Center of mass with one object moving

AI Thread Summary
The discussion focuses on calculating the center of mass (COM) for a system with two objects on a stick, where one object is stationary and the other can slide. The correct formula for the COM is highlighted, emphasizing the need to account for the initial position of the moving mass. The integration of the velocity function over time is necessary to determine the position of the moving object accurately. Participants confirm the validity of the approach while acknowledging the importance of understanding the conceptual aspects of the problem. The conversation also includes a light-hearted reference to a character from DotA 2.
DrummingAtom
Messages
657
Reaction score
2

Homework Statement



Two objects are on a stick where one is stuck in place and the other one can slide. Both objects have the same mass and the stick has negligible mass.

Homework Equations



COM = \frac{m_1 x_1 + m_2x_2}{m_1+m_2}
v_1 = t

The Attempt at a Solution



COM = \frac{m_1 \int v_1 + m_2x_2}{m_1+m_2}

Would this be correct assuming that the starting point for measuring x1 and x2 would be the left side? Thanks
 

Attachments

  • COM.jpg
    COM.jpg
    3.9 KB · Views: 477
Physics news on Phys.org
I would be a little careful here. You need to know the initial position of the moving mass:
\frac{m_1\left (\int v_1(t) \,dt +x_1(0) \right ) + m_2x_2}{m_1+m_2}
where x_1(t) is a function of time and x_1(0) is the constant of integration.
 
Last edited:
RoshanBBQ said:
I would be a little careful here. You need to know the initial position of the moving mass:
\frac{m_1\left (\int v_1(t) \,dt +x_1(0) \right ) + m_2x_2}{m_1+m_2}
where x_1(t) is a function of time and x_1(0) is the constant of integration.

Thanks for responding. Ahh yes good point. Adding your part, would that be correct? I just made up the problem which is why I'm a little unsure of the conceptual part.
 
DrummingAtom said:
Thanks for responding. Ahh yes good point. Adding your part, would that be correct? I just made up the problem which is why I'm a little unsure of the conceptual part.

I believe it is all right. The equation gives the center of mass as a function of time.
 
Ok cool. Thanks for your help. By the way.. Roshan from DotA? :wink:
 
DrummingAtom said:
Ok cool. Thanks for your help. By the way.. Roshan from DotA? :wink:

Yes. DotA 2 now.
 
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