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Vertical Circular Motion

mmsphyschem by mmsