A Moment of Inertia/physical pendulum Problem

  • Thread starter Thread starter dominatesean
  • Start date Start date
  • Tags Tags
    Moment Pendulum
AI Thread Summary
The discussion revolves around calculating the moment of inertia of a 14-cm-long wrench swinging on a hook with a period of 0.90 seconds and stretching a spring by 2.8 cm. Participants emphasize the need for a visual figure to clarify the wrench's center of mass (CM) positioning, as the CM being 14 cm from the hook raises confusion given the wrench's length. The mass of the wrench can be derived from the spring constant and the stretch, leading to the equation I = m*g*L*T^2/2π for moment of inertia. Suggestions include writing the equation of motion and using the radius of gyration to express the mass moment of inertia. The conversation highlights the importance of clear diagrams in solving physics problems effectively.
dominatesean
Messages
8
Reaction score
0

Homework Statement


The 14-cm-long wrench in the figure swings on its hook with a period of 0.90 s. When the wrench hangs from a spring of spring constant 380 , it stretches the spring 2.8 cm
What is the wrench's moment of inertia about the hook?

cm of wrench is also 14 cm according to pic

Homework Equations


I = m*g*L*T^2/2pi

(How I found the mass)
Fsp = -ks
Fg = mg

The Attempt at a Solution


I=1.08571428*9.8*.14*.028^2/2 pi
Not the right answer not sure what to do or where to go
 
Last edited:
Physics news on Phys.org
It would really help if we had the figure; that has a lot of the information in it.

From the second sentence you should be able to compute the mass of the wrench.

It is pretty hard to understand how the CM can be 14 cm from the hook when the wrench is 14 cm long. That is why we really need to see the figure.
 
According to the figure you posted, that 14 cm is the distance from the pivot to the CM, so this is a very, very strange "14 cm long wrench."

Now write the equation of motion, and use the definition of the radius of gyration in order to express the mass moment of inertia. From the period information you should be able to find the radius of gyration. From there, you can finally go back to the MMOI itself.
 
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 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
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...

Similar threads

Replies
10
Views
9K
Replies
3
Views
7K
Replies
42
Views
3K
Replies
11
Views
4K
Replies
4
Views
2K
Replies
14
Views
5K
Replies
11
Views
7K
Back
Top