How to calculate weight of object swinging a radius?

  • Thread starter Thread starter dishguy55
  • Start date Start date
  • Tags Tags
    Radius Weight
Click For Summary
To calculate the weight developed by a 20 lbs object swinging from a 30-inch arm, one must consider the forces acting on the object during circular motion. The discussion emphasizes the importance of angular velocity, with the formula w = 2π/t being relevant, where t is the period of rotation. The user also notes that RPM should be converted to seconds for accurate calculations. Understanding Newton's second law is crucial for relating the forces to the weight measured in pounds. Overall, the conversation focuses on the need for proper unit conversion and the application of physics principles to solve the problem.
dishguy55
Messages
4
Reaction score
0

Homework Statement


20 lbs. on the end of a 30 inch arm. Dropping from 2oclock position to 5oclock position.

How much wieght will be developed if this was dropped on a scale measuring lbs.


Homework Equations


I am unsure how to calculate this.


The Attempt at a Solution


If there is a formula I can try to calculate myself.
I just don't know where to start.
 
Physics news on Phys.org
Hi there...
Let me see if I get this straight:, by
...if this was dropped on a scale measuring lbs.
did you mean that at that juncture, on this sort of pendulum, the thread/rope(whatever) is suddenly cut, and the object is dropped on a weight measure?
If that's the case, then you must consider what sort of forces act upon an object moving in such a structure; Essentially(and this is quite a hint), you need to think about circular motion...
Give it a try, it's bound to work...
Good luck,
Daniel
 
Ok. i researched that and found this formula.

w=2*3.14/t.

w in my angular velocity. that is what i need to know.

t is the period for 1 rotation.

Is the rotation in rpm.

I know my rpm is 19.

If so then w would be .33

If my angular velocity is .33

How does that convert to lbs?
 
Well, let me simply rearrange what you wrote, if you don't mind...
You're correct in positing that: \omega = \frac{2\pi}{t}, Where t is the time of a single rotation(however, in SI units, this is ALWAYS measured in seconds); Therefore, RPMs won't do.
My advice is(based on my personal experience) is to try and convert any problem to metric units(SI based), and if the solution explicitly requires Imperial/non-standard representation, only then(after having the proper resolution), should you turn it back to whatever's necessary(but that's my opinion, and you're of course free to do what you're comfortable with).
As for those pesky Lbs, think about what a force suggests, its equation, and what do Lbs represent in this case(Newton's 2nd law would be helpful).
You're on the right track though,
Daniel
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
4
Views
2K
  • · Replies 44 ·
2
Replies
44
Views
7K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
11
Views
2K
  • · Replies 28 ·
Replies
28
Views
3K