Uniform circular motion question

In summary: Homework Statement In pictureHomework Equationsf=mav^2/rThe Attempt at a SolutionΣFy=-mg - Fn = m(-v^2/r)-55*9.8 - Fn = m(-v^2/r)v= 16π/4.5 since circumference = 2πr divide by the rotations time taken per second = 11.17m/s-Fn = 55(-15.6) + 55*9.8Fn = -319What am i doing wrong here??In summary, the student is trying to solve a homework problem, but is
  • #1
isukatphysics69
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1. Homework Statement

In picture

Homework Equations


f=ma
v^2/r

The Attempt at a Solution


ΣFy=-mg - Fn = m(-v^2/r)
-55*9.8 - Fn = m(-v^2/r)
v= 16π/4.5 since circumference = 2πr divide by the rotations time taken per second = 11.17m/s
-Fn = 55(-15.6) + 55*9.8
Fn = -319

What am i doing wrong here??
 

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  • #2
isukatphysics69 said:
View attachment 224234 1. Homework Statement
In picture

Homework Equations


f=ma
v^2/r

The Attempt at a Solution


ΣFy=-mg - Fn = m(-v^2/r)
-55*9.8 - Fn = m(-v^2/r)
v= 16π/4.5 since circumference = 2πr divide by the rotations time taken per second = 11.17m/s
-Fn = 55(-15.6) + 55*9.8
Fn = -319

What am i doing wrong here??
When the rider is in the top position, all forces are in the same direction - down.
Gravity acts down, the ring pushes down (it has to push down, since the rider is below the ring at that point) and the Nett Force is also down (towards the centre, of course, which is below/down from the rider)
Since all three forces (two acting plus one resultant) are in the same direction, it is just a simple addition of magnitudes.
Weight Force + Action Force = Resultant force
(Action Force is the force that the ring pushes with.)

BTW: when you get the answer, just say xxx.xx N down, rather than +xxx.xx N or - xxx.xxN. Otherwise you have to define what direction + or - mean.
 
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  • #3
Ok i guess they just wanted magnitude so the answer was just 320N rounded.. now i am on the second part which is
Find the force that the ring pushes on person at the bottom of the ride.. so now the normal force is positive, mg is still negative, so
ΣFy = FN -mg = m(v^2/r)
FN =
PeterO said:
When the rider is in the top position, all forces are in the same direction - down.
Gravity acts down, the ring pushes down (it has to push down, since the rider is below the ring at that point) and the Nett Force is also down (towards the centre, of course, which is below/down from the rider)
Since all three forces (two acting plus one resultant) are in the same direction, it is just a simple addition of magnitudes.
Weight Force + Action Force = Resultant force
(Action Force is the force that the ring pushes with.)
Thank you! got it
 
  • #4
isukatphysics69 said:
View attachment 224234 1. Homework Statement
In picture

Homework Equations


f=ma
v^2/r

The Attempt at a Solution


ΣFy=-mg - Fn = m(-v^2/r)
-55*9.8 - Fn = m(-v^2/r)
v= 16π/4.5 since circumference = 2πr divide by the rotations time taken per second = 11.17m/s
-Fn = 55(-15.6) + 55*9.8
Fn = -319

What am i doing wrong here??
I think the error is in the line
ΣFy=-mg - Fn = m(-v^2/r)
It should be
ΣFy=-mg - Fn = -m(v^2/r)

It is the Right Hand side as a whole that is negative, not just the component v. The negative sign is just a reference to direction of the quantity, not its components.
 
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  • #5
isukatphysics69 said:
-Fn = 55(-15.6) + 55*9.8 = -319
Fn = -319

What am i doing wrong here??

-Fn = -319. What is Fn?
 
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  • #6
ehild said:
-Fn = -319. What is Fn?
i see i made a stupid mistake, thank you
 

1. What is uniform circular motion?

Uniform circular motion is a type of motion in which an object moves in a circular path at a constant speed. This means that the object covers equal distances in equal time intervals, resulting in a constant velocity.

2. What causes uniform circular motion?

The object in uniform circular motion experiences a centripetal force, which is directed towards the center of the circular path. This force is responsible for keeping the object moving in a circular path at a constant speed.

3. Can an object in uniform circular motion have a changing speed?

No, an object in uniform circular motion cannot have a changing speed. This is because the object's velocity is always tangent to the circular path and therefore, the speed cannot change. The only way the speed can change is if there is a change in the radius of the circular path.

4. What is the relationship between the radius of the circular path and the speed of the object in uniform circular motion?

The speed of the object in uniform circular motion is directly proportional to the radius of the circular path. This means that as the radius increases, the speed also increases and vice versa.

5. How is uniform circular motion different from linear motion?

In uniform circular motion, the object moves along a circular path at a constant speed, while in linear motion, the object moves along a straight path at a constant speed. Additionally, in uniform circular motion, the direction of the object's velocity is constantly changing, while in linear motion, the direction remains constant.

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