Force and Torque exerted by Ferris wheel

In summary, pmk69 has posted a calculation inquiring about the force and torque exerted by a Ferris wheel with a diameter of 30m, speed of 10 RPM, and a mass of 35000kgs. After calculating the force and torque, it was found that the force is 575143.38 Newtons and the torque is 8623125 Nm. It is advised to define all variables and show units when using equations in future calculations.
  • #1
pmk69
9
0
Hi all,

I'm new to this physics forum. Please tell me that my calculation whether is it correct or not. The diameter of Ferris wheel 30m. speed 10 RPM. Mass of the wheel is 35000kgs. calculating the force and torque exerted by the wheel.

Force = ma; a = v^2/r; 15.7^2/15= 16.43 m/s^2;
therefore, Force = F = 35000 kgs x 16.43 m/s^2 = 575143.38 Newtons;
Torque = Iw = I x alpha = I x (a/r); I = MR^2 = 35000 x 15^2 = 7875000 kg-m^2;
alpha = a/r = 16.43/15 = 1.095 rad/sec;
So, Torque = 7875000 x 1.095 = 8623125 Nm

thanks,
pmk69
 
Physics news on Phys.org
  • #2


Hi pmk69,

Welcome to the physics forum! Your calculation looks correct to me. It's always important to double check your units and make sure they are consistent throughout the calculation. In this case, you used meters for distance and seconds for time, which is consistent with the SI units used in physics. Great job!

Just a few things to keep in mind for future calculations:

1. Make sure to define all your variables and explain any assumptions you are making. For example, in this case, it would be helpful to specify that "a" is the centripetal acceleration and "w" is the angular velocity.

2. When using equations, it's important to show the units for each variable and the final result. This helps to ensure that the units are consistent and also makes it easier for others to follow your calculation.

Overall, great work on your calculation and welcome to the physics community! Keep asking questions and exploring the world of physics.
 
  • #3


Hello pmk69,

Your calculation for the force and torque exerted by the Ferris wheel seems to be correct. The force exerted by the wheel on its passengers is equal to the mass of the wheel multiplied by its centripetal acceleration, which you have calculated correctly using the formula a = v^2/r. The torque exerted by the wheel is equal to its moment of inertia (I) multiplied by its angular acceleration (alpha). You have correctly calculated the moment of inertia using the formula I = MR^2, where M is the mass of the wheel and R is its radius. Your calculation for the angular acceleration (alpha) is also correct, as it is equal to the linear acceleration (a) divided by the radius (r). Multiplying these two values together gives you the torque exerted by the wheel.

One thing to note is that the units for torque should be in Nm (Newton-meters), not just N (Newtons). This is because torque is a vector quantity, with both magnitude and direction. So the full units for torque should be kg-m^2/s^2, which simplifies to Nm.

Overall, your calculation seems to be correct and follows the correct formulas for calculating force and torque in rotational motion. Keep up the good work!
 

1. What is force and torque exerted by a Ferris wheel?

Force and torque are two fundamental concepts in physics that describe the movement and rotation of objects. Force is the push or pull that an object experiences, while torque is the turning or twisting force that causes an object to rotate. In the context of a Ferris wheel, the force and torque exerted are the result of the wheel's rotation and the weight of the passengers.

2. How is force and torque calculated on a Ferris wheel?

The force and torque exerted by a Ferris wheel can be calculated using basic physics equations. The force can be determined by multiplying the mass of the passengers by the acceleration due to gravity, while the torque can be calculated by multiplying the force by the distance from the center of rotation to the point where the force is applied.

3. What factors affect the force and torque exerted by a Ferris wheel?

Several factors can influence the force and torque exerted by a Ferris wheel. These include the weight of the passengers, the speed and direction of the wheel's rotation, and the distance between the center of rotation and the point where the force is applied. Additionally, the design and construction of the Ferris wheel can also impact the force and torque exerted.

4. How does the force and torque exerted by a Ferris wheel affect the passengers?

The force and torque exerted by a Ferris wheel can have a significant impact on the experience of the passengers. The force experienced by the passengers can cause them to feel weightless or experience a change in direction, while the torque can cause them to lean or feel a twisting sensation. The intensity of these effects can vary depending on the speed and direction of the wheel's rotation.

5. Are there any safety concerns related to the force and torque exerted by a Ferris wheel?

While Ferris wheels are designed to be safe, the force and torque exerted by them can pose potential risks to passengers. Excessive force or torque can cause the wheel to malfunction or lead to passenger discomfort or injury. Therefore, it is essential for Ferris wheels to be regularly inspected and maintained to ensure that the force and torque exerted are within safe limits.

Similar threads

  • Introductory Physics Homework Help
Replies
22
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
3
Replies
97
Views
3K
  • Introductory Physics Homework Help
Replies
7
Views
238
  • Introductory Physics Homework Help
Replies
5
Views
982
  • Introductory Physics Homework Help
Replies
2
Views
844
  • Introductory Physics Homework Help
2
Replies
44
Views
5K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
824
Back
Top