How Do I Determine the Center of Mass in This Situation?

AI Thread Summary
To determine the center of mass in the given situation, the calculations indicate that the center of mass is 3.5 length units from each end of the rod, which has a uniform mass distribution of 0.4 kg. The initial result of 2.35 kg for mass P_B does not match the answer hint, which suggests rounding to 2.40 kg. The accuracy of the calculations is deemed sufficient to warrant only two significant figures. Concerns about the diagram's accuracy and labeling have been raised, suggesting potential issues with the setup. The discussion emphasizes the importance of verifying calculations and understanding rounding conventions in physics problems.
Memo
Messages
35
Reaction score
3
Homework Statement
See the photo below. A uniform rod AC with a mass of 0.4kg, hang m1=5kg. Find m2 for the system to be in equilibrium.
Answer hint: m2=2.4(kg)
Relevant Equations
M=F*d*sina
387331626_3165611777081502_1178316719802908685_n.jpg


OA*P_A = OH*P + OB*P_B
⇔ OA*P_A = ¾*OA*P + 2OA*P_B
⇔ 5*10 = ¾*0.4*10 + 2*P_B
→P_B=23.5 →m2=2.35 (kg)

My result doesn't match the answer hint, and when I change the rod's centre of mass to OH=1, it fits the answer hint. How do I determine the centre of mass in this situation?
 
Physics news on Phys.org
Your answer matches the hint to 2 significant figures which the accuracy of the calculation. Your method is correct.
 
kuruman said:
Your answer matches the hint to 2 significant figures which the accuracy of the calculation. Your method is correct.
Thanks for reaching out. Normally, that would be considered incorrect. I meant, the figure gives out nicely, why would my teacher round it.
 
Memo said:
Thanks for reaching out. Normally, that would be considered incorrect. I meant, the figure gives out nicely, why would my teacher round it.
Because you have to eyeball the distances and the accuracy of the calculations does warrant anything more than two sig figs. Ask your teacher and see what answer you get.
 
The details of the diagram make me suspicious. The labels O, H, P and the down arrow at P seem to have been added later. The vertical divisions on the rod are unevenly spaced.
I suspect it has been inaccurately recycled.
 
Memo said:
How do I determine the centre of mass in this situation?
The center of mass is located 3.5 length units from each end of the rod.
The mass is evenly distributed along the rod because:
"A uniform rod AC with a mass of 0.4kg".

The answer hint is simply rounding 2.35 kg to 2.40 kg.
The error is only 2.1 %.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top