Maximum angle reached by a cube placed inside a spinning cylinder

In summary: If you have two numbers ##x## and ##y## such that ##x^2 + y^2 = 1##, then the point ##(x,y)## lies on the unit circle in the xy-plane. Then the polar angle ##\phi## of the point satisfies ##\cos \phi = x## and ##\sin \phi = y##.
  • #1
Samuel N
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Misplaced Homework Thread moved to the Schoolwork forums
I am trying to solve a problem where we have to find the maximum angle before a cube starts sliding when said cube is placed inside a spinning hollow cylinder (the cylinder is placed horizontally). The radius of the cylinder is 0.4 m, the coefficient of static friction between the cube and the cylinder is 0.75 and the period of the spinning cylinder is 2 revolutions/second. I tried to solve the problem but I end up with a transcendental equation that can't seem to be solved:
μω2R = μgsin(θ) - gcos(θ).
Any help would be appreciated.
Thanks
 
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  • #2
Welcome, Samuel!

Is that cube mass-less and dimension-less?
 
  • #3
Hi, thank you.
The dimensions of the cube are not mentioned so I am guessing that they are negligible. As for the mass, it is not mentioned either in my homework question but the m terms cancel out when I try to solve the problem which means theta probably does not depend on the mass of the cube.
 
  • #4
Could you show us your work so far, as well as any diagram?
 
  • #5
I think we must assume the cylinder rotates slowly, and that the cube does not roll, but slides down the internal surface.

This seems to be yet another problem of the; slope = arctangent(friction coefficient).
 
  • #6
I have written the equation I have so far in my original post, but sure I can take a pdf scan of my work and send it here give me a minute or two.
 
  • #7
Baluncore said:
I think we must assume the cylinder rotates slowly, and that the cube does not roll, but slides down the internal surface.

This seems to be yet another problem of the; slope = arctangent(friction coefficient).
Yes, this is what I thought too but the problem I have is that I have a sum of a sine and a cosine and I can't isolate theta
 
  • #8
here is what I have so far(I used a system of axis tilted by the angle theta):
 

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  • #9
Samuel N said:
Yes, this is what I thought too but the problem I have is that I have a sum of a sine and a cosine and I can't isolate theta
You may need Newton's method to solve the equality.

That assumes the cylinder is rotating slowly enough for the cube to slide.
Tangential force = m * g * Sin( a )
Radial force = ( m * r * w*w ) + ( m * g * Cos( a ) )
First slip will be when tangential force = u * radial force.
I think r*w*w is too great, so the cube will not slide.
The cube will be held against the wall by the rotation.
 
  • #10
Baluncore said:
You may need Newton's method to solve the equality.

That assumes the cylinder is rotating slowly enough for the cube to slide.
Tangential force = m * g * Sin( a )
Radial force = ( m * r * w*w ) + ( m * g * Cos( a ) )
First slip will be when tangential force = u * radial force.
I think r*w*w is too great, so the cube will not slide.
The cube will be held against the wall by the rota
Thanks I managed to solve it using Newton's method. I forgot I could just send the constant to the right side of the equation and solve it by finding the zeros. Turns out I said 2 revolutions per second but I misread it and it was actually 0.5 revolutions per second. I get a theta of about 0.887 rads
 
  • #11
Samuel N said:
I managed to solve it using Newton's method.
1668978606782.png


You can solve this with a trick. Rearrange the equation as $$\frac{\mu_s \omega^2 R}{g} = \sin \theta - \mu_s \cos \theta$$ Plug in ##\mu_s = .75 = 3/4##. Manipulate to $$\frac{3 \omega^2 R}{5g} = \frac{4}{5}\sin \theta - \frac{3}{5} \cos \theta$$ Define the angle ##\phi## by ##\sin \phi = \frac{3}{5}## and ##\cos \phi =\frac{4}{5}##. Use a trig identity to write the right-hand side in terms of a single trig function.
 
