Moment of intertia and net work problem

AI Thread Summary
A ballerina spins at an initial speed of 1.5 rev/s with extended arms and increases her angular speed to 4.0 rev/s after drawing her arms in, resulting in a moment of inertia of 0.88 kg m². To calculate the net work done to increase her angular speed, the correct approach involves using the change in kinetic energy formula, which is derived from the moment of inertia and angular velocity. The work done is calculated as the difference between the final and initial kinetic energy, leading to a result of approximately 175.82 joules. The discussion clarifies the correct equations to use for calculating both moment of inertia and net work. Overall, the calculations and methodology presented are validated as accurate.
keylostman
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A ballerian spins initially at 1.5rev/s when arms are extended. She then draws in her arms to her body and her moment of intertia is .88kg m^2 and her angular speed increases to 4.0rev/s. Determine the network she did to increase her angular speed?

To find moment of inertia was pretty simple i used IW=I2W2 for and solved for I, I(1.5rev/s * 2PI rad/1rev)=(.88kg m^2)(4.0rev/s * 2PI rad/1rev) and i got 2.3kg M^2. Ok this was fairly easy. But for Net WORK, how do i calculate it ? I am not to sure which equation to use... Is it Sum of Net WORK= I2W2-I1W1, where W is squared?
 
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Find the change in her kinetic energy.
 
I2W2-I1W1, where W is squared?

This equation i use correct ?
 
No. (It's almost right.) You used the correct equation for KE in your other thread.
 
1/2IW^2 ? I thought Work done would be final work - intital work,
 
keylostman said:
1/2IW^2 ?
Right.
I thought Work done would be final work - intital work,
The work done will equal the change in energy.
 
so its then .5 * .88 * (4.0 * 2pi) ?
 
W = (1/2) (.88)(25.13)^2 - (1/2)(2.3)(9.42)^2

W =((1 / 2) * .88 * (25.13^2)) - ((1 / 2) * 2.3 * (9.42^2)) =
W= 175.820576 joules

this is what i did, and i believe it rights, makes sense
 
Looks good.
 
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