Acceleration of a ferris wheel

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In uniform circular motion, a Ferris wheel does not experience tangential acceleration, as it maintains a constant speed along its circular path. The acceleration acting on a rider is centripetal, directed towards the center of the wheel, and is calculated using the formula mv²/r. While gravity acts downward, it has a tangential component only at the top and bottom of the wheel, which is balanced by the normal force from the seat. The forces acting on the rider, including gravity and the normal force, must sum to equal the centripetal force required for circular motion. Therefore, the net force is always directed towards the center, confirming the absence of tangential acceleration in this scenario.
nashsth
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Hi everyone I am confused about the acceleration of a ferris wheel undergoing uniform circular motion. By definition, the wheel undergoing uniform circular motion can't have a tangential acceleration, but I am confused as to why. I attached a diagram below to show you what I mean. Could you please clarify this confusion?

Thanks! :-)

https://www.flickr.com/photos/132657700@N04/22089785996/in/dateposted-public/
 
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Welcome to PF!

The acceleration due to gravity is separate, always in the same direction (only tangential twice per revolution) and so not part of the tangential acceleration due to uniform circular motion (which, as you say, is by definition zero).
 
I assume that blue dot is, say, a person riding that Ferris wheel undergoing uniform circular motion. Does gravity have a tangential component? Sure (except at top and bottom). But gravity is not the only force acting on the person. The net force will be centripetal.
 
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Hi, could you please expand on what you said? I didn't understand when you said the acceleration due to gravity is separate. :-(
 
nashsth said:
Hi, could you please expand on what you said? I didn't understand when you said the acceleration due to gravity is separate. :-(
Whether it is separate or not really is an issue of what the specific problem is asking you or specific explanation is referring to. There is nothing beyond that: it is either separate or it isn't, because someone wants it to be or doesn't.
 
nashsth said:
I attached a diagram below to show you what I mean.
The force from the wheel on the blue mass has a tangential component, which cancels the tangential component of gravity.
 
nashsth said:
Hi everyone I am confused about the acceleration of a ferris wheel undergoing uniform circular motion. By definition, the wheel undergoing uniform circular motion can't have a tangential acceleration, but I am confused as to why. I attached a diagram below to show you what I mean. Could you please clarify this confusion?

Thanks! :-)

https://www.flickr.com/photos/132657700@N04/22089785996/in/dateposted-public/
We follow Newton's second law to the letter. What are all the forces on the object (the blue dot)? First is gravity, always downward and equal in magnitude to mg, two is the force by whatever seat or harness is in contact with the object. This could include a normal force + friction. These forces must be together equal to the mass m times the acceleration. What is the acceleration? Since the object is moving in uniform circular motion, the acceleration is always toward the center and equal in magnitude to mv^2/r. There is no tangential acceleration in uniform circular motion. So the vector sum of the forces listed above must be equal to the mass times the acceleration. End of story.
 
Chandra Prayaga said:
We follow Newton's second law to the letter. What are all the forces on the object (the blue dot)? First is gravity, always downward and equal in magnitude to mg, two is the force by whatever seat or harness is in contact with the object. This could include a normal force + friction. These forces must be together equal to the mass m times the acceleration. What is the acceleration? Since the object is moving in uniform circular motion, the acceleration is always toward the center and equal in magnitude to mv^2/r. There is no tangential acceleration in uniform circular motion. So the vector sum of the forces listed above must be equal to the mass times the acceleration. End of story.
To further elaborate on what I wrote above, the normal force and the friction will keep adjusting all the time so that their sum, along with gravity, always gives the m tiomes the acceleration toward the center.
 
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