Calculating Power of Ferris Wheel Motor Despite Friction

  • Thread starter Bearbull24.5
  • Start date
  • Tags
    Power Wheel
In summary: 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 wrongWhat did you get for the starting w? ..24434051looks a bit weird.You get a negative value for the difference in the energies because the work of the friction decreases the energy. You need to input a positive power value to get the same result.
  • #1
Bearbull24.5
50
0

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
 
Physics news on Phys.org
  • #2
The angular momentum values are not given. But you need angular velocities, so determine them from the given rev/hour values.

ehild
 
  • #3
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?
 
  • #4
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
 
  • #5
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.
 
  • #6
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
 
  • #7
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
 
  • #8
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
 
  • #9
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
 
  • #10
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
 

Related to Calculating Power of Ferris Wheel Motor Despite Friction

What is the "Power of a Ferris Wheel"?

The "Power of a Ferris Wheel" refers to the amount of energy required to operate a ferris wheel and keep it in motion. It is typically measured in kilowatts (kW) or horsepower (hp).

How is the power of a ferris wheel determined?

The power of a ferris wheel is determined by its size, weight, and design. Larger and heavier ferris wheels require more power to overcome inertia and keep them moving, while those with more efficient designs may require less power.

What factors can affect the power of a ferris wheel?

The power of a ferris wheel can be affected by various factors such as wind resistance, friction in the bearings, and the weight of the passengers. Temperature and weather conditions can also impact the power needed to operate a ferris wheel.

How is the power of a ferris wheel used?

The power of a ferris wheel is used to operate the motor that turns the wheel and to power any other features of the ride, such as lights and music. It is also used to overcome resistance and keep the wheel in motion, providing a smooth and enjoyable ride for passengers.

How does the power of a ferris wheel compare to other machines?

The power of a ferris wheel is relatively low compared to other machines, as it is designed to provide a gentle and slow-moving ride. For example, a large ferris wheel may require around 80-100 kW of power, while a typical car engine can produce over 100 kW of power.

Similar threads

  • Introductory Physics Homework Help
Replies
6
Views
425
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
6K
  • Introductory Physics Homework Help
Replies
3
Views
5K
  • Introductory Physics Homework Help
Replies
19
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
3K
  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
5K
  • Introductory Physics Homework Help
Replies
3
Views
1K
Back
Top