  • #12
TSny said:
View attachment 317481

You can solve this with a trick. Rearrange the equation as $$\frac{\mu_s \omega^2 R}{g} = \sin \theta - \mu_s \cos \theta$$ Plug in ##\mu_s = .75 = 3/4##. Manipulate to $$\frac{3 \omega^2 R}{5g} = \frac{4}{5}\sin \theta - \frac{3}{5} \cos \theta$$ Define the angle ##\phi## by ##\sin \phi = \frac{3}{5}## and ##\cos \phi =\frac{4}{5}##. Use a trig identity to write the right-hand side in terms of a single trig function.
I don't understand the part where you define the angle as phi. If you don't mind, could you explain further please?
 
  • #13
Samuel N said:
I don't understand the part where you define the angle as phi. If you don't mind, could you explain further please?

If you have two numbers ##x## and ##y## such that ##x^2 + y^2 = 1##, then the point ##(x,y)## lies on the unit circle in the xy-plane.

Then the polar angle ##\phi## of the point satisfies ##\cos \phi = x## and ##\sin \phi = y##.

So, if we have two numbers ##x## and ##y## satisfying ##x^2 + y^2 = 1##, we can let ##\phi## be the angle that satisfies ##x = \cos \phi## and ##y = \sin \phi##.

In our case, we have the numbers ##x = 4/5## and ##y = 3/5## which satisfy ##x^2 + y^2 = 1##. So, we can let ##4/5 = \cos \phi## and ##3/5 = \sin \phi##.
 
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  • #14
TSny said:
In our case, we have the numbers ##x = 4/5## and ##y = 3/5## which satisfy ##x^2 + y^2 = 1##. So, we can let ##4/5 = \cos \phi## and ##3/5 = \sin \phi##.
Okay I think I get it now. I can substitute the fractions and then use the product trig identities to get a single trig function.

Edit: It worked thank you so much. I get the same value as I did with Newton's method but negative. I guess it makes sense because it can happen on both sides of the cylinder. How would I find the periodicity of the angle (pi*k + theta)?
 
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  • #15
Samuel N said:
Okay I think I get it now. I can substitute the fractions and then use the product trig identities to get a single trig function.

Edit: It worked thank you so much. I get the same value as I did with Newton's method but negative. I guess it makes sense because it can happen on both sides of the cylinder.
Ok, good.
 
  • #16
TSny said:
You can solve this with a trick. Rearrange the equation as $$\frac{\mu_s \omega^2 R}{g} = \sin \theta - \mu_s \cos \theta$$ Plug in ##\mu_s = .75 = 3/4##. Manipulate to $$\frac{3 \omega^2 R}{5g} = \frac{4}{5}\sin \theta - \frac{3}{5} \cos \theta$$ Define the angle ##\phi## by ##\sin \phi = \frac{3}{5}## and ##\cos \phi =\frac{4}{5}##. Use a trig identity to write the right-hand side in terms of a single trig function.
Why the 5 in the denominator?
Sorry, I don't get it.
 
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  • #17
Lnewqban said:
Why the 5 in the denominator?
Sorry, I don't get it.
The equation was divided through by 5 so that the coefficients of ##\sin \theta## and ##\cos \theta## would have magnitudes of ##4/5## and ##3/5##. The squares of these add to 1 which goes along with the identity ##\cos^2\phi + \sin^2 \phi = 1##. Then we can make the substitution ##\cos \phi = 4/5## and ##\sin \phi = 3/5##.

In general, if we have the expression ##a \sin \theta \pm b \ cos \theta##, we could write it as

##\sqrt{a^2 + b^2}\left( \frac{a}{\sqrt{a^2+b^2}} \sin \theta \pm \frac{b}{\sqrt{a^2+b^2}} \cos \theta \right)##

Then let ##\cos \phi =\frac{a}{\sqrt{a^2+b^2}}## and ##\sin \phi = \frac{b}{\sqrt{a^2+b^2}}##.

