Uniform circular motion question

[isukatphysics69]

1. Homework Statement
In picture

Homework Equations

f=ma
v^2/r

The Attempt at a Solution

ΣF_y=-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??

 

[PeterO]

View attachment 224234 1. Homework Statement
In picture

Homework Equations

f=ma
v^2/r

The Attempt at a Solution

ΣF_y=-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.

 

[isukatphysics69]

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
ΣF_y = FN -mg = m(v^2/r)
FN =

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

 

[PeterO]

View attachment 224234 1. Homework Statement
In picture

Homework Equations

f=ma
v^2/r

The Attempt at a Solution

ΣF_y=-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
ΣF_y=-mg - Fn = m(-v^2/r)
It should be
ΣF_y=-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.

 

[ehild]

-Fn = 55(-15.6) + 55*9.8 = -319
Fn = -319

What am i doing wrong here??

-Fn = -319. What is Fn?

 

[isukatphysics69]

-Fn = -319. What is Fn?

i see i made a stupid mistake, thank you