How Is Velocity and Acceleration Affected as a Carousel Slows Down?

[studentofphy]

Homework Statement

A carousel takes 1.5 min to complete one revolution while rotating at a constant rate. A person rides on the carousel platform at a distance 3.2m from the center.

(a) From its state of constant rotation, the carousel then uniformly slows to a stop in time (delta t). Produce a diagram qualitatively indicating the rider's position, velocity, and acceleration at an instant during which the carousel slows.

(b) Produce expressions for the time dependence of the rider's speed and acceleration during the time while the carousel slows.

Homework Equations

a=(4(pi^2)r)/(t^2)

v=(2pi*r)/t

The Attempt at a Solution

(a) When the carousel slows down the magnitude of the acceleration and the magnitude of the velocity are going to decrease. When it finally stops rotating, the acceleration and velocity will go to zero. Intuition tells me these statements are true, but I don't know how to represent these ideas qualitatively.

(b)I have attached a image of my approach for this part.

 

[gneill]

There are two accelerations to consider: centripetal and linear. You'll want to show the directions of both on your diagrams, as well as their resultant (they are vectors that you can add for a net acceleration).

For (b), start by assuming some initial speed for the rider, say Vo. Write an expression that would give velocity V(t) if the velocity goes from Vo to 0 over time Δt. Hint: it's a linear (straight line) relationship for which you know the start and end points.

 

[studentofphy]

There are two accelerations to consider: centripetal and linear. You'll want to show the directions of both on your diagrams, as well as their resultant (they are vectors that you can add for a net acceleration).

For (b), start by assuming some initial speed for the rider, say Vo. Write an expression that would give velocity V(t) if the velocity goes from Vo to 0 over time Δt. Hint: it's a linear (straight line) relationship for which you know the start and end points.

Did you have a chance to look at the diagrams I have attached? I am showing the initial velocity as Vi and then 2 more periods that represent slowing and stopped. Is it close to what you are describing?

I am confused on how it can be linear if it is moving in a circle. Can you elaborate on that point?

 

[gneill]

Did you have a chance to look at the diagrams I have attached? I am showing the initial velocity as Vi and then 2 more periods that represent slowing and stopped. Is it close to what you are describing?

Yes, I saw your attachment. It might be better if you were to number the positions and designate the vectors at those locations with subscripts referring to those numbers, like $a_{c_2}$ for the centripetal acceleration at point 2, rather than using Δ (which is usually reserved to mean "change in" some variable). You'll want to place vectors for the velocity and accelerations at each point.

I am confused on how it can be linear if it is moving in a circle. Can you elaborate on that point?

An object moving in a circle has some speed at any point along its path. While the overall motion may be circular, at any given instant in time that speed is associated with a velocity vector that points in the direction of motion which is tangent to the circle at that point. The speed is just the magnitude of that velocity vector. Speed is a scalar value that does not have any particular direction associated with it.

The problem states that the rotation rate uniformly slows to a stop over some period of time. That means the speed of the rider also drops uniformly. "Uniformly" is physics-speak for a linear relationship. So the speed is dropping linearly from its initial value down to zero.

 

[Nickie]

The equation for the acceleration of the rider during the slowing down phase can be derived using the equation for centripetal acceleration, which is given by a=(4(pi^2)r)/(t^2). As the carousel slows down, the time (t) will increase, therefore decreasing the acceleration. Similarly, the velocity of the rider can be calculated using the equation v=(2pi*r)/t. As t increases, the velocity will decrease. These relationships can be plotted on a graph, with time on the x-axis and acceleration/velocity on the y-axis. The resulting graph will have a negative slope, indicating a decrease in both acceleration and velocity as time increases.

Overall, the rider's position, velocity, and acceleration can be represented on a diagram as follows:

Position: The rider will move in a circular path with a radius of 3.2m, centered at the center of the carousel. As the carousel slows down, the rider's position will remain the same, but the velocity and acceleration vectors will decrease in magnitude.

Velocity: The rider's velocity vector will be tangent to the circular path at all times, with a magnitude given by v=(2pi*r)/t. As the carousel slows down, the magnitude of the velocity vector will decrease.

Acceleration: The rider's acceleration vector will always point towards the center of the carousel, with a magnitude given by a=(4(pi^2)r)/(t^2). As the carousel slows down, the magnitude of the acceleration vector will decrease.

Image:

(b) Expressions for the time dependence of the rider's speed and acceleration during the time while the carousel slows can be written as:

Velocity: v(t) = (2pi*r)/t

Acceleration: a(t) = (4(pi^2)r)/(t^2)