Maximizing Angular Frequency for Stable Uniform Circular Motion with Two Blocks

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Homework Help Overview

The problem involves two identical blocks connected by a string on a spinning turntable, with the goal of determining the maximum angular frequency that prevents either block from sliding. The inner block is positioned 1 cm from the center, while the outer block is 6 cm away. The coefficient of static friction is given as µs = 0.75.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the forces acting on each block, including tension and friction, and how these relate to centripetal acceleration. There are attempts to derive equations for angular frequency and tension, with some questioning the signs and relationships between forces.

Discussion Status

Participants are actively exploring the relationships between tension, friction, and angular frequency. Some have successfully derived expressions for angular frequency, while others are struggling with the calculations for tension. There is no explicit consensus on the final approach, but several productive lines of reasoning are being pursued.

Contextual Notes

Participants note the horizontal nature of the turntable and the assumption that the blocks have equal mass, which simplifies some calculations. There are indications of confusion regarding algebraic manipulations and unit conversions.

nahya
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here is the problem:
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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 1 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 = 0.75, and the string is taut.

What is the maximum angular frequency such that neither block slides?
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the sum of all forces on each block = mass * centripetal acceleration.
f1/f2 = friction on the outer/inner blocks, respectively
m1/m2 = mass of the outer/inner blocks, respectively
R1/R2 = radius of the outer/inner blocks, respectively
T = tension

so, f1 + T = m1 * a; f2 + T = m2 * a, which can also be written as...
f1 + T = m1 * ω^2 * R1; f2 + T = m2 * ω^2 * R2

solving for ω^2, i got...
ω^2 = (F1 + T)/(m1 * R1);
ω^2 = (F2 + T)/(m2 * R2).

that's how far I've gone, and I'm pretty sure I've gone the wrong direction.
will anyone please direct me to a right approach?
thanks.
 
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nahya said:
so, f1 + T = m1 * a; f2 + T = m2 * a

It should be f_2 -T =m_2 a_2
the direction of the friction force of the inner one is opposite to that of the tension force.

Find the relation between T and mg, and you can find \omega
 
Last edited:
At block1, f1 and T are both acting in the same direction, radially inwards, so (f1 + T) is correct. But at block2, f2 is acting radially inwards but T is acting radially outwards!

Apart from the wrong sign, your approach looks ok so far.
 
oh, i see.
how can i relate T and mg? I'm not given a lot of information for the problem.
instead, can i add the two equations to get rid of T?
doing that, i got...
F1 + F2 = ω^2 * (m1*R1 + m2*R2)
i'm stuck there again...
 
You've got rid of the T, that's right.

Now, can you relate the friction forces with the coefft of friction and the normal reaction of the the blocks ?
 
well, the friction forces = fric. coeff. * the normal force.
the normal force, however, depends on the mass of the block, doesn't it?
 
Last edited:
The turntable is horizontal (I expect!) so the normal reaction is the weight of each mass = mg. (Acting vertically downwards)

At the moment you don't need to know m1 or m2, simply that they are equal masses.
They will cancel out.
 
yes, the turntable is horizontal =)
for ω^2, i got...
0.75(m1+m2)g
--------------
m1*R1 + m2*R2
maybe i forgot my algebra, but is it possible to cancel m1 and m2 there?
 
m1 = m2!

just cancel out top and bottom :)
 
  • #10
omg !
bleh ! >_<
i'm an idiot.
it's 2 am, after all... ^^

thanks, guys.

edit:

i got 1.5mg = m(R1+R2)ω^2, got 1.4491 for ω, but it's still wrong!
did i do something wrong?

edit:

bleh. i had to change the radius from cm to m.
 
Last edited:
  • #11
okay so i have the same exact problem, and i was able to find the angular frequency, however i am having trouble finding the tension! i tried plugging back in the angular frequency to the equations that we previously determined and nothing is working! can you tell me what you did in order to find the T?
 
  • #12
If you got the angular frequency OK, then you should be able to get the tension by using one of the earlier equations. You should check your working. If you can't find the error, then post your working and someone will check it out.
 
  • #13
so from the previos problems we found that f1 + T=m1*w^2*R1 and f2-T= m2*w^2*R2. there were two unknowns in the equations, tension and the angular frequency. i know that to solve for the angular frequency we had to solve for w^2 get rid of the tension and simplify. so now to solve for the tension shouldn't i be able to use those previous equations to find T? I'm probably doing it wrong since everything i do comes up with a wrong answer. but what i did was solve for f so that f=m1*w^2*R1-T and f=m2*w^2*R2+T then i set those equal to each other and solved for T so that T= (m1*w^2R1-m2*w^2*R2)/2 and that still doesn't work...can you help me derive the correct equation to solve for the tension?
 
  • #14
I don't see anything wrong with your working, except that R1 and R2 should be the other way arouind. i.e. T = (1/2)w²(m2R2 - m1R1)

1) do you have the value for m1/2 ?
2) do you have the answer ?
 

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