Circular Motion: Ferris Wheel Dynamics

AI Thread Summary
The discussion focuses on the dynamics of a Ferris wheel, specifically calculating the difference in apparent weight of a passenger at the highest and lowest points. Participants clarify that the difference can be expressed as Fnb - Fnt = xW, where x is a computed value. For part b, it is suggested that the relationship v^2/r = g can be used to find the time for one revolution. In part c, the apparent weight at the lowest point can be determined using the equation Fnb = mv^2/r + mg, with mass canceling out in the calculations. Overall, the conversation emphasizes understanding the relationships between forces and apparent weight in circular motion.
Mehta29
Messages
16
Reaction score
0
I'm on the right track, but I'm stuck here...

The radius of a Ferris wheel is 5 m and it makes one rev in 10 sec

a Find the difference b/w the apparent weight of a passenger at the highest and lowest points, expressed as a fraction of his weight, W

b What would the time for one rev be if teh apparent weight at the top were zero?
c What would be the apparent weight at the low point??

I have at the top that

mv^2/r = mg - Fn

and at the bottom

mv^2/r = Fn -mg

I really don't undestand what a is asking...(Fnbottom - Fntop)/ W ??
 
Physics news on Phys.org
Mehta29 said:
I really don't undestand what a is asking...(Fnbottom - Fntop)/ W ??
Yes you do. It's exactly what you said. Write your answer as Fnb - Fnt = _______W
 
but I am confused as if anything else would be needed...liek that blank befor e the W...would i need to expand any further or would i just keep it
Fnbottom - Fntop = xW

b would just be v^2/r = g? and then t = 2pir/v

and c would be Fnb = mv^2/r + mg...but how would i eliminate m?
 
Last edited:
Mehta29 said:
but I am confused as if anything else would be needed...liek that blank befor e the W...would i need to expand any further or would i just keep it
Fnbottom - Fntop = xW

b would just be v^2/r = g? and then t = 2pir/v

and c would be Fnb = mv^2/r + mg...but how would i eliminate m?
Solve your earlier top and bottom equations for Fn. Take the difference between the two. Your result will be of the form
Fnb - Fnt = mA where A is a number that can be computed from the given information. You can do that. Once you have that form, multiply and divide by g
mA = mgA/g = WA/g
You can calculate A/g to express your answer as
Fnb - Fnt = (A/g)W with A/g replaced by a number.
 
i got A and B...but how would i manage part c?

im not seein anyway to cancel out the mass...
 
Mehta29 said:
i got A and B...but how would i manage part c?

im not seein anyway to cancel out the mass...
What is your understanding of "apparent weight". All the forces in the problem are proportional to mass. It will cancel out.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top