Reaction Force of Rod at 90 and 180 Degrees

In summary, you are trying to find the hinge reaction for a uniform rod that is rotating about a horizontal axis. You use conservation of energy to calculate the velocity and acceleration of the center of mass.
  • #1
Kishlay
91
3

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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  • #2
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.
 
  • #3
okay will try it once again
 
  • #4
by your method i have got v=3g/2 which does not matches the ans..
 
  • #5
please help me...!
 
  • #6
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:
  • #7
v represents the velocity of centre of mass
 
  • #8
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.
 
  • #9
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...
 

FAQ: Reaction Force of Rod at 90 and 180 Degrees

1. What is the reaction force of a rod at 90 degrees?

The reaction force of a rod at 90 degrees is the force exerted on the rod perpendicular to its length when a load is applied at a 90 degree angle to the rod. This force is equal in magnitude and opposite in direction to the load applied.

2. How does the reaction force of a rod at 90 degrees affect its stability?

The reaction force of a rod at 90 degrees plays a crucial role in the stability of the rod. It helps to keep the rod in place and prevents it from toppling over when a load is applied at a perpendicular angle.

3. What happens to the reaction force of a rod at 90 degrees if the load is increased?

If the load applied at a 90 degree angle to the rod is increased, the reaction force will also increase in order to maintain equilibrium. This is known as the law of action and reaction, where every action has an equal and opposite reaction.

4. What is the difference between the reaction force of a rod at 90 degrees and 180 degrees?

The main difference between the reaction force of a rod at 90 degrees and 180 degrees is the direction of the force. At 90 degrees, the force is perpendicular to the rod, while at 180 degrees, the force is parallel to the rod.

5. How is the reaction force of a rod at 90 degrees calculated?

The reaction force of a rod at 90 degrees can be calculated using the formula F = W x tan(θ), where F is the reaction force, W is the applied load, and θ is the angle between the rod and the load. This formula applies to a rod that is fixed at one end and free to rotate at the other end.

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