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Blocks on turntable

  • Thread starter huskydc
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Two identical blocks are tied together with a string and placed along the same radius of a turntable that is spinning about its center. The inner block is 4 cm from the center and the outer block is 5 cm from the center. The coefficient of static friction between the turntable and the blocks is µs = 0.79, and the string is taut.

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What is the maximum angular frequency such that neither block slides?
 

James R

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How far have you got with this so far?
 
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well, since i know what r and µ is, i was thinking about solving for velocity(tangential) first, but my question is that for r, do i account for two blocks' position?

that is, what should I use for r? the farthest block out (r =5cm)? i'm clueless with physics.
 
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The key is that neither block slides and the string is taught. That means you can take advantage of a balance of forces between the blocks and the table. Start by drawing the blocks from the side view and write down all the forces that you know are there. Then, you can start filling in forces because you know they must be balanced so that it doesn't move.
 
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ok i got!!! if you're curious, i got omega = 13.12
 
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It would seem that husky and I are either in the same physics class, or using the same homework system. :) I'm still having trouble with this one (this and the accel plane are killing me).

This is my problem:
Two identical blocks are tied together with a string and placed along the same radius of a turntable that is spinning about its center. The inner block is 2 cm from the center and the outer block is 6 cm from the center. The coefficient of static friction between the turntable and the blocks is µs = 0.8, and the string is taut.

What is the maximum angular frequency such that neither block slides?

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I think I'm just struggling with what the angular frequency actually IS. This is what I think I understand: when the turntable is rotating, the static friction force is pushing each block toward the center of the turntable. The tensions in the string between the two blocks point at each other:

[ 1 ] ---T---> ---- <---T--- [2]

I have no idea how to get started on this though. I'm not sure what I'm actually solving for.

Thank you so much!

A
 
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I've been working on this one again all morning. :\ In the vertical direction, the only forces are the normal force and the weight, which cancel each other out. In the horizontal direction, are the only two forces the static friction (pointing toward the center of the turntable for both blocks) and the tension (pointing outward for block 1 and inward for block 2)?

I know I need to set up 2 equations w/two unknowns, the angular frequency and the tension, but I'm not sure how to get started with that. I'm completely lost. :( I know when I find those equations and set them equal to ma, the a is the centripetal force, so a=v^2/r.

If the two equations are:

(block one) sigma F = T - f = ma
(block two) sigma F = -T - f = ma

How can I figure out what T and f are? I don't have the mass either. I think I'm missing something very key here. :(

A
 

Pyrrhus

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Try this approach for circular motion:

[tex] a_{radial} = R \dot{\theta}^{2} [/tex]

so

[tex] \sum F_{x} = mR \dot{\theta}^{2} [/tex]

From the FBDs you will have 2 unknowns and 2 equations, you can solve this.
 
What is the capital R in those equations?

I'm sorry. I'm so lost. :(
 

Pyrrhus

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R is for radius, i got a question does the string passes throught the centroid the turntable? or are the two block tied together such as the string doesn't pass throught the centroid of the turntable?
 
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Is R1 = 2, and R2 = 6, are those the two radii? I don't understand where I find the theta from for each equation, since I don't have an angle? I have it all drawn out on a whiteboard, but I think there's something I'm missing. Or is the theta in those equations the angular frequency, and thus what I'm supposed to find?

My roommate is digging out his old physics textbook for me, hoping it has a better definition of angular frequency. I think that's what I don't understand.
 
The string only connects the two blocks, it is not attached to the turntable in any way.
 

Pyrrhus

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Yes that theta is what you need to find.

inner block:

[tex] -T + \mu mg = mR_{1} \dot{\theta}^2 (1)[/tex]

outer block:

[tex] T + \mu mg = mR_{2} \dot{\theta}^2 (2)[/tex]

ok now add 1 and 2.
 
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Pyrrhus

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You know the arclength equation?

