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

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

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James R

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How far have you got with this so far?

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that is, what should I use for r? the farthest block out (r =5cm)? i'm clueless with physics.

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- #5

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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?

-----

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

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?

-----

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 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

- #8

Pyrrhus

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[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.

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What is the capital R in those equations?

I'm sorry. I'm so lost. :(

I'm sorry. I'm so lost. :(

- #10

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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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.

- #12

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The string only connects the two blocks, it is not attached to the turntable in any way.

- #13

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.

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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- #14

Pyrrhus

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[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]

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- #16

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Did anyone confirm this? I actually got a different answer, 39.35 rad/sec.huskydc said:ok i got!!! if you're curious, i got omega = 13.12

- #17

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]

[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.

- #19

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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.Aki Yamaguchi said:

- #20

OlderDan

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The problem says the blocks are on the sameCyclovenom 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?

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.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.

- #21

Doc Al

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Another form for the centripetal acceleration is [itex]\omega^2 R[/itex]; this may prove more useful.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.

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.If the two equations are:

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

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

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.How can I figure out what T and f are? I don't have the mass either.

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- #22

Pyrrhus

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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]

- #23

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Nevermind!Jelfish said:Did anyone confirm this? I actually got a different answer, 39.35 rad/sec.

- #24

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(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

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

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