Rectangular Disk with Circular Hole Period Question

AI Thread Summary
A circular disk with a rectangular hole, having a radius of 0.620 m and mass of 0.470 kg, is suspended at its perimeter and oscillates as a pendulum. The moment of inertia is given as I_p = 1.60E-1 kgm², with the center of mass located 0.120 m from the circle's center. To find the period of oscillation, the correct formula involves the distance from the pivot to the center of mass, not just the distance from the center of the circle. The discussion emphasizes the importance of returning to first principles and understanding the derivation of formulas for deeper comprehension. Ultimately, clarifying the correct distance for calculations is crucial for solving the problem accurately.
JohnnyCollins
Messages
4
Reaction score
0

Homework Statement


A circular disk with a rectangular hole has a radius of 0.620 m and mass of 0.470 kg. It is suspended by a point on its perimeter as shown in the figure. The moment of inertia about this point is I_p = 1.60E-1 kgm2. Its center of mass is located at a distance of s=0.120 m from the center of the circle as shown. If the disk is allowed to oscillate side to side as a pendulum, what is the period of oscillations


Homework Equations



ω=2∏f
f=1/T
I_g=1/2mr^2

The Attempt at a Solution


I'm lost on this question, any help would be greatly appreciated!
 
Physics news on Phys.org
Since you are stuck - go back to first principles ... start by drawing a free-body diagram for an arbitrary angular displacement and work out the torques. Fortunately you have been given the moment of inertia and the center of mass. Make an approximation for small angles and solve for T.
 
Last edited:
Simon Bridge said:
Since you are stuck - go back to first principles ... start by drawing a free-body diagram for an arbitrary angular displacement and work out the torques. Fortunately you have been given the moment of inertia and the center of mass. Make an approximation for small angles and solve for T.

So i took your advice and used the formula T=2∏√I/(mgd) and used 0.120 m for my distance, but I still ended up with the wrong answer. Am I on the right track ?
 
d, in your formula, is the distance from the pivot to the center of mass.
0.120m is the distance from the center of the circle to the center of mass.
spot the difference.
 
Simon Bridge said:
d, in your formula, is the distance from the pivot to the center of mass.
0.120m is the distance from the center of the circle to the center of mass.
spot the difference.

Got it, thanks a lot for your help!
 
No worries.

When you get stuck - go back to first principles and do a derivation.
It's amazing how naive you can be about this and still get results - just "describe the system in math and then fiddle the math" is very powerful.

In the process you may discover that you could have used a short-cut ... but having gone back over the basics deepens your understanding of the process.
Happy hacking.
 
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