Ferris Wheel work done by net force

[PsychonautQQ]

My textbook says that if you are on a ferris wheel that is rotating, the total work done by all the forces acting on your is zero. How is that possible? You are moving so isn't work being done to you?

 

[Legaldose]

Yes you are moving, but if you get on the Ferris Wheel at the same location that you get off, then the total displacement is 0 because you ended up in the same place, therefore the net work done is 0.

 

[PsychonautQQ]

the net work isn't a function of time? like when you're halfway around the Ferris wheel work has been done right?

 

[Legaldose]

Yes, but when the ferris wheel stops and you get off, you are in the same place that you started, the net displacement is 0, so the net work done is 0. For the first half of the turn, the ferris wheel will do some work on you, but the second half, it will do the exact same amount of negative work, cancelling out the original work it did.

 

[Tanya Sharma]

My textbook says that if you are on a ferris wheel that is rotating, the total work done by all the forces acting on your is zero. How is that possible? You are moving so isn't work being done to you?

Assuming ferris wheel is moving at constant speed ,the net force acting on the body will be centripetal (i.e towards the center) .The velocity of the body at any instant is tangential .Since the velocity and the force are orthogonal(perpendicular) to each other ,the net work done given by ∫F.ds will be zero.

Yes, but when the ferris wheel stops and you get off, you are in the same place that you started, the net displacement is 0, so the net work done is 0. For the first half of the turn, the ferris wheel will do some work on you, but the second half, it will do the exact same amount of negative work, cancelling out the original work it did.

The reasoning is incorrect ...

Consider the first loop which the person completes .The person starts from rests and acquires some velocity after one loop . According to this reasoning,the work done will be zero .But that is not the case .There will be some net tangential force acting on the body ,doing work,responsible for the body to acquire some velocity .According to work kinetic energy theorem ,the change in kinetic energy is the net work done by all the forces on the person.

 

[PsychonautQQ]

So if there is a change in kinetic energy wouldn't that mean that there IS work done...? And yet you said that the direction of motion and the force are orthogonal and therefore the work done is zero..?

 

[Tanya Sharma]

So if there is a change in kinetic energy wouldn't that mean that there IS work done...?

The person in a ferris wheel starts from rest and gradually acquires a speed .Till the time the person starts moving with the constant speed ,work will be done .Why ? because there will be a tangential force increasing the speed of person .Since there is a force(tangential) in the direction of displacement(tangential),work will be done.

But after some time ,when the person is rotating with a constant speed ,the force will be entirely radial .Hence work done will be zero.

So,I guess,the book assumes that the ferris wheel is moving with a constant speed.

And yet you said that the direction of motion and the force are orthogonal and therefore the work done is zero..?

Force and direction of motion will be orthogonal , when the person is moving with a constant speed.There is no contradiction in what i have said.

 

[NihalSh]

My textbook says that if you are on a ferris wheel that is rotating, the total work done by all the forces acting on your is zero. How is that possible? You are moving so isn't work being done to you?

hey PsychonautQQ!

The textbook assumes the condition that the ferris wheel is rotating at constant speed, it does not consider initial acceleration and final deceleration. So, without going into individual forces that are acting (which could be far complex, e.g. friction, gravity, tension, etc) we can say that since rotational kinetic energy isn't changing, the total work done must be zero by all the forces.

$$K_{rotational}=\frac{1}{2}I.ω^2$$

$ω$ is angular velocity and $I$ is rotational inertia.

I hope it clears your doubt