Moment of inertia and rotational kinetic energy

In summary, the problem involves finding the net work required to accelerate a merry-go-round from rest to a rotation rate of 1.00 revolutions per 8.00 seconds. Using the change in kinetic energy equation, the answer is calculated to be 14211.7J. Alternatively, one could also use kinematics equations and solve for work, keeping the solution in terms of variables to get a more elegant result.
  • #1
pb23me
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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


[itex]\Delta[/itex]KE=Wnet=1/2(Iw2)=14211.7J
[itex]\tau[/itex]=([itex]\Delta[/itex]w/[itex]\Delta[/itex]t)


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?
 
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  • #2
pb23me said:
[itex]\Delta[/itex]KE=Wnet=1/2(Iw2)=14211.7J
[itex]\tau[/itex]=([itex]\Delta[/itex]w/[itex]\color{red}{\Delta}[/itex]t)
You used the wrong variable! :-p It should be Δθ, not Δt

[itex]\tau[/itex] = (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 = [itex]\tau[/itex]·θ (definition of work, assuming uniform torque)
  • [itex]\tau[/itex] = (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:)
 
  • #3
thank you:smile:
 

FAQ: Moment of inertia and rotational kinetic energy

What is moment of inertia?

Moment of inertia is a measure of an object's resistance to changes in its rotational motion. It is the rotational analog of mass in linear motion.

How is moment of inertia calculated?

Moment of inertia is calculated by multiplying the mass of an object by the square of its distance from the axis of rotation. The formula is I = mr^2, where I is the moment of inertia, m is the mass, and r is the distance from the axis of rotation.

What is the relationship between moment of inertia and rotational kinetic energy?

Rotational kinetic energy is directly proportional to the moment of inertia. This means that as the moment of inertia increases, so does the rotational kinetic energy. This relationship can be represented by the formula KE = 1/2 Iω^2, where KE is the rotational kinetic energy, I is the moment of inertia, and ω is the angular velocity.

How does the distribution of mass affect moment of inertia?

The distribution of mass affects moment of inertia because it determines how far the mass is from the axis of rotation. Objects with a larger moment of inertia have more mass located farther from the axis of rotation, while objects with a smaller moment of inertia have more mass located closer to the axis of rotation.

What are some real-life examples of moment of inertia and rotational kinetic energy?

Some real-life examples of moment of inertia and rotational kinetic energy include a spinning top, a rotating merry-go-round, a rolling ball or wheel, and a swinging pendulum. In each of these examples, the moment of inertia and rotational kinetic energy play a crucial role in the object's motion.

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