# Solving Acceleration Vector for Man on Ferris Wheel

• armolinasf
In summary: The derivative of v=3+.5t isn't 3. Come on, you can do better than that. And the acceleration due to the velocity, 9/7 m/s^2 is perpendicular to the path and the other component is tangential. Add them as vectors.In summary, the man is sitting at the bottom of a ferris wheel that is accelerating. The acceleration is perpendicular to the path and the magnitude is 9/7 m/s^2.
armolinasf

## Homework Statement

There is a man sitting at the lowest point of a ferris wheel that is just beginning to move it has velocity 3m/s and is accelerating at a constant rate of .5m/s^2. find the magnitude and direction of the acceleration vector.

## The Attempt at a Solution

I know that acceleration has two components, perpendicular and tangential, one controlling direction and the other magnitude.

So if the acceleration is a constant 5m/s^2 then its components would be perpendicular .5sin(0) and parallel .5cos(0). this gives acceleration = arctan(1)= 45 degrees.

But this would mean that the vector is pointing away from the center of ferris wheel, which should not be the case if this is uniform circular motion. But its not he's his speed is increasing. But if its increasing what's preventing him from being flung off the wheel?

I think the acceleration is tangential. .5sin(0)=0. How are you getting 45 degrees out of this again?

angle=arctan(y/x)=arctan(.5/.5)

armolinasf said:
angle=arctan(y/x)=arctan(.5/.5)

.5*sin(0)=0, .5*cos(0)=.5. I don't think the quotient is .5/.5.

Wow, I can't believe I miss some of these mistakes. But would still mean that there is no perpendicular component correct?

armolinasf said:
Wow, I can't believe I miss some of these mistakes. But would still mean that there is no perpendicular component correct?

All of the acceleration due to the motion of the ferris wheel is towards the center of the wheel. Isn't it? Isn't it all perpendicular? I think what you are missing here (and that I have been ignoring since I got fixated on looking at more surface problems) is that the motion is circular, not linear. You can't even answer any questions about the magnitude of the acceleration unless you know the radius of the ferris wheel.

Last edited:
So perpendicular accl would be equal to 3^2/14, if the diameter=14?

armolinasf said:
So perpendicular accl would be equal to 3^2/14, if the diameter=14?

Not if the diameter is 14. It would be if the RADIUS were 14. Specifying units would be helpful here. And you still have a tangential acceleration to worry about if you want the total acceleration.

Alright I think I am starting to find out where I am getting confused. perp accl would be equal 9/7 and tangential would be equal to the derivative of the magnitude of velocity. in this case v=3+.5t, the derivative would be 3 and that would be the tangential acceleration.

armolinasf said:
Alright I think I am starting to find out where I am getting confused. perp accl would be equal 9/7 and tangential would be equal to the derivative of the magnitude of velocity. in this case v=3+.5t, the derivative would be 3 and that would be the tangential acceleration.

The derivative of v=3+.5t isn't 3. Come on, you can do better than that. And the acceleration due to the velocity, 9/7 m/s^2 is perpendicular to the path and the other component is tangential. Add them as vectors.

## 1. What is acceleration vector?

Acceleration vector is a mathematical representation of an object's change in velocity over time. It includes both the magnitude (speed) and direction of the object's acceleration.

## 2. How does acceleration vector relate to a man on a Ferris wheel?

On a Ferris wheel, the man is constantly changing direction and speed, which means he is experiencing acceleration. The acceleration vector would show the magnitude and direction of his change in velocity as he moves around the wheel.

## 3. Why is it important to solve for the acceleration vector in this scenario?

Solving for the acceleration vector allows us to understand and analyze the motion of the man on the Ferris wheel. It can help us make predictions about his speed and direction at different points on the wheel, and also determine the forces acting on him.

## 4. How do you calculate the acceleration vector for a man on a Ferris wheel?

To calculate the acceleration vector, you would need to know the man's position and velocity at different points on the wheel. Then, you can use the equation a = (vf - vi)/t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time. This will give you the acceleration vector for each point on the wheel.

## 5. Can the acceleration vector change over time for a man on a Ferris wheel?

Yes, the acceleration vector can change over time for a man on a Ferris wheel. This is because the man's speed and direction are constantly changing as he moves around the wheel. The acceleration vector will reflect these changes and may vary at different points in time.

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