Force of Gravity on a Man on a Ferris Wheel

[cheerspens]

Homework Statement

A man sitting on the edge of his seat on a Ferris Wheel has a mass of 50.0 kg. The Ferris Wheel has a radius r=30m and the ferris wheel completes a single revolution every 20 seconds. Find the force between the man and the chair.

Homework Equations

F_NET=ma
F_g=mg

\tau=2\pir/v
a=v²/r

The Attempt at a Solution

I found the force of gravity on the man to be 490N. I think the period is 0.05rev/sec so I set up the \tau equation to be 0.05=2\pi(30)/v. I solved for V however and get 3769.91 m/s. It seems like too big of a number. This then gives me a very large acceleration of 473740.71m/s².
I need to solve for a to plug into my F_NET equation in order to find the force on the chair on man.
What am I doing wrong to get these large numbers?

 

[rock.freak667]

The period is the time taken for one revolution, so T = ?

then you can use ω=2π/T

Your main error lies in the periodic time.

 

[cheerspens]

But I don't have a velocity so how can I solve for T? I thought I had to solve for velocity first?

 

[cheerspens]

Force and Circular Motion??

Homework Statement

A man sitting on the edge of his seat on a Ferris Wheel has a mass of 50.0 kg. The Ferris Wheel has a radius r=30m and the ferris wheel completes a single revolution every 20 seconds. Find the force between the man and the chair.

Homework Equations

F_NET=ma
F_g=mg

\tau=2\pir/v
a=v²/r

The Attempt at a Solution

I set up F_NET to be F_GP-F_CP=ma. I found the force of gravity on the man to be 490N and we know the person's mass so the equation is now:
490-F_CP=(50)a
So in order to solve for the force between the chair and the person (F_CP) you have to solve for the acceleration (a).
I think the period is 0.05rev/sec so I set up the \tau equation to be 0.05=2\pi(30)/v. I solved for V however and get 3769.91 m/s. It seems like too big of a number. This then gives me a very large acceleration of 473740.71m/s².
I need to solve for a to plug into my F_NET equation in order to find the force on the chair on man.
What am I doing wrong to get these large numbers?

 

[kuruman]

Try finding the speed directly. How much distance does he cover in 20 seconds?

As for the force between the chair and the man, it looks like not enough information is given. That force depends on where the man is. For example at the "12 o' clock" position the force is different from the "6 o' clock" position.

 

[cheerspens]

Would it me 0.05rev/sec? And that is velocity?
(I also just attached the diagram)

 

[kuruman]

You have posted duplicate threads. I will refrain from posting here until the admins merge the threads.

 

[Filip Larsen]

It may help to think (or read in your textbook) about if there is a simple relationship between the centripetal acceleration, angular speed and radius for a steady circular motion. This should give you the [STRIKE]horizontal [/STRIKE]centripetal force acting on the man which, when vectorially added to the vertical gravity force acting on the man should give you the combined force acting on the man.

Edit: I didn't know a Ferris wheel is a "vertical" wheel until a diagram was added (english being my 2nd language and all) and for that wheel the centripetal acceleration is of course not horizontal in general.

 

[rock.freak667]

But I don't have a velocity so how can I solve for T? I thought I had to solve for velocity first?

You can use T to get v.

One revolution takes 20 seconds, so the period is what ?

 

[cheerspens]

0.05rev/sec??

 

[rock.freak667]

0.05rev/sec??

Periodic time is the time taken for one revolution. So T is ? (they gave it to you)

Then you can find ω.

 

[berkeman]

(two threads merged. please do not multiple post.)

 

[cheerspens]

Periodic time is the time taken for one revolution. So T is ? (they gave it to you)

Then you can find ω.

Oh that's what I forgot and what had me confused. Well that solves everything. Thank you!