one need to use the components of u and v in the kinematic equations. For this one need the angles at the beginning and the end of the motion.

It seems that you measured the angle above the horizontal that the cylinder came onto the floor, $$\beta$$. Did you also measure the angle of the ramp, $$\alpha$$? If not can you still get it?

i have the angle of the ramp, so i know the angle at which the cylinder fell once it rolled off the ramp
but i don't know the angle of the cylinder's path when it hits the floor

^ for the equation that i came up with..
all i did was find t in terms of a, and substitute that in for the first equation
then i squared the first equation and set it equal to the second equation
then i solve for a

i wrote the equation wrong up there.
it's supposed to be:

You equation is correct.

So if you want to use it to solve for the acceleration components in the x- and y-directions you need to use the velocity components of vi (measured speed at the top) for your vi in your equation and the d (distance to the table) and h (height that the cylinder fell through)
distances. The x velocity component will be

$$v_{ix} = u\cos(\alpha)$$

and the y-velocity component will be

$$v_{iy} = u\sin(\alpha)$$

where $$\alpha$$ is the inclination of the ramp to the horizontal.

If you measured the angle of the photogate at the bottom, $$\theta$$, the angle of the velocity of the cylinder at the bottom, $$\beta$$, would be just the complement of this angle $$\beta = 90^o - \theta$$. Which would make that calculations considerably easier. Without this angle the measurement at the bottom is of no use.

Last edited:
ok thanks a lot for your help, andrevdh
really appreciate it

## 1. What is air resistance?

Air resistance, also known as drag, is a force that opposes the motion of an object as it moves through the air. It is caused by the collision of air particles with the object's surface.

## 2. How is air resistance calculated?

The force of air resistance is calculated using the equation F = 1/2 * ρ * v^2 * A * Cd, where ρ is the density of air, v is the velocity of the object, A is the cross-sectional area of the object, and Cd is the drag coefficient. The drag coefficient depends on the shape and surface characteristics of the object.

## 3. What factors affect air resistance?

The amount of air resistance experienced by an object depends on several factors, including the speed and direction of the object, the shape and size of the object, the density of the air, and the surface characteristics of the object.

## 4. How does air resistance affect an object's motion?

Air resistance can significantly slow down an object's motion, especially at high speeds. It also creates a force in the opposite direction of the object's motion, causing it to slow down or change direction.

## 5. How can air resistance be reduced?

Air resistance can be reduced by making an object more streamlined, reducing its surface area, or by using materials with lower drag coefficients. Additionally, reducing the speed and direction of an object can also decrease the amount of air resistance it experiences.

• Introductory Physics Homework Help
Replies
15
Views
505
• Introductory Physics Homework Help
Replies
8
Views
1K
• Introductory Physics Homework Help
Replies
27
Views
1K
• Introductory Physics Homework Help
Replies
13
Views
1K
• Introductory Physics Homework Help
Replies
39
Views
2K
• Introductory Physics Homework Help
Replies
8
Views
206
• Introductory Physics Homework Help
Replies
2
Views
2K
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
56
Views
2K