Reaction Force of Rod at 90 and 180 Degrees

AI Thread Summary
The discussion revolves around calculating the hinge reaction force of a uniform rod rotating about a horizontal axis when released from rest. Participants analyze the energy conservation equation and the relationship between linear and angular velocities, noting discrepancies in calculated values. The conversation highlights the importance of correctly identifying forces acting on the rod, particularly in the y-direction, and emphasizes the need for clarity in defining positive directions for forces and accelerations. There is a focus on using torque to relate angular acceleration to the forces acting on the rod. The thread concludes with a suggestion to refine calculations and equations to accurately determine the hinge reaction.
Kishlay
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Homework Statement




a uniform rod of mass m and length L is free to rotate in the vertical plane about a horizontal axis passing through its end. the rod is released from rest in the position shown by slightly displacing it clockwise. Find the hinge reaction at the axis of rotation at the instant the rod turns through
(a) 90* (b) 180*

Homework Equations


i am not getting the answer by the following attempt, any help will be highly appreciated...!


The Attempt at a Solution


i have conserved energy as shown in the following equation:-
mgL/2 = mv2/2 + Iω2/2

and also ω=2v/L
so i have further got v2=3gL/7
 

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If I denotes the moment of inertia about the fixed axis (that is, about the end of the rod) then the total kinetic energy is Iω2/2 instead of Mvc2/2 + Iω2/2.
 
okay will try it once again
 
by your method i have got v=3g/2 which does not matches the ans..
 
please help me...!
 
What velocity have you calculated? Does v represent the velocity of the center of mass or the velocity of the far end of the rod or the velocity of some other point?

Note that v = 3g/2 is inconsistent in terms of units. You have velocity on the left and acceleration on the right.

Can you show more detail of how you got this result?
 
Last edited:
v represents the velocity of centre of mass
 
Kishlay said:
v represents the velocity of centre of mass

OK. Since v = 3g/2 cannot be correct (because the units don't check), you must have made a mistake. But there is no way for us to know where you made the mistake unless you show your work.
 
mgl/2=Iω2/2
mgl/2= 1/2x1/3ml2x(2v/l)2
m,1/2 and also l2 gets canceled
then
gl=1/3x4v2
v=(3gl/4)1/2
 
  • #10
That's correct.
 
  • #11
but i have not got the hinge reaction
 
  • #12
i have got v
 
  • #13
Yes, that's right. What concepts are you going to use to get the hinge reaction?
 
  • #14
H= hinge reaction
Hx=2mv2/l
and for y direction
Hy=mg
 
  • #15
Kishlay said:
H= hinge reaction
Hx=2mv2/l

OK, but since I don't know which direction you chose for the positive x direction, I don't know whether Hx is toward the left or toward the right.

and for y direction.
Hy=mg
This is not correct. ƩFy = May.
Does the center of mass have any acceleration in the y direction?
 
  • #16
TSny said:
OK, but since I don't know which direction you chose for the positive x direction, I don't know whether Hx is toward the left or toward the right.


Hx is towards right
 
  • #17
TSny said:
This is not correct. ƩFy = May.
Does the center of mass have any acceleration in the y direction?

whats the problem with this??
 
  • #18
Kishlay said:
Hx is towards right

In what direction is the x component of the acceleration of the center of mass when the rod is horizontal?
 
  • #19
TSny said:
In what direction is the x component of the acceleration of the center of mass when the rod is horizontal?

right
 
  • #20
Kishlay said:
whats the problem with this??

In the equation ƩFy = May, what are the forces that appear on the left side and what are the signs for these forces?
 
  • #21
TSny said:
In the equation ƩFy = May, what are the forces that appear on the left side and what are the signs for these forces?

gravitational
May=Mg-Hy
 
  • #22
The force of gravity is not the only force acting on the rod that has a y component.
 
  • #23
Kishlay said:

May=Mg-Hy

In what direction is the positive y direction?
 
  • #24
upwards...
 
  • #25
So, Mg acts upward??
 
  • #26
he he he... sorry downwards
 
  • #27
So, how would you write your equation?
 
  • #28
TSny said:
In what direction is the x component of the acceleration of the center of mass when the rod is horizontal?

Kishlay said:
right

The center of mass is moving in a circle. When the rod is horizontal, what is the direction of the centripetal acceleration of the center of mass?
 
  • #29
May=Mg-Hy
 
  • #30
I need to quit for now. It's 1 AM here. You essentially have an equation that you can use to determine Fy once you determine ay. Try to relate ay to the angular acceleration ##\alpha## of the rod and think about how to get ##\alpha##. Hint: torque.
 
  • #31
Kishlay said:
May=Mg-Hy

That will work if you're taking your positive y-direction downward.
 
  • #32
ok fine will work on it...
 

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