What Is the Maximum Speed to Keep the Flywheel's Wheels on Track?

[Kmaster]

A flywheel of mass 42.7 kg is supported by an axle 1.65 m long, with two small wheels. The axle's wheels roll on an elevated track, part of which has a curve of radius, R = 15.9 m. If the flywheel has angular momentum 75.50 kg·m2/s, what is the maximum speed at which the system can take the curve without a wheel lifting from the track?

Any help with solving this problem would be greatly appreciated. thanks.

 

[berkeman]

You need to show us your work in order for us to help you (PF rules). That's why there is a Homework Posting Template for all new threads here in the homework forums. Please show us your work on #2 and #3, and we'll try to help guide you to figuring out the solution.

 

[Andrew Mason]

I think we are supposed to assume that the radius of the wheels is negligible compared to the radius of the flywheel, but it does not give us either.

If you do as Berkeman suggests and state your approach to this, we will have a better idea of how to help you.

AM

 

[Kmaster]

Thank you for the directions. Sorry I didn't read the posting rules.

Equations:
Linear: F=ma p=mv
Rotational τ=I*α L= I*ω
cicrular: a=v^2/R

The way the drawing is set up, the wheels are much smaller than the fly wheel. also, the outer radius (where one of the wheels is)= R+1/2axle= 1.9+1.65/2=2.725m and the inner radius =R-1/2axle=1.075m

I tried to combine the equations using force=torque. mv^2/R=Iα And solved I=mv^2/R/α =L/ω After that I said the distance traveled around the curve/v=time=ω/α. The I got stuck and I'm fairly certain the process doesn't really go anywhere.

The hint provided is:The rate at which the angular momentum must change going around the curve equals the maximum net torque which the weight of the flywheel can provide.

Thank you for any help.

 

[Andrew Mason]

Thank you for the directions. Sorry I didn't read the posting rules.

Equations:
Linear: F=ma p=mv
Rotational τ=I*α L= I*ω
cicrular: a=v^2/R

The way the drawing is set up, the wheels are much smaller than the fly wheel. also, the outer radius (where one of the wheels is)= R+1/2axle= 1.9+1.65/2=2.725m and the inner radius =R-1/2axle=1.075m

I tried to combine the equations using force=torque. mv^2/R=Iα And solved I=mv^2/R/α =L/ω After that I said the distance traveled around the curve/v=time=ω/α. The I got stuck and I'm fairly certain the process doesn't really go anywhere.

The hint provided is:The rate at which the angular momentum must change going around the curve equals the maximum net torque which the weight of the flywheel can provide.

Thank you for any help.

Another useful equation is:

\vec{\tau} = \frac{d\vec{L}}{dt}

You have two mutually perpendicular angular momenta - one provided by the flywheel and the other provided by the flywheel/cart going around the track. The angular momentum of the flywheel is pointing in the radial direction so it keeps changing. What is the direction of the change of the angular momentum of the flywheel in a small time interval dt? (what is the direction of the angular momentum at time t and the direction at the time t+dt? what is the difference?) That is the direction of the torque vector. What is the magnitude of that torque vector? What torque is opposing it? When are the two torques equal?

AM

 

[Kmaster]

Thank you very much. My teacher derived some equation and the problem was solved.