Reaction on the particle in terms of angular velocity

In summary: F = Fw + FrF = mg + mω2rF = m (g + ω2r)F = m (g + 0.4 ω2)Therefore, the reaction force of the table on the mass can be expressed as m (g + 0.4 ω2).
  • #1
moenste
711
12

Homework Statement


A particle is attached by means of a light inextensible string to a point 0.4 m above a smooth horizontal table. The particle moves on the table in a circle of radius 0.3 m with angular velocity ω. Find the reaction on the particle in terms of ω. Hence find the maximum angular velocity for which the particle can remain on the table

Answers: m (g - 0.4 ω2), √5g/2

2. The attempt at a solution
At first we find the hypotenuse: 0.42 + 0.32 = 0.52 m. Then we find sin α = 0.6 and cos α = 0.8.

The maximum angular velocity is:
R cos α = mg (vertical)
R sin α = mω2r (horizontal)

R = mg / 0.8 and R = mω2r / 0.6
mg / 0.8 = mω2r / 0.6
0.6 mg = 0.8 mω2r
g = 0.4 ω2
ω = √5g/2 or 5 rad s-1 (if g = 10).

Though I don't know where to start with "Find the reaction on the particle in terms of ω." Isn't it R sin α = mω2r → R = 0.5 mω2?

Update:
We have R = 0.5 mω2 and R = mg / 0.8
0.5 mω2 = mg / 0.8
0.4 mω2 = mg
0 = mg - 0.4 mω2
0 = m (g - 0.4 ω2)

It does fit the answer, but I don't quite understand how zero (0) represents "the reaction on the particle". Any ideas please?
 
Physics news on Phys.org
  • #2
Start by drawing a diagram of the situation labeling all the forces.
 
  • Like
Likes moenste
  • #3
gleem said:
Start by drawing a diagram of the situation labeling all the forces.
Did that. This is what I have:
416fdbf474c1.jpg
 
  • #4
I am not sure what "reaction" means in this context. From the wording, I guess that it is the normal force exerted by the table. It goes to zero when the mass is just about ready to lift off the surface. You need to write Newton's 2nd law in the horizontal and vertical direction and include the normal force in the vertical direction and the components of the tension in the vertical and radial directions.

On edit: When I posted, the drawing was not available. You need to include the normal force in the vertical direction, not just the component of the tension.
 
  • Like
Likes moenste
  • #5
kuruman said:
I am not sure what "reaction" means in this context. From the wording, I guess that it is the normal force exerted by the table. It goes to zero when the mass is just about ready to lift off the surface.
If it goes to zero isn't this correct then:
We have R = 0.5 mω2 and R = mg / 0.8
0.5 mω2 = mg / 0.8
0.4 mω2 = mg
0 = mg - 0.4 mω2
0 = m (g - 0.4 ω2)
?

kuruman said:
On edit: When I posted, the drawing was not available. You need to include the normal force in the vertical direction, not just the component of the tension.
You mean this kind of normal force?
 
  • #6
What is the meaning of your last equation? for the record.
 
  • Like
Likes moenste
  • #7
gleem said:
What is the meaning of your last equation? for the record.
The vertical and horizontal components of force are equally strong and therefore are equal to zero. But again, I am not sure about this part, though it does fit the answer.
 
  • #8
moenste said:
The vertical and horizontal components of force are equally strong and therefore are equal to zero.

Which means?
 
  • #9
gleem said:
Which means?
This (0 = m (g - 0.4 ω2)) is the reaction on the particle in terms of ω?
 
  • #10
moenste said:
The vertical and horizontal components of force are equally strong and therefore are equal to zero.

What I would say is that the vertical component of the tension in the string (i.e. the lifting component of the tension) is equal to the force of gravity resulting in no reaction of the table on the mass (i.e. the mass merely touches the table.

moenste said:
his (0 = m (g - 0.4 ω2)) is the reaction on the particle in terms of ω?

It is the condition giving the angular frequency of rotation in which the mass just lifts off from the table.
 
  • Like
Likes moenste
  • #11
To finish up how would you write the expression for the reaction force of the table on the mass as a function of ω?
 
  • Like
Likes moenste
  • #12
gleem said:
To finish up how would you write the expression for the reaction force of the table on the mass as a function of ω?
F = mg
F = mω2r

mg = mω2r → mg - mω2r
?
 
  • #13
Look at your drawing. Is the entire force F opposing the weight mg? If not, how much of F opposes mg? Same thing in the horizontal direction. What piece of F provides the centripetal acceleration?
 
