Dumbbell In a Cylindrical Cavity

Consider the following problem:

At any general time tt, the configuration of the system is as shown in the figure below:

  • Let the reaction force on the mass AA be N1N_1. It is directed along AOAO and towards OO.

  • Let the reaction force on the mass BB be N2N_2. It is directed along BOBO and towards OO.

  • Let the tension developed on the massless rod be TT. The tension acts on mass AA along ABAB directed towards BB and acts on mass BB along BABA directed towards AA.

From the diagram, it is apparent that:

ϕ+θ=π4\phi + \theta = \frac{\pi}{4}

Let the coordinates of point mass AA be:

xA=rcosθx_A = r \cos{\theta} yA=rsinθy_A = -r\sin{\theta}

Let the coordinates of point mass BB be:

xB=rsinθx_B = -r\sin{\theta} yB=rcosθy_B = -r\cos{\theta}

The following equations can be obtained by drawing appropriate free body diagrams for each mass. I have left out the free body diagrams from this analysis.

mx¨A=N1cosθTcosϕm\ddot{x}_A = -N_1 \cos{\theta} - T\cos{\phi} my¨A=N1sinθmgTsinϕm\ddot{y}_A = N_1 \sin{\theta}-mg - T\sin{\phi} mx¨B=N2sinθ+Tcosϕm\ddot{x}_B = N_2 \sin{\theta} + T\cos{\phi} my¨B=N2cosθmg+Tsinϕm\ddot{y}_B = N_2 \cos{\theta}-mg + T\sin{\phi}

The acceleration expressions can be found by differentiating xAx_A, yAy_A, xBx_B and yBy_B twice with respect to time. Note that the system is released from rest, so:

θ(0)=0\theta(0) = 0 θ˙(0)=0\dot{\theta}(0) = 0

And let:

θ¨(0)=α\ddot{\theta}(0) = \alpha

Also, when the rod is just released, one can compute the following results:

ϕ=π4\phi = \frac{\pi}{4} x¨A=0\ddot{x}_A = 0 y¨A=rα\ddot{y}_A = - r \alpha x¨B=rα\ddot{x}_B = -r\alpha y¨B=0\ddot{y}_B = 0

Plugging these all these expressions into the equations above, and simplifying gives:

0=N1T20 = -N_1 -\frac{T}{\sqrt{2}} rα=mgT2 - r \alpha = -mg -\frac{T}{\sqrt{2}} rα=T2 - r \alpha = \frac{T}{\sqrt{2}} 0=T2+N2mg0 = \frac{T}{\sqrt{2} }+ N_2 - mg

Solving for N2N_2 gives:

N2=3mg2\boxed{N_2 = \frac{3mg}{2}}

The above is the reaction force computed when the system is just released from rest.

Note by Karan Chatrath
2 weeks, 2 days ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

@Talulah Riley Here you go.

Karan Chatrath - 2 weeks, 2 days ago

Log in to reply

Nice solution. I've responded to your report btw; I don't know if you got a notification for the response @Karan Chatrath

Krishna Karthik - 2 weeks, 2 days ago

Log in to reply

@Karan Chatrath Thank you so much for the note.

Talulah Riley - 2 weeks, 2 days ago

Log in to reply

@Karan Chatrath About the report, I solved the original Irodov problem (which doesn't have friction) correctly using energy conservation; but solving for tension numerically as per the equation above gives a tension of 800 Newtons at maximum.

Krishna Karthik - 2 weeks, 2 days ago

Log in to reply

@Karan Chatrath I've reposted the problem now.

Krishna Karthik - 2 weeks, 2 days ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...