Minimum force required to rotate a lamina

Click For Summary
SUMMARY

The discussion focuses on determining the minimum force required to rotate a lamina about points A, B, and C. Two approaches were explored: the first involved calculating the moment of inertia (IC) and static friction, while the second considered the torque produced by forces FA and FB about point C. The second approach yielded an incorrect answer, highlighting the complexity of friction distribution and its impact on rotational dynamics. The participants emphasized the need for a deeper understanding of internal constraint forces and uniform friction assumptions in rigid body dynamics.

PREREQUISITES
  • Understanding of static friction and its role in rotational motion.
  • Familiarity with moment of inertia calculations for laminae.
  • Knowledge of torque and its relationship to forces in rotational systems.
  • Basic principles of rigid body dynamics, including internal constraint forces.
NEXT STEPS
  • Study the derivation of moment of inertia for various shapes, particularly triangular laminae.
  • Learn about the principles of torque and static equilibrium in rigid body dynamics.
  • Research the effects of friction distribution on rotational motion in laminae.
  • Explore the concept of internal constraint forces and their implications in rigid body mechanics.
USEFUL FOR

Students and professionals in physics and engineering, particularly those focusing on mechanics, rotational dynamics, and friction in rigid bodies.

  • #31
PhysicsBoi1908 said:
how do we conclude that magnitude of friction is constant
The weight per unit area is constant. When the lamina moves, all parts go into kinetic friction, so when it is about to move all parts must be at max static friction.
 
  • Like
Likes   Reactions: PhysicsBoi1908
Physics news on Phys.org
  • #32
Ah! Thank you very much.
 
  • #33
haruspex said:
I see a way.
Drop a perpendicular from C to meet AB at D. Consider the torques required to rotate the two smaller triangles about C.

It is very clever! If you don't explicitly cut the lamina there will be an internal force between the two parts. The result is that the torque you have to apply at A/B is slightly more than calculated and the torque you have to apply at B/A is slightly less than expected. But these two discrepancies will cancel exactly, and their sum will just be the same as if the lamina were not actually connected, except at the hinge (i.e. no internal force between the two parts).
 

Similar threads

To find the centre of gravity of a lamina
  • · Replies 2 ·
Replies
2
Views
2K
To find the resultant of forces on a lamina
  • · Replies 5 ·
Replies
5
Views
3K
How Do You Calculate the Moment of Inertia for Combined Square Laminas?
  • · Replies 3 ·
Replies
3
Views
3K
To find the tipping point of a lamina
  • · Replies 2 ·
Replies
2
Views
1K
Tetrahedron with 3 points fixed, and force applied to 4th
  • · Replies 11 ·
Replies
11
Views
2K
Calculating acceleration of gravity and inertia of irregular lamina
  • · Replies 10 ·
Replies
10
Views
5K
Moment of Inertia perpendicular to lamina
  • · Replies 1 ·
Replies
1
Views
2K
Undergrad Resolve moment of inertia at an angle
  • · Replies 2 ·
Replies
2
Views
2K
Euler's Equations, a freely rotating lamina
  • · Replies 2 ·
Replies
2
Views
7K
Moment of inertia of a square lamina through a diagonal
  • · Replies 9 ·
Replies
9
Views
23K