Why Is It Harder to Rotate A Rod with a Mass in the Center?

  • Thread starter Thread starter sagebum
  • Start date Start date
  • Tags Tags
    Inertia
AI Thread Summary
The discussion centers on the difficulty of rotating a rod with a mass in the center compared to a rod with two masses at the ends, given that the total mass is equal. Participants debate the implications of moment of inertia and torque, with some suggesting that the rod with the mass in the center should be easier to rotate due to its lower moment of inertia. However, confusion arises regarding the work required to achieve the same angular velocity for both setups. Clarifications indicate that the mass at the center does not contribute to rotational velocity, leading to the conclusion that it may indeed be harder to rotate the rod with the mass in the center. The conversation emphasizes the need for a clearer understanding of rotational dynamics and the role of moment of inertia.
sagebum
Messages
5
Reaction score
0

Homework Statement


Which is harder to rotate?
A rod with a mass (m1) in the center, or a rod with two masses (m2)at the ends.
Assume mass m1 = 2*m2, so the total masses of rod+masses are equal.
I am rotating the masses at the center.

My professor said that the one with the mass in the center is harder to rotate, but I don't understand how he explained it. I actually thought the one with the mass in the center was easier to rotate.

Can someone explain why the one with the mass in the center is harder to rotate?
Is it because if we were to rotate each to the same angular velocity, it would take more work to get the first rod to that angular velocity?

Relevant equations.
I think I'd use Kr = I \omega^{2}
and if they go up to the same angular velocities, it requires more work?
It doesn't seem right, because if the I of the first rod is lower, so the Kr would be lower, so that means the work required to get it to that speed was lower?
I don't see why the one with the mass in the center is easier to rotate.
 
Physics news on Phys.org
assuming that you are rotating about an axis that goes thru the center then it would harder to rotate the m2 case because it has larger moment of inertia. unless i have misunderstood the setup.
torque = moment of inertia x angular acceleration
you may have mis-heard your prof.. or there are something else in the system?
 
sagebum said:

Homework Statement


Can someone explain why the one with the mass in the center is harder to rotate?
Is it because if we were to rotate each to the same angular velocity, it would take more work to get the first rod to that angular velocity?

It wouldn't take any energy to rotate the rod with m1 in the centre, because the velo of the point mass at the centre would be zero.

Perhaps you misunderstood your professor, as mjsd says?
 
Yeah, I thought what u said was the answer too, I'll have to ask my professor again, maybe i switched up what he said.

Thanks for helping
 
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