  • Like
Likes moenste
  • #14
moenste said:
F = mg
F = mω2r

mg = mω2r → mg - mω2r
?

Just a note about notation that can help a bit. You used the same symbol in F in two different expressions. This is true only for the situation where the rotation is just fast enough to lift the mass off the table. You should first write the general case in which you would distinguish between the weight and the lifting forces of the rotation.

Ie. Fr = mrω2 and

Fw = mg

Then apply the condition that the rotation is just fast enough to lift the mass of the table i.e. Fw = Fr
Make sure you distinguish the forces first.
 
  • Like
Likes moenste
  • #15
kuruman said:
Look at your drawing. Is the entire force F opposing the weight mg? If not, how much of F opposes mg? Same thing in the horizontal direction. What piece of F provides the centripetal acceleration?
F cos α opposes mg (vertical) and F sin α opposes mω2r.

gleem said:
Just a note about notation that can help a bit. You used the same symbol in F in two different expressions. This is true only for the situation where the rotation is just fast enough to lift the mass off the table. You should first write the general case in which you would distinguish between the weight and the lifting forces of the rotation.

Ie. Fr = mrω2 and

Fw = mg

Then apply the condition that the rotation is just fast enough to lift the mass of the table i.e. Fw = Fr
Make sure you distinguish the forces first.
Where Fw is weight and Fr = radius / radian?
 
  • #16
As gleem wrote, it is better to think of it as Fw = mg which is equal to (as you said) F cosα. If you put these two together what do you get? Likewise, put together Fr = mω2r and Fr = F sin α.
 
  • Like
Likes moenste
  • #17
kuruman said:
As gleem wrote, it is better to think of it as Fw = mg which is equal to (as you said) F cosα. If you put these two together what do you get? Likewise, put together Fr = mω2r and Fr = F sin α.
If we put Fr = mω2r and Fw = mg together we have F cos α = mg and F sin α = mω2r
mg / cos α = mω2r / sin α
mg sin α = mω2r cos α
g sin α = ω2r cos α
 
  • #18
Can you solve this for ω which is one of the questions in the problem?

Knowing ω, can you find F, which is the other question?
 
  • Like
Likes moenste
  • #19
kuruman said:
Can you solve this for ω which is one of the questions in the problem?

Knowing ω, can you find F, which is the other question?
g sin α = ω2r cos α
ω = √(g sin α) / (r cos α)

Isn't it what've I found in the first place?
moenste said:
At first we find the hypotenuse: 0.42 + 0.32 = 0.52 m. Then we find sin α = 0.6 and cos α = 0.8.

The maximum angular velocity is:
R cos α = mg (vertical)
R sin α = mω2r (horizontal)

R = mg / 0.8 and R = mω2r / 0.6
mg / 0.8 = mω2r / 0.6
0.6 mg = 0.8 mω2r
g = 0.4 ω2
ω = √5g/2 or 5 rad s-1 (if g = 10).
 
  • #20
Right. Just to make things neater what is sin α / cos α ? Also, what about F as a function of ω?
 
  • Like
Likes moenste
  • #21
kuruman said:
Right. Just to make things neater what is sin α / cos α ? Also, what about F as a function of ω?
Since sin α / cos α = tan α, so: ω = √(g tan α) / r

I actually don't quite understand what does "F as a function of ω" mean... Isn't it Fr = mω2r? We have F on the left side and ω on the right side.
 
  • #22
moenste said:
I actually don't quite understand what does "F as a function of ω" mean... Isn't it Fr = mω2r? We have F on the left side and ω on the right side.
No. We have the radial component Fr on the left side not the magnitude F. You are looking for F and you already know how Fr is related to F. So ...
 
  • Like
Likes moenste
  • #23
kuruman said:
No. We have the radial component Fr on the left side not the magnitude F. You are looking for F and you already know how Fr is related to F. So ...
Hm, Fr = F sin α and we have ω = √(g tan α) / r
sin α = Fr / F
ω = √(g tan α) / r = √(g sin α) / (r cos α) = √(g (Fr / F) / (r cos α) = √(gF) / (Fr r cos α)
?
 
  • #24
Leave ω as ω. You are looking for an equation that looks like F = Some expression that involves g, m, α, ω and r. That's what is meant by "the reaction force in terms of ω". Take another look at the radial equation, rewrite Fr in terms of F and solve for F.
 
