Rotational Motion & Rotational Kinetic Energy

In summary: I think you're also trying to use w = w(initial)+at, but those are just derivatives, you need derivatives of the position and velocity vectors, w(initial), w(t), a(t)In summary, the children walk towards the center of the platform until they are 0.50 m from the center. At this point, the platform's rotational speed increases to 8.21 rev/min.
  • #1
robert6774
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Four children stand at the edge of a circular horizontal platform that is free to rotate about a vertical axis. Each child has a mass of 35 kg and they are at positions that are a quarter-circle from each other. The platform has a moment of inertia equal to 500 kg*m^2 and a radius of 2.0 m. The system is initially rotating at 6.0 rev/min. The children walk towards the center of the platform until they are 0.50 m from the center.

m= 35 kg
I= 500 kg*m^2
r1= 2.0 m
w(initial)= 6.0 rev/min or 0.628 rad/s (correct me if I'm wrong)
r2= 0.50 m

(a) What is the rotational speed of the platform when the children are at the 0.50 m positions?

(b) What is the change in kinetic energy of the system?

Here are a few equations I might need to use:

w= d(theta)/dt

a= dw/dt

w= w(initial)+at

(theta)= w(initial)t + 1/2at^2

2a(theta)= w^2 - w(initial)^2

s=r(theta)

a= w^2r

K= 1/2 Iw^2

I don't think I should bring time as a variable into the picture because it would make it more things more complicated. I don't really know where to start. The only thing I've really done so far is write down the givens and draw a diagram. Thats a lot of equations but I know I don't need them all.

I tried using a= w^2r but I ended up with weird units. Maybe I don't understand the equation. So it would be (6.0)^2(2.0)= 72 rev/min*m. What?

The thing is I feel as though I'm missing one too many variables. When I try to utilize an equation, I can't solve it. For example, when I try to use w= w(initial)+at I'm missing both the acceleration and the time. When using 2a(theta)= w^2 - w(initial)^2 I'm missing teh acceleration and theta.

Because my only lead with a= w^2 isn't working, I'm stuck.

Any help and guidance would be very much appreciated, thank you.
 
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  • #2
welcome to pf!

hi robert6774! welcome to pf! :smile:

(have a theta: θ and an omega: ω and an alpha: α and try using the X2 and X2 icons just above the Reply box :wink:)
robert6774 said:
Four children stand at the edge of a circular horizontal platform that is free to rotate about a vertical axis. Each child has a mass of 35 kg and they are at positions that are a quarter-circle from each other. The platform has a moment of inertia equal to 500 kg*m^2 and a radius of 2.0 m. The system is initially rotating at 6.0 rev/min. The children walk towards the center of the platform until they are 0.50 m from the center.…

I tried using a= w^2r …

you don't need the force or acceleration or time

find ωf by using conservation of https://www.physicsforums.com/library.php?do=view_item&itemid=313" ), applied to the children-and-platform :smile:
 
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  • #3
Oh wow now I feel stupid. One of those "oh...duh" moments. Anyway thank you very much for your help.

One thing though, is my conversion from 6.0 rev/min to 0.628 rad/sec correct? I tried using the latter in the conservation of angular momentum equation but I ended up getting a slower second velocity which doesn't make sense.
 
  • #4
hi robert6774! :smile:

(just got up :zzz: …)

6.0 rev/min = 6.0/60 rev/sec = 0.1 rev/sec = 0.1*2π rad/sec = 0.628 rad/sec seems ok

how did it come out slower?
 
  • #5
I'm still getting the wrong answer. The key says its 12 rev/min but I'm getting 8.21 rev/minute.

Here's my work:

Conservation of angular momentum (of children and platform) gives:

I[tex]\omega[/tex][tex]_{}1[/tex]+m[tex]\omega[/tex][tex]_{}1[/tex]r[tex]_{}1[/tex] = I[tex]\omega[/tex][tex]_{}2[/tex]+m[tex]\omega[/tex][tex]_{}2[/tex]r[tex]_{}2[/tex]

Solve for [tex]\omega[/tex][tex]_{}2[/tex],

[tex]\omega[/tex][tex]_{}2[/tex]= ((I+mr[tex]_{}1[/tex])[tex]\omega[/tex]1)/(I+mr[tex]_{}2[/tex])

Plug in values:

((500+(4)(35)(2.0))(6.0))/(500+(4)(35)(0.5)) = 4680/570 = 8.21 rev/min

Where did I go wrong?

I'm pretty new to the Latex stuff so this attempt at making the equations look pretty may not go so well, bear with me please.
 
  • #6
hi robert6774! :smile:

(please don't mix latex and text in the same line, it's very difficult to read :redface:)

you're using mωr for angular momentum, it has to be radius "cross" momentum, = r x mv = mωr2
 

1. What is rotational motion?

Rotational motion is the movement of an object around an axis or center point. This type of motion is different from linear motion, which is movement in a straight line.

2. What factors affect rotational motion?

The factors that affect rotational motion include the mass of the object, the distance from the axis of rotation, and the speed of rotation. Other factors such as air resistance and friction may also play a role.

3. What is rotational kinetic energy?

Rotational kinetic energy is the energy an object possesses due to its rotational motion. It is dependent on the object's mass, speed of rotation, and distance from the axis of rotation.

4. How is rotational kinetic energy calculated?

The formula for calculating rotational kinetic energy is KE = 1/2 * I * ω^2, where KE is the kinetic energy, I is the moment of inertia, and ω is the angular velocity. The moment of inertia can be calculated using the shape and mass distribution of the object.

5. How does rotational motion relate to real-world applications?

Rotational motion is prevalent in many real-world applications, such as the rotation of wheels on a car, the spinning of a top, and the movement of a propeller on an airplane. Understanding rotational motion is essential in designing and optimizing these types of systems.

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