How Is the Coefficient of Friction Calculated for an Amusement Park Ride?

AI Thread Summary
The discussion focuses on calculating the coefficient of friction necessary to keep riders from falling in a 3-meter cylinder spinning at 5 radians per second. The user initially calculated the velocity as 15 m/s and derived the centripetal force as 75M, equating it to the force of friction. However, it was clarified that in this scenario, the centripetal force acts as the normal force, not the frictional force. The user acknowledges the mistake and indicates a better understanding of the problem. The conversation highlights the importance of correctly identifying forces in physics problems.
chicagobears
Messages
8
Reaction score
0

Homework Statement


In an amusement park ride, people are spun at 5 radians per sec in a 3 m cylinder. What is the coefficient of friction to prevent people from falling down?


Homework Equations


Centripetal Force= (mv^2)/r
Force of friction=(mu)(mg)
Velocity=(r)(ω)

The Attempt at a Solution


V=(3 m)(5 rad/sec)=15 m/s
Centripetal Force=(M(15^2))/3=75M
Centripetal force=force of friction...
75M=9.8(mu)M
75/9.8=(mu)=7.653?
 
Physics news on Phys.org
No need to answer. I get this now...
 
Is it a vertical cylinder of radius 3 m? (A 3m cylinder suggests the diameter is 3.)
With the people above the floor of the cylinder?
If so, the centripetal force is the normal force, not the force of friction.
 
yea that was my mistake...
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging 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...
View full post »

Similar threads

A problem from a physics national competition
Replies
13
Views
1K
Force at the Bottom of a Circular Amusement Park Ride
Replies
3
Views
2K
Circular Motion Questions (energies, forces, angular velocities, etc.)
Replies
2
Views
1K
Circular Motion With Constant Angular Acceleration
Replies
3
Views
2K
Circular Motion of a Cyclist and a Car going around a bend in the road
Replies
6
Views
4K
Direction of friction in circular motion
Replies
5
Views
3K
Circular Motion — Object on a rotating conic surface....
Replies
20
Views
5K
Amusement Rotor Ride Explanation
Replies
6
Views
10K
Circular motion of a mass on a string on an inclined plane
Replies
14
Views
4K
Circular Motion of an amusement park ride
Replies
1
Views
3K

Hot Threads

Recent Insights

Back
Top