Circular Motion Problem: Conservation of Energy and Newton's Second Law

[inky]

Homework Statement

Please see attachment my problem.

Homework Equations

Law of conservation of energy
Newton's Second law for circular motion

The Attempt at a Solution

Please see attachment.
I didn't get the answer same as solution. Please help me.

 

[ehild]

R2 and R1 are vectors, of opposite directions. You calculated their magnitude. Determine the vector difference.

ehild

 

[inky]

Thank you for quick reply. This is the old exam question. But I need magnitutude of R2-R1. Please see the exam solution.

 

[ehild]

Oops. Take care of the direction of R1 and the condition that theta should be less than 33.7 degrees. Is then cosθ lower or higher than 2/3? Just think of the case when θ=0. What is the direction of R1?

ehild

 

[inky]

If Θ=0, direction of R1 is toward the centre at any pointl. I didn't get idea as well.

 

[inky]

I confused one thing is in solution,
R1=mgcos(theta)-mv[2]/a

I got
R1=mv[2]/a-mgcos(theta)

Could you explain for this? Thanks a lot.

 

[ehild]

It is

R1=mgcos(theta)-mv^2/a,

because R1 points upward, away from the centre. The direction of the force R1 is wrong as shown in the figure. Just think: What is the direction of R1 at the top, so as the bead do not fall down?

As R1 is the magnitude of the force, it can not be negative. But your formula for R1 results in a negative number.



ehild

 

[inky]

No. This is vertical circular motion. For bead threaded onto the wire , direction of normal force is toward the centre at any point. If the particle is moving outside the loop, normal force is outward.

 

[ehild]

It is a bead on a wire. The normal force can be both outward and inward. How can the bead be in rest at the top, if two downward forces act on it?

Imagine you have this loop of wire, and you hold a bead to it at the top, just below the wire so the normal force acts downward. Release the bead, what happens?

ehild

 

[inky]

It is a bead on a wire. The normal force can be both outward and inward. How can the bead be in rest at the top, if two downward forces act on it?

Imagine you have this loop of wire, and you hold a bead to it at the top, just below the wire so the normal force acts downward. Release the bead, what happens?

ehild

According to the mechanics textbooks, only we consider one normal force. Not two normal force. Please see Application of Mathematics or Mechanics about vertical circular motion.

 

[inky]

According to the mechanics textbooks, only we consider one normal force. Not two normal force. Please see Application of Mathematics or Mechanics about vertical circular motion.

At the top two down forces acting on it. Normal force and weight. For roller coaster, at the top , normal force and weight directions are downward as well.

 

[ehild]

Yes, but the roller coaster has some speed at the top. If it does not move, it falls down.

The centripetal force is the resultant of normal force and gravity.

mv^2/R = N +G. N=mv^2/R-G, which can be either positive or negative, depending on v.

Why don't you calculate the value R1 with your formula for a small angle, for example 10 degrees, instead of sticking to the statement that "For bead threaded onto ("on" in the original text) the wire , direction of normal force is toward the centre at any point"? The bead has a hole, and the wire goes through it. The wire can exert force both toward and away the centre, as it can be in contact with opposite walls of that hole in the bead.

I have to leave now. Think about what I have said.

ehild

 

[inky]

Yes, but the roller coaster has some speed at the top. If it does not move, it falls down.

The centripetal force is the resultant of normal force and gravity.

mv^2/R = N +G. N=mv^2/R-G, which can be either positive or negative, depending on v.

Why don't you calculate the value R1 with your formula for a small angle, for example 10 degrees, instead of sticking to the statement that "For bead threaded onto ("on" in the original text) the wire , direction of normal force is toward the centre at any point"? The bead has a hole, and the wire goes through it. The wire can exert force both toward and away the centre, as it can be in contact with opposite walls of that hole in the bead.

I have to leave now. Think about what I have said.

ehild

Dear ehild,
Thanks a lot for explaining patiently. When is velocity negative? I didn't see any velocity is negative in my textbook. We consider force only. If the force is same as acceletatin direction, usually we take positive.

 

[ehild]

Velocity is a vector, it can have any directions. It is speed, the magnitude of velocity, that can not be negative.

You are right, the force is considered positive with respect to the acceleration if it has the same direction as that of the acceleration.

Now, about normal force: The (centripetal) acceleration points toward the centre. If the bead is at the top of the circle, gravity G=mg points also towards the centre so it is positive. The centripetal force is the sum of gravity G and the normal force R1. All these forces are vectors, they act along the vertical direction now. You found that R1=mv²/a-mgcos(theta). Calculate, R1 for theta =0, 30° and 40°, please, and tell me what you got. ehild

 

[inky]

Velocity is a vector, it can have any directions. It is speed, the magnitude of velocity, that can not be negative.

You are right, the force is considered positive with respect to the acceleration if it has the same direction as that of the acceleration.

Now, about normal force: The (centripetal) acceleration points toward the centre. If the bead is at the top of the circle, gravity G=mg points also towards the centre so it is positive. The centripetal force is the sum of gravity G and the normal force R1. All these forces are vectors, they act along the vertical direction now. You found that R1=mv²/a-mgcos(theta). Calculate, R1 for theta =0, 30° and 40°, please, and tell me what you got.


ehild

I got these values.
R(40)>R(30)>R(0)

 

[ehild]

It is all right that R1 increases with the angle, but is it positive or negative?

You wrote that

R1= mv₁²/a-mgcos(θ),

and

1/2 mv₁²=mg(1-cos(θ)),

right?

So

R1=mg(2-3 cos(θ)).

If θ=0, R1=mg(2-3)=-mg negative.
If θ=30°, R1=mg(2-2.598)= -0.598 mg, negative.
If θ=40°, R1=mg(2-2.298)= -0.298 mg, negative.

You need the magnitude of the normal force. It can not be negative. Your formula is wrong.
The magnitude of the normal force is the negative of the values above. The correct form is

R1=|mg(2-3 cos(θ))| =mg(3 cos(θ)-2)

ehild

 

[inky]

It is all right that R1 increases with the angle, but is it positive or negative?

You wrote that

R1= mv₁²/a-mgcos(θ),

and

1/2 mv₁²=mg(1-cos(θ)),

right?

So

R1=mg(2-3 cos(θ)).

If θ=0, R1=mg(2-3)=-mg negative.
If θ=30°, R1=mg(2-2.598)= -0.598 mg, negative.
If θ=40°, R1=mg(2-2.298)= -0.298 mg, negative.

You need the magnitude of the normal force. It can not be negative. Your formula is wrong.
The magnitude of the normal force is the negative of the values above. The correct form is

R1=|mg(2-3 cos(θ))| =mg(3 cos(θ)-2)

ehild

If R is negative, the particle will fall out from the circular loop.

Thank you very much ehild. Now I got a lot of knowledge. Really thanks.

 

[ehild]

Well, this one will not fall down, as it is a bead, (a small ball with a hole through the middle) and the wire goes through the hole.

ehild