Solving Rotor Ride Physics for a Monkey: Velocity, Friction, and More

[runningirl]

Homework Statement

A rotor ride has a radius of 10 m. The ride is in its experimental phase, so the engineers have decided to use a monkey instead of a person. The mass of the monkey is 30 kg. The sides of the ride are designed so that the maximum friction riders will experience is 0.4*Fn.

a) The rider turns so that it completes 10 rotations every 60 sec.
i) Determine monkey's velocity
ii) Determine normal force
iii) Will the monkey stay in position or will it slide down the wall?

b) Determine the max velocity that will cause the monkey to stick motionless to the wall when the floor drops away.

Homework Equations

The Attempt at a Solution

i) 1 rotation=6 s
so v=2*pi*r/T
=(2*pi*10)/6

ii) F=ma
30*a

a=v^2/r

((2*pi*10)/6)^2/10

how can i solve iii and c?

 

[PhanthomJay]

Homework Statement

A rotor ride has a radius of 10 m. The ride is in its experimental phase, so the engineers have decided to use a monkey instead of a person. The mass of the monkey is 30 kg. The sides of the ride are designed so that the maximum friction riders will experience is 0.4*Fn.

a) The rider turns so that it completes 10 rotations every 60 sec.
i) Determine monkey's velocity
ii) Determine normal force
iii) Will the monkey stay in position or will it slide down the wall?

b) Determine the max velocity that will cause the monkey to stick motionless to the wall when the floor drops away.

Homework Equations

The Attempt at a Solution

i) 1 rotation=6 s
so v=2*pi*r/T
=(2*pi*10)/6

yes

ii) F=ma
30*a

a=v^2/r

((2*pi*10)/6)^2/10

That's the centripetal acceleration, so what's the centripetal (normal) force?

how can i solve iii and c?

For iii, what forces are acting in the vertical direction? For (b), look first in the vertical direction and then in the radial direction. directions.

 

[runningirl]

yes That's the centripetal acceleration, so what's the centripetal (normal) force? For iii, what forces are acting in the vertical direction? For (b), look first in the vertical direction and then in the radial direction. directions.

oh, for ii it should be times 30. i forgot to type it in :)

for b, would i have to make fn=0 and then do:
0+mg=mv^2/r?
and then for part iii... it wouldn't slip because... fn>mg?

 

[PhanthomJay]

oh, for ii it should be times 30. i forgot to type it in :)

yes, OK, that gives the centripetal force, which is the normal force of the wall acting on the monkey in the centripetal (horizontal) direction.

for b, would i have to make fn=0 and then do:
0+mg=mv^2/r?

No, this is a rotor ride moving in a horizontal circle, not a roller coaster. Look in the verical direction...the monkey's weight acts vertically down, and what other force acts vertically up? Then use Newton's laws in that direction to solve for that upward force and ultimately the required normal force in the horizontal direction.

and then for part iii... it wouldn't slip because... fn>mg?

There are 2 normal forces in this problem. One is the vertical normal contact force between the floor and the monkey (before the floor drops away). The other normal force is the horizontal contact force of the wall on the monkey. Focus on the latter for the normal force to use in the problem.

 

[runningirl]

for iii)
Ff=Fn*.4
which is less than mg...
meaning that it would slide down.

though i don't get the part of the min. velocity.

also, there was another part to the question:

Should we worry when a person weighing more than the monkey gets on the ride? why or why not?

 

[PhanthomJay]

that upward force is fn.. so why would i need to use Newton's laws?
and even if i do find that upward force, how would that help me find the normal force in the horizontal direction?

i guess to solve part b i would need to figure our part iii and compare that to something?

When you look at the forces on the monkey in the vertical direction, assume the floor is not there, and that the friction force between the wall and monkey is the upward force. The friction force is a function of the normal (centripetal) force perpendicular to it.

 

[PhanthomJay]

for iii)
Ff=Fn*.4
which is less than mg...
meaning that it would slide down.

Yes, where Fn is mv^2/r

though i don't get the part of the min. velocity.

What value of Fn will make the monkey NOT slip? Then calculate the min veleocity

also, there was another part to the question:

Should we worry when a person weighing more than the monkey gets on the ride? why or why not?

Instead of using m =30 kg, let m = m, then see what happens with your equations...

 

[runningirl]

Yes, where Fn is mv^2/r What value of Fn will make the monkey NOT slip? Then calculate the min veleocity
Instead of using m =30 kg, let m = m, then see what happens with your equations...

maximum friction force=.4(Fn)
so.. i'd have .4(Fn)=mg and then find Fn for that and then say that any value greater than that would make it not slip.

THANKS SO MUCH! :)
you were so helpful!

 

[runningirl]

okay, what do you mean set m=m?
how would that help me find out whether the person would slip or not?

thanks!

 

[Doc Al]

okay, what do you mean set m=m?
how would that help me find out whether the person would slip or not?

Rather than use a particular numeric value for the mass, such as 30 kg, just call the mass 'm' and solve things symbolically. As PhanthomJay says: see what happens.

 

[runningirl]

Rather than use a particular numeric value for the mass, such as 30 kg, just call the mass 'm' and solve things symbolically. As PhanthomJay says: see what happens.

okay, that's what i did.. and i got...m to cancel out?!
.4(10.47^2(m))/10)=9.8m?!
i just did what i did for the previous part of the problem.

 

[Doc Al]

okay, that's what i did.. and i got...m to cancel out?!

Yes, the mass cancels out. So what does that tell you?

 

[runningirl]

um... I'm not sure.
that it doesn't matter what the weight is?

 

[Doc Al]

um... I'm not sure.
that it doesn't matter what the weight is?

Exactly. As long as the minimum speed is met, the mass doesn't matter.