  • Like
Likes moenste
  • #25
kuruman said:
Leave ω as ω. You are looking for an equation that looks like F = Some expression that involves g, m, α, ω and r. That's what is meant by "the reaction force in terms of ω". Take another look at the radial equation, rewrite Fr in terms of F and solve for F.
Maybe something like Fr = F sin α, F = Fr / sin α, since Fr = mω2r we have F = mω2r / sin α. I can substitute ω = √(g tan α) / r and get F = (mg tan α) / sin α, but in that case I don't have ω and r...
 
  • #26
F = mω2r / sin α is what you're looking for. It is an expression that gives the reaction force F in terms of ω.
 
  • Like
Likes moenste
  • #27
kuruman said:
F = mω2r / sin α is what you're looking for. It is an expression that gives the reaction force F in terms of ω.
Yes, but in that case we have F = mω2r / sin α and if we plug in r (0.3) and sin α (0.6) we have F = 0.5 mω2 and not [m (g - 0.4 ω2)] like in the answer.
 
  • #28
I see what's going on now. So far what you have called F is the tension in the string. The solution for F in the previous post is the solution for the tension when ω is at the threshold value and the mass is about to lift off, i.e. when the normal force N is zero. To answer the first question correctly, you need to include the normal force N in addition to the tension F and the weight mg in your diagram as I suggested in post #4. The reaction force that the problem is asking you to find is N and goes to zero when ω is at the threshold value. So you need to rewrite the vertical equation to include N. The horizontal equation remains unchanged.
 
  • Like
Likes moenste
  • #29
kuruman said:
So far what you have called F is the tension in the string.
Yes.

kuruman said:
The solution for F in the previous post is the solution for the tension when ω is at the threshold value and the mass is about to lift off, i.e. when the normal force N is zero.
"Hence find the maximum angular velocity for which the particle can remain on the table." as required.

kuruman said:
To answer the first question correctly, you need to include the normal force N in addition to the tension F and the weight mg in your diagram as I suggested in post #4. The reaction force that the problem is asking you to find is N and goes to zero when ω is at the threshold value. So you need to rewrite the vertical equation to include N. The horizontal equation remains unchanged.
Hm, how would it look like? I mean shall I just write "N = normal reaction" just under F cos α on the graph?

F cos α + N = mg?
F sin α = mω2r
 
  • #30
kuruman said:
I see what's going on now. So far what you have called F is the tension in the string. The solution for F in the previous post is the solution for the tension when ω is at the threshold value and the mass is about to lift off, i.e. when the normal force N is zero. To answer the first question correctly, you need to include the normal force N in addition to the tension F and the weight mg in your diagram as I suggested in post #4. The reaction force that the problem is asking you to find is N and goes to zero when ω is at the threshold value. So you need to rewrite the vertical equation to include N. The horizontal equation remains unchanged.
It gets F cos α + N = mg and F sin α = mω2r so F = (mg - N) / cos α and F = mω2r / sin α
(mg - N) / cos α = mω2r / sin α
(mg - N) sin α = mω2r cos α
ω = √ [(mg - N) tan α] / mr
 
Last edited:
  • #31
Almost there. You need to solve for N not ω, that is have N by itself on the left side of the equation.
 
  • Like
Likes moenste
  • #32
kuruman said:
Almost there. You need to solve for N not ω, that is have N by itself on the left side of the equation.
Hm, it indeed gets to N = m (g - 0.4 ω2). And that is the reaction on the particle in terms of ω?
 
  • #33
Yes it is. IMO it would have been clearer to have asked for the normal force in terms of ω.
 
  • Like
Likes moenste

FAQ: Reaction on the particle in terms of angular velocity

1. What is angular velocity?

Angular velocity is a measure of the rate at which an object rotates or moves around a fixed axis. It is commonly represented by the symbol ω and is measured in radians per second (rad/s).

2. How does angular velocity affect particles?

Angular velocity affects particles by determining the speed and direction of their rotational motion. The greater the angular velocity, the faster the particle will rotate around the axis.

3. What factors influence the angular velocity of a particle?

The angular velocity of a particle can be influenced by factors such as the distance from the axis of rotation, the mass of the particle, and the force acting on the particle.

4. What is the relationship between angular velocity and linear velocity?

Angular velocity and linear velocity are related by the radius of rotation. The linear velocity of a particle at a certain distance from the axis of rotation is equal to the product of its angular velocity and the distance from the axis.

5. How is angular velocity measured?

Angular velocity can be measured using various instruments such as a tachometer or an accelerometer. It can also be calculated using the formula ω = Δθ/Δt, where Δθ is the change in angle and Δt is the change in time.

Back
Top