Normal force of a mass sliding down a quarter circle

[Jamest39]

Homework Statement

A 22-gram mass is released from rest at position A on a stationary frictionless surface whose shape is that of a quarter circle of radius R = 0.66 m.
(b) Determine the magnitude of the normal force that acts on the mass when it is at position B (it is at position B when the imaginary radii form a 48° on the top sector of the quarter circle and a 42° angle on the bottom half).

Homework Equations

N = mgcosθ

The Attempt at a Solution

In order to use the equation above to determine the normal force, you need would need an angle between the normal force vector and the mg vector but couldn't find any way to do so. The answer was given to us as 0.48 N, and I plugged that into the equation above to get θ = 77.14°, but I don't know any way I would be able to get that angle.

 

[RedDelicious]

Is this the full question?

I notice you went from the question to part (b). Please include all information, if you haven't.

 

[Jamest39]

Is this the full question?

I notice you went from the question to part (b). Please include all information, if you haven't.

Yeah, question (a) was "Determine the speed of the mass at the instant when the mass is at position B."
mgh = (1/2)mv^2
v = sqrt(2gh)
h = R sin θ So,
v = sqrt(2*9.8*0.66*sin48°) = 3.10 m/s

 

[RedDelicious]

Yeah, question (a) was "Determine the speed of the mass at the instant when the mass is at position B."
mgh = (1/2)mv^2
v = sqrt(2gh)
h = R sin θ So,
v = sqrt(2*9.8*0.66*sin48°) = 3.10 m/s

That's what I thought.

Do you think that might be related to part b in any way? Keep in mind this is circular motion. Any forces that might affect the normal force?

 

[Jamest39]

That's what I thought.

Do you think that might be related to part b in any way? Keep in mind this is circular motion. Any forces that might affect the normal force?

The centripetal acceleration? a = v^2/r

 

[RedDelicious]

The centripetal acceleration? a = v^2/r

Yep. Now you just need to use Newton's 2nd law again but with all the forces this time.

 

[Jamest39]

Yep. Now you just need to use Newton's 2nd law again but with all the forces this time.

So ΣF = ma,
and the forces acting on it are gravity and the normal force. But how can I express mg in that formula when its at that position on the curve?

 

[RedDelicious]

So ΣF = ma,
and the forces acting on it are gravity and the normal force. But how can I express mg in that formula when its at that position on the curve?

You do it the same way you did it the first time. The first time you set it equal to the normal force because it was the only force acting on it, but this time it isn't, so you don't, but its component is still the same.

 

[Jamest39]

You do it the same way you did it the first time. The first time you set it equal to the normal force because it was the only force acting on it, but this time it isn't, so you don't, but its component is still the same.

I can set ΣF = ma equal to the magnitude of the normal force?

 

[RedDelicious]

I can set ΣF = ma equal to the magnitude of the normal force?

Before you get there you're going to need the correct velocity.

You set PE = KE, which would be fine if it were true. What that implies is that all of the potential energy it initially had has been converted into kinetic energy, which we know is not true because you found its height at point B and it definitely wasn't zero, meaning it has potential energy that you're not accounting for. Starting from the basic fact that the conservation of energy tells us that, in an isolated system, the initial energy equals the final energy.

[tex]Ei=E_f\\

mgh_i+\frac{1}{2}mv_i^2=mgh_f+\frac{1}{2}mv_f^2[/tex]

or in your case specifically

[tex]mgR=mgRsin\left(\theta \right)+\frac{1}{2}mv_f^2[/tex]