Magnitude of acceleration in circular motion

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Homework Statement


A ferris wheel with radius 14.0m is turning about a horizontal axis through its center. The linear speed of a passenger on the rim is constand and equal to 7.69m/s. What is the magnitude of the passenger's acceleration as she passes through the lowest point in her circular motion?

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The Attempt at a Solution


Because the speed is constant I thought that the magnitude of the acceleration would be zero and only the direction of the acceleration would have a value, but zero isn't the right answer so there's a concept to answering this question that I'm completely missing and I have no idea how to go about finding it. Does the fact that the person is at the lowest point make a difference? All of this is very new to me and I'm just very confused.
 

Answers and Replies

  • #2
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If you tie a rock to a string and swing it in a circle - like the ferris wheel - doesn't it apply tension to the string? What causes the tension?

What are the relevant equations you have at your disposal?
 
  • #3
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I'm given atan=(dlvl)/dt
 
  • #4
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I'm given atan=(dlvl)/dt
I have no idea what you have written. That said, what about formulas for centripetal acceleration such as V^2/R or R*omega^2 where omega is angular speed in radians/second?
 
  • #5
haruspex
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Because the speed is constant I thought that the magnitude of the acceleration would be zero
Acceleration is a change in velocity. (Linear acceleration is a change in linear velocity, angular acceleration is a change in angular velocity.)
Velocity is a vector. The speed is the magnitude of the velocity.
If the speed stays constant but the direction changes then the velocity changes, hence the acceleration is not zero. Acceleration is also a vector. When the acceleration is at right angles to the velocity the speed stays constant. If an object moves in a circle at constant speed its acceleration is towards the centre of the circle.
 

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