What is the maximum coefficient of friction for a composite body on a slope?

AI Thread Summary
The discussion focuses on determining the maximum coefficient of friction (μ) for a composite body on a slope, emphasizing the importance of the rod's position relative to the pipe's axis. The participants analyze torque calculations involving normal and friction forces, noting that the rod must be positioned to the extreme right for maximum μ. There is a correction regarding the conditions under which the system should slip without rotating. The conversation also explores calculating μ as a function of the angle θ and maximizing it with respect to a variable φ. The key takeaway is that proper positioning and understanding of forces are crucial for accurate calculations of μ.
Rikudo
Messages
120
Reaction score
26
Homework Statement
A thin rod of mass m is welded on inner surface of a thin-walled pipe of mass M parallel to the axis of the pipe. The composite body thus formed is placed on a uniform slope as in the figure. Inclination of the slope is adjustable. Find range of coefficient of friction between the pipe and the slope for which the body slides down without rotation.
Relevant Equations
Torque
1657948211713.png


Here is the hint that the book gave me:

"For the maximum value of μ, the rod must be to the extreme right i.e. horizontally rightwards of the axis of the pipe"

I think what it meant is the same as this:
1657948709558.png

Note: in the calculation below, ##r## is the distance from the center of pipe to the CoM of the system and ##r = mR/(m+M)##

So, the torque calculation (with the system's CoM as origin) for maximum μ will be :
##N r = f R##
##Nr=NμR##
##m/(m+M)=μ##

But, the book says that it is ##m/\sqrt{M(M+2m)}##

Where did I do wrong?
 
  • Like
Likes Delta2
Physics news on Phys.org
I would start with this picture
torque wrt CoM: torque from normal force = Nrsinθ = torque from friction force = μN*L
Find L as an expression of r and θ
1657953100131.png
 
Last edited by a moderator:
  • Informative
  • Like
Likes Delta2 and Rikudo
drmalawi said:
The weight of the rod will also produce a torque
This will be neutralized by torque produced by the pipe's weight. Hence, in the torque equation, we only use friction and normal force.
 
  • Like
Likes malawi_glenn
Rikudo said:
This will be neutralized by torque produced by the pipe's weight. Hence, in the torque equation, we only use friction and normal force.
Yeah I forgot that it should only slip, not rotate at all.
Look at the pic above, try to calculate µ as a function of θ
 
@Rikudo now picture is correct...
 
AH! Now I got it. Thanks!
 
  • Like
Likes malawi_glenn
Rikudo said:
"For the maximum value of μ, the rod must be to the extreme right i.e. horizontally rightwards of the axis of the pipe"

I think what it meant is the same as this:
View attachment 304258
For information, the hint says:
"For the maximum value of μ, the rod must be to the extreme right i.e. horizontally rightwards of the axis of the pipe"​

But in your diagram, the rod (red dot) is not at the extreme right, and therefore not at the same horizontal level (i.e. not at the same height) as the axis of the pipe.
 
  • Like
Likes Lnewqban and malawi_glenn
Steve4Physics said:
But in your diagram, the rod (red dot) is not at the extreme right, and therefore not at the same horizontal level (i.e. not at the same height) as the axis of the pipe.
That's OK. One can find ##\mu## at some angle ##\varphi## below the horizontal line and then maximize ##\mu## with respect to ##\varphi##. Just a little bit more work, but that's what I did to explore what ##\mu## looks like as a function of ##\varphi.##
 
Back
Top