Thin rods (H) rotating about an axis

  • Thread starter Thread starter deadpenguins
  • Start date Start date
  • Tags Tags
    Axis Rotating
AI Thread Summary
The discussion centers on calculating the angular speed of a rigid body shaped like an "H" made of three identical thin rods, rotating about a horizontal axis along one leg. Participants emphasize the need to calculate the moment of inertia for each rod using the perpendicular and parallel axis theorems, noting that one leg can be disregarded as it acts as the pivot. The moment of inertia for a rod pivoted at one end is mentioned as a key factor in the calculations. Additionally, the third rod is considered to behave like a point mass at the end of a rod due to its falling motion. Ultimately, the conversion from potential energy to rotational kinetic energy is highlighted as essential for determining the angular speed.
deadpenguins
Messages
3
Reaction score
0

Homework Statement


A rigid body is made of three identical thin rods, each with length L=0.600m, fastened together in the form of a letter H. The body is free to rotate about a horizontal axis that runs along the length of one of the legs of the H. The body is allowed to fall from rest from a position in which the plane of the H is horizontal. What is the angular speed of the body when the plane of the H is vertical?


Homework Equations


Not really sure



The Attempt at a Solution


I am not really sure where to start since all that is given is L. I think that each rod must have its moment of inertia calculated separately about the axis using the perpendicular and parallel axis theorems, but we have not reviewed these subjects extensively and I am not sure of how I would go about doing this.
 
Physics news on Phys.org
Yes you are sensing the right solution.

As I read it the plane of the H rotates about one leg. Since one leg is already the pivot it can be disregarded. (r=0 after all.)

So that means then you have two more moments to account for.

Do you know the moment of inertia for a rod pivoted about 1 end? It's easily looked up or determined.

Then you just need to figure the moment of the 3rd rod. But since it is just falling and not rotating won't it simply act like a point mass at the end of a rod?

After you have the moment of inertia you know how to figure the potential energy to rotational kinetic energy don't you?
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top