Calculating Power of Ferris Wheel Motor Despite Friction

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To calculate the power of the motor maintaining the Ferris wheel's rotation despite friction, it's essential to convert the given revolutions per hour into angular velocities in radians per second. The work done by the motor can be determined using the change in kinetic energy formula, which requires accurate values for initial and final angular velocities. The discussion highlights the importance of correctly calculating these velocities and emphasizes that the power output must be a positive value, reflecting the energy lost to friction. Ultimately, the power calculation should yield a value in watts, suitable for the scale of a Ferris wheel.
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Homework Statement


A motor keep a Ferris wheel (with moment of inertia 6.22 × 107 kg · m2) rotating at 14 rev/hr. When the motor is turned off, the wheel slows down (because of friction) to 9.3 rev/hr in 28 s.
What was the power of the motor that kept
the wheel rotating at 14 rev/hr despite fric-
tion?
Answer in units of W.

Homework Equations


W=(1/2)Iw(f)^2-(1/2)w(i)^2
P=Tau*w

w= angular momentum

The Attempt at a Solution



I really want to use the first equation to solve for work but I am unsure if I can use the angular momentum values that are given or if I need to convert them
 
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The angular momentum values are not given. But you need angular velocities, so determine them from the given rev/hour values.

ehild
 
I meant angular velocities, its been a long day.

So I would calculate each of those values by 2 pi and stick them into the top formula?
 
Bearbull24.5 said:
So I would calculate each of those values by 2 pi and stick them into the top formula?

You need to calculate the revolutions per second then multiply by 2pi to get the angular velocity.

By applying your first equation, you get the change of KE.

The change of KE equals to the work done. You have to determine the power.
Power is work divided by time.

ehild
 
I ended up with -1848547.342 for work. This is what I did

((.5)*(6.22*10^7)*(.01623156^2))-((.5)*(6.22*10^7)*(..24434051^2))

After dividing by the 28s it gave me a very large negative number that was wrong.
 
I thought maybe I only had to do half of the first equation since it did not ask for the change in power from the 2 speeds. So I did W=(1/2)Iw(i)^2 and then divided it by the 28s. This was also wrong
 
What did you get for the starting w? ..24434051looks a bit weird.

You get negative value for difference of the energies as it is the work of the friction: it decreases the energy. Using the motor, its keeps the KE constant if it yields the same power that is lost because of friction. So you need to input a positive power value. Start with recalculating the w-s.

A few kW is not a big power. That of a simple heater or water boiler is about 1 kW. And a Ferris Wheel is usually quite big.

ehild
 
What did you get for the starting w? ..24434051looks a bit weird.

You get negative value for difference of the energies as it is the work of the friction: it decreases the energy. Using the motor, its keeps the KE constant if it yields the same power that is lost because of friction. So you need to input a positive power value. Start with recalculating the w-s.

A few hundred watt is not a big power. That of a simple heater or water boiler is about 1 kW=1000 W. And a Ferris Wheel is usually quite big.

ehild
 
I took my starting revolutions per hour (14rev/hr) and converted it to be .0038888 rev/sec. I then multiplied it by 2pi to get .24434051
 
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Bearbull24.5 said:
I took my starting revolutions per hour (14rev/hr) and converted it to be .0038888 rev/sec. I then multiplied it by 2pi to get .24434051

It is 0.02443... You missed a zero.

ehild
 
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