Help finding Centripital force angles

[burban864]

I'm stumped on my hw...can anyone help?

The question is long, but as follows: a merry-go-round has circular platform that is 1m from the central axis at its inner edge and is 5m from the central axis at its outer edge. The ride takes 10s for 1 rotation. A passenger holds himself to the surface with a pair of very stickey shoes and is most comfortable when he orients his body length along the line of the net force on him. Determine the angle his body makes to the vertical A) 1m from the axis, B) 3m from the axis, and c) 5m from the axis.

Any ideas...just looking to understand how to set this one up equationwise.
Thanks

 

[Nenad]

well, there are two forces action on the passanger, assuming no air resistance. There is a centripital force and a tagential force.

$$F_{centripital} = \frac {mv^2}{r}$$

$$F_{tangential} = mgsin \theta$$

the tangential force is due to gravity and the centripital one is due to the circular motion. To solve your problem you have to draw some free body diagrams and use vector addition to get the result. Split up your vectors into component form and get your resultant that way. The person will be facing in the angle of this vector. If you need more help, pm me.

Regards,

Nenad

 

[GSXR750]

Hi there,

I can definitely help you with this problem. First, let's start by understanding what centripetal force is. Centripetal force is the force that keeps an object moving in a circular path. In this case, the passenger on the merry-go-round is experiencing centripetal force as they rotate around the central axis.

To find the angle that the passenger's body makes with the vertical at different distances from the axis, we can use the equation Fc = mv^2/r, where Fc is the centripetal force, m is the mass of the passenger, v is the velocity, and r is the distance from the central axis.

We know that the ride takes 10 seconds for 1 rotation, so the velocity can be calculated by dividing the circumference of the circular platform (2πr) by the time (10 seconds). This gives us a velocity of 2πr/10 = 0.2πr m/s.

Now, we can plug in the given distances (1m, 3m, and 5m) into the equation and solve for the angle. I'll show you how to do it for the first distance (1m) and you can follow the same steps for the other distances.

Fc = mv^2/r
Fc = m(0.2πr)^2/1
Fc = m(0.04π^2r^2)
Fc = 0.04π^2mr^2

We also know that the net force on the passenger is equal to the centripetal force, so we can set the net force equal to the centripetal force and solve for the angle.

Net force = mg*cosθ
0.04π^2mr^2 = mg*cosθ

We can rearrange this equation to solve for the angle:

cosθ = 0.04π^2mr^2/mg
θ = cos^-1(0.04π^2mr^2/mg)

Now, we can plug in the values for mass (let's say the passenger weighs 50kg), gravity (9.8 m/s^2), and the given distance (1m) to solve for the angle:

θ = cos^-1(0.04π^2*50*1^2/(50*9.8))
θ = cos^-1(0.0128)
θ = 84.3 degrees

So