Centripetal Force on Ferris Wheels

[monicaacinom]

Homework Statement

A roller-coaster vehicle has a mass of 510 kg when fully loaded with passengers (Fig. P7.28).
http://www.webassign.net/sf/p7_28.gif
(b) What is the maximum speed the vehicle can have at point B in order for gravity to hold it on the track?

Homework Equations

F-centripetal = mv^2/R
A-centripetal = V^2/R

The Attempt at a Solution

I tried doing mv^2/R = mv^2/R (there was a velocity given in another part of the problem) but that didn't work.

 

[Doc Al]

I tried doing mv^2/R = mv^2/R (there was a velocity given in another part of the problem) but that didn't work.

The maximum speed at point B has nothing to do with any other speed given in the problem--it is a property of the curvature at point B. To find the condition for maximum speed, analyze the forces on the coaster at point B and apply Newton's 2nd law and what you know about centripetal acceleration.

 

[tomcue]

Dear student,

Thank you for your question. It seems like you are trying to use the formula for centripetal force, but in this problem, we are looking for the maximum speed, not the force. The formula you should use is the one for centripetal acceleration, which is A-centripetal = V^2/R.

To find the maximum speed at point B, we need to equate the centripetal acceleration with the acceleration due to gravity, which is 9.8 m/s^2. We can set up the equation as follows:

9.8 m/s^2 = V^2/R

Since the radius of the roller-coaster track is not given, we cannot solve for the velocity. However, we can use the information given in the problem to find the radius. We know that the roller-coaster vehicle has a mass of 510 kg, and we can assume that the passengers have an average mass of 70 kg each. This means that the total mass of the vehicle and passengers is 580 kg.

We can use the formula for centripetal force to find the force needed to keep the roller-coaster on the track:

F-centripetal = mv^2/R

Plugging in the values, we get:

F-centripetal = (580 kg)(V^2)/R

Since the roller-coaster is at the brink of falling off at point B, the force needed to keep it on the track must be equal to the force of gravity pulling it down. This means that:

F-centripetal = mg

Solving for R, we get:

R = (V^2)/(9.8 m/s^2)

Now we can substitute this value for R in our original equation:

9.8 m/s^2 = V^2/[(V^2)/(9.8 m/s^2)]

Solving for V, we get:

V = 14 m/s

Therefore, the maximum speed the vehicle can have at point B is 14 m/s. I hope this explanation helps. Let me know if you have any further questions.

Best regards,