Reaction on the particle in terms of angular velocity

[moenste]

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?

 

[gleem]

Start by drawing a diagram of the situation labeling all the forces.

 

[moenste]

Start by drawing a diagram of the situation labeling all the forces.

Did that. This is what I have:

 

[kuruman]

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.

 

[moenste]

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)
?

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?

 

[gleem]

What is the meaning of your last equation? for the record.

 

[moenste]

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.

 

[gleem]

The vertical and horizontal components of force are equally strong and therefore are equal to zero.

Which means?

 

[moenste]

Which means?

This (0 = m (g - 0.4 ω²)) is the reaction on the particle in terms of ω?

 

[gleem]

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.

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.

 

[gleem]

To finish up how would you write the expression for the reaction force of the table on the mass as a function of ω?

 

[moenste]

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
?

 

[kuruman]

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?

 

[gleem]

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. F_r = mrω² and

F_w = mg

Then apply the condition that the rotation is just fast enough to lift the mass of the table i.e. F_w = F_r
Make sure you distinguish the forces first.

 

[moenste]

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.

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. F_r = mrω² and

F_w = mg

Then apply the condition that the rotation is just fast enough to lift the mass of the table i.e. F_w = F_r
Make sure you distinguish the forces first.

Where F_w is weight and F_r = radius / radian?

 

[kuruman]

As gleem wrote, it is better to think of it as F_w = mg which is equal to (as you said) F cosα. If you put these two together what do you get? Likewise, put together F_r = mω²r and F_r = F sin α.

 

[moenste]

As gleem wrote, it is better to think of it as F_w = mg which is equal to (as you said) F cosα. If you put these two together what do you get? Likewise, put together F_r = mω²r and F_r = F sin α.

If we put F_r = mω²r and F_w = 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

 

[kuruman]

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?

 

[moenste]

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?

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).

 

[kuruman]

Right. Just to make things neater what is sin α / cos α ? Also, what about F as a function of ω?

 

[moenste]

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 F_r = mω²r? We have F on the left side and ω on the right side.

 

[kuruman]

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 F_r on the left side not the magnitude F. You are looking for F and you already know how F_r is related to F. So ...

 

[moenste]

No. We have the radial component F_r on the left side not the magnitude F. You are looking for F and you already know how F_r is related to F. So ...

Hm, F_r = F sin α and we have ω = √(g tan α) / r
sin α = F_r / F
ω = √(g tan α) / r = √(g sin α) / (r cos α) = √(g (F_r / F) / (r cos α) = √(gF) / (F_r r cos α)
?

 

[kuruman]

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 F_r in terms of F and solve for F.

 

[moenste]

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 F_r in terms of F and solve for F.

Maybe something like F_r = F sin α, F = F_r / sin α, since F_r = 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...

 

[kuruman]

F = mω²r / sin α is what you're looking for. It is an expression that gives the reaction force F in terms of ω.

 

[moenste]

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.

 

[kuruman]

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.

 

[moenste]

So far what you have called F is the tension in the string.

Yes.

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.

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

 

[moenste]

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