# Circular motion of hoop and mass

#### HardestPart

1. The problem statement, all variables and given/known data

A mass M of 6.00E-1 kg slides inside a hoop of radius R=1.40 m with negligible friction. When M is at the top, it has a speed of 5.23 m/s. Calculate the size of the force with which the M pushes on the hoop when M is at an angle of 31.0°.

I have no idea from where to begin

I dont want you to do my homework,Can someone give me a hint?

#### rock.freak667

Homework Helper
Start by drawing a free body diagram. The use conservation of energy.

#### HardestPart

When I do the diagram
the total forces work on the mass is :
mg=mv^2/r

the low of energy is:
Mgh=1/2mv^2r

But I cant see how the angle is related to the sloution?

#### ideasrule

Homework Helper
Can you use the conservation of energy to figure out the speed of the mass at 31 degrees? If so, do that first, and make sure you can get a numerical value.

Also, mg is not the only force on the block. N, the normal force, provides part of the centripetal acceleration.

#### HardestPart

But I dont know which force I have to do to components
I need cos ans sin right?
How can I do that?

#### HardestPart

I tried to find the speed of the mass at 31 degrees by:
N+mgcosa=mv^2/r
but i dont know what to but instead of N?

#### HardestPart

I tried to slove the question by :
N+mgcosa=mv^2/r
but i dont know what to put instead of N!?

#### rock.freak667

Homework Helper
I tried to slove the question by :
N+mgcosa=mv^2/r
but i dont know what to put instead of N!?
N is what you want to find.

#### HardestPart

I tried to fins the speed at angle 31 by:
I/2mv^2+mg2R=1/2mv^2+mgRsina
all m goes together
v^2+4gR=v^2+2gRsina
the answer i get from solving it i plugg into this:
N+mgcosa=mv^2/R
N=mv^2/Rsina-mgcosa
I get N=18.27N
but it is a wrong answer
Can you tell me where i went wrong?

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