Moment of inertia and rotational kinetic energy

AI Thread Summary
To calculate the net work required to accelerate a merry-go-round from rest to a rotation rate of 1.00 revolutions per 8.00 seconds, the change in kinetic energy equation was used, resulting in 14,211.7 J. Initially, there was confusion regarding the use of torque, as the time to reach the final angular velocity was not known. It was pointed out that the correct variable for torque calculations should be Δθ instead of Δt. Alternative methods, such as using kinematics, were suggested to derive the work done, emphasizing the work-energy theorem as a simpler approach. Overall, the discussion focused on the correct application of physics equations to solve the problem efficiently.
pb23me
Messages
204
Reaction score
0

Homework Statement


A merry-go-round has a mass of 1640 kg and a radius of 7.50 m. How much net work is required to accelerate it from rest to a rotation rate of 1.00 revolutions per 8.00s? Assume it is a solid cylinder.


Homework Equations


\DeltaKE=Wnet=1/2(Iw2)=14211.7J
\tau=(\Deltaw/\Deltat)


The Attempt at a Solution

I tryed to get the answer by finding net torque first, but i didnt think i could find the net torque because i don't have the time that the merry go round goes from zero to w=.785rad/s.Then i just used the change in kinetic energy equation to get 14211.7J could i have found this answer the other way?
 
Physics news on Phys.org
pb23me said:
\DeltaKE=Wnet=1/2(Iw2)=14211.7J
\tau=(\Deltaw/\color{red}{\Delta}t)
You used the wrong variable! :-p It should be Δθ, not Δt

\tau = (ΔW)/(Δθ)​

The Attempt at a Solution

I tryed to get the answer by finding net torque first, but i didnt think i could find the net torque because i don't have the time that the merry go round goes from zero to w=.785rad/s.Then i just used the change in kinetic energy equation to get 14211.7J could i have found this answer the other way?

Yes. You could use kinematics if you wanted to. (But the work-energy theorem, which you ended up using in the end, is much easier.)

If you wanted to use kinematics, use the following equations:
  • W = \tau·θ (definition of work, assuming uniform torque)
  • \tau = (Newton's second law in angular terms)
  • ωf2 - ωi2 = 2αθ (one of your angular kinematics equations).
Combine the equations and solve for W. :smile: (Hint: if you keep everything in terms of variables, and solve for W before substituting in specific numbers, you'll get a pleasing result! :wink:)
 
thank you:smile:
 
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...
Back
Top