Thus, ##a \sin \theta \pm b \ cos \theta = \sqrt{a^2 + b^2}\left( \cos\phi\sin \theta \pm \sin \phi \cos \theta \right) = \sqrt{a^2 + b^2} \sin (\theta \pm \phi)##
 
  • #18
I have a question about this problem: How do we know that the cube slides and does not stay stuck on the inner wall for an entire revolution?

Suppose we consider the FBD of the cube when it is at the 3 o' clock position. The relevant equations are
$$\begin{align} & N=m\omega^2 R \nonumber \\ & f_{\text{s}}=mg \nonumber \end{align}$$The maximum force of static friction is $$f_{\text{s}}^{\text{max}}=\mu_s N=\frac{3}{4}m\omega^2 R.$$ We need to compare the number multiplying the mass on the right hand side against ##g##. $$\frac{3}{4}\omega^2 R=(0.75)\times (4\pi~\text{rad}/\text{s})^2\times (0.4~\text{m})=47.3~\text{m}/\text{s}^2\approx 4.7g.$$There is more than enough friction at the 3 o' clock position to keep the cube from sliding. Also, at the 12 o' clock position, $$N+mg=m\omega^2 R \implies N=m(\omega^2 R-g) >0$$which means that, unless I missed something, the cube makes it all the way around without sliding or falling off.
 
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  • #19
kuruman said:
unless I missed something, the cube makes it all the way around without sliding or falling off.

There was a mistake in the statement of the problem where it said
Samuel N said:
the period of the spinning cylinder is 2 revolutions/second.
It should have been worded as "the period of the spinning cylinder is 2 seconds/revolution". So, ##\omega = \pi## rad/s.
 
  • #20
TSny said:
There was a mistake in the statement of the problem where it said

It should have been worded as "the period of the spinning cylinder is 2 seconds/revolution". So, ##\omega = \pi## rad/s.
That would do it, but how are we supposed to know? The pdf posted by the OP also shows ω = 2π rad/s.
 
  • #21
kuruman said:
That would do it, but how are we supposed to know? The pdf posted by the OP also shows ω = 2π rad/s.
Yes, it was a bit confusing. But he made a correction in post #10.
 
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  • #22
TSny said:
The equation was divided through by 5 so that the coefficients of ##\sin \theta## and ##\cos \theta## would have magnitudes of ##4/5## and ##3/5##.
Now i see it.
Very clever.
Thank you!
 
  • #23
Samuel N said:
Turns out I said 2 revolutions per second but I misread it and it was actually 0.5 revolutions per second.
Ah, that does explain why I found no root.
 

Related to Maximum angle reached by a cube placed inside a spinning cylinder

What is the maximum angle reached by a cube placed inside a spinning cylinder?

The maximum angle reached by a cube placed inside a spinning cylinder depends on the size and speed of the cylinder, as well as the dimensions of the cube. It can range from 0 degrees to a maximum of 90 degrees.

How does the size of the cylinder affect the maximum angle reached by the cube?

The size of the cylinder plays a significant role in determining the maximum angle reached by the cube. A larger cylinder will allow for a larger range of motion and therefore a higher maximum angle, while a smaller cylinder will limit the maximum angle that can be reached.

What happens if the cube is not perfectly centered inside the cylinder?

If the cube is not perfectly centered inside the cylinder, it may affect the maximum angle that can be reached. The cube may experience more friction or resistance on one side, causing it to rotate at a different rate and potentially limiting the maximum angle it can reach.

Can the maximum angle be calculated using a specific formula?

Yes, there are mathematical formulas that can be used to calculate the maximum angle reached by a cube placed inside a spinning cylinder. These formulas take into account the dimensions and speed of the cylinder, as well as the dimensions of the cube.

Is the maximum angle reached by the cube affected by the material of the cylinder?

Yes, the material of the cylinder can affect the maximum angle reached by the cube. A smoother material may allow for less friction and a higher maximum angle, while a rougher material may cause more resistance and limit the maximum angle that can be reached.

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