[tex] ds = R d \theta [/tex]

Well when R is constant such as in the circular motion you can get

[tex] \frac{ds}{dt} = R \frac{d \theta}{dt} [/tex]

[tex] v = R \dot{\theta} [/tex]

where [itex] \frac{d \theta}{dt} = \dot{\theta} [/itex] is the angular speed also known as angular frequency (in oscillatory movements, vibrations)

for the normal component of acceleration:

[tex] a_{radial} = \frac{v^2}{R} = R \dot{\theta}^{2} [/tex]
 
How can I get those masses to cancel out? I'm ending up w/three unknowns. Tension, mass, and theta. :\
 
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huskydc said:
ok i got!!! if you're curious, i got omega = 13.12
Did anyone confirm this? I actually got a different answer, 39.35 rad/sec.
 

Pyrrhus

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[tex] -T + \mu mg = mR_{1} \dot{\theta}^2 (1)[/tex]

[tex] T + \mu mg = mR_{2} \dot{\theta}^2 (2)[/tex]

After adding

[tex] 2 \mu mg = mR_{1} \dot{\theta}^2 + mR_{2} \dot{\theta}^2 [/tex]

[tex] 2 \mu g = R_{1} \dot{\theta}^2 + R_{2} \dot{\theta}^2 [/tex]
 
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hey aki, do you have a instant messenger or something, b/c i can walk you through it if you want.
 
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Aki Yamaguchi said:
How can I get those masses to cancel out? I'm ending up w/three unknowns. Tension, mass, and theta. :\
Don't forget about the friction force. That has a factor of m inside it. Get rid of the Tension by solving for T for both equations and setting them equal.
 

OlderDan

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Cyclovenom said:
R is for radius, i got a question does the string passes throught the centroid the turntable? or are the two block tied together such as the string doesn't pass throught the centroid of the turntable?
The problem says the blocks are on the same radius, so the string is only on one side of center.

Cyclovenom said:
Yes that theta is what you need to find.

inner block:

[tex] -T - \mu mg = mR_{1} \dot{\theta}^2 (1)[/tex]

outer block:

[tex] T - \mu mg = mR_{2} \dot{\theta}^2 (2)[/tex]

ok now add 1 and 2.
You might want to change your sign convention. The RHS of these equations are positive quantities, so the LHS of each equation should be net positive.
 

Doc Al

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Aki Yamaguchi said:
I know I need to set up 2 equations w/two unknowns, the angular frequency and the tension, but I'm not sure how to get started with that. I'm completely lost. :( I know when I find those equations and set them equal to ma, the a is the centripetal force, so a=v^2/r.
Another form for the centripetal acceleration is [itex]\omega^2 R[/itex]; this may prove more useful.

If the two equations are:

(block one) sigma F = T - f = ma
(block two) sigma F = -T - f = ma
Realize that while both blocks have the same angular speed, they have different accelerations. And mind your sign convention: both the acceleration and the friction point towards the center of the turntable.

How can I figure out what T and f are? I don't have the mass either.
T will end up being one of your unknowns, along with the angular speed. f will be the maximum static friction force, which is [itex]\mu N[/itex]. (What's N?) You don't need the mass.
 
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Pyrrhus

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You are right Olderdan, i made a mistake in my sign convention.

inner block:
[tex] -T + \mu mg = mR_{1} \dot{\theta}^2 (1)[/tex]

outer block:
[tex] T + \mu mg = mR_{2} \dot{\theta}^2 (2)[/tex]
 
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Jelfish said:
Did anyone confirm this? I actually got a different answer, 39.35 rad/sec.
Nevermind!
 
When I solve equation 1 for T, and stick that into equation two, I can't seem to solve it. All of the masses cancel out (each term has a mass, so dividing by m cancels them all out), but then when I try to solve it, it doesn't work.

(w = theta)

1) T - umg = m R1 w^2
2) -T - umg = m R2 w^2

so from equation 1, T = umg + m R1 w^2

sticking that into equation 2:

- (umg + m R1 w^2) - umg = m R2 w^2

Since each of those terms has a mass, if you divide the entire thing by mass, all of the mass cancels out (which I thought would be good). I'm then left with one unknown, w=theta.

However, when I try to solve for that, it doesn't seem to work. Even if I enter it into my TI-89 and hit "solve" it just comes up with "false" ??

Thank you so much for trying to help me with this. I'm sorry I'm so bad at it. :\

A
 
huskydc: my aim name is forestkelp, if you're still around
 

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