Moment of Inertia problem from MCAT practice book.

  • Thread starter MCAT35
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Homework Statement



Experiement 1:

One student sits on a stool that rotates freely. He holds a 5-kg mass in each hand. Initially, the student has a angular velocity of 5 radians/sec with his arms in his lap.

Question:

In exp 1, with his arms outstretched, the student drops the weights. This will cause the angular velocity of the student to:

answer: remain the same

can someone explain to me why "moment of inertia would not change for the system when the student drops the weight" since Inertia is proportional to mass x r^2 wouldn't a decrease in mass after the weight drop decrease the moment of inertia?



Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
tiny-tim
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Welcome to PF!

Hi MCAT35! Welcome to PF! :smile:
can someone explain to me why "moment of inertia would not change for the system when the student drops the weight" since Inertia is proportional to mass x r^2 wouldn't a decrease in mass after the weight drop decrease the moment of inertia?
Forget moment of inertia … it's only a means to an end, and that end is angular momentum …

angular momentum (about an axis) will be constant so long as there is no external torque (about that axis) …

so is there any external torque if

i] he drops the weights
ii] he throws the weights downward
iii] he throws the weights outward
iv] he throws the weights sideways? :wink:
 
  • #3
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ok, that makes sense,

Thanks Tim and I'm glad to be here! Such an impressive forum!

Gonna try and answer this don't know if they are right. But no external torque about the axis of which he is spinning right?

i] he drops the weights (no, if he spins horizontally, this is an vertical force?)
ii] he throws the weights downward (same logic as 1?)
iii] he throws the weights outward (intuitively, I think this should increase angular momentum.)
iv] he throws the weights sideways? (yes?)

let me know lol..
 
  • #4
tiny-tim
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Hi MCAT35! :smile:

Correct except iii]

(maybe you misunderstood my question? … i meant radially outward)

the change in momentum is radial, so the distance of its line of action from the axis is zero, ie the moment of momentum (same thing as angular momentum) is zero :wink:
 
  • #5
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Hi MCAT35! :smile:

Correct except iii]

(maybe you misunderstood my question? … i meant radially outward)

the change in momentum is radial, so the distance of its line of action from the axis is zero, ie the moment of momentum (same thing as angular momentum) is zero :wink:
aha, cool got it. Radially outward = no lever arm (line of action) therefore external torque = 0. ic now, thanks so much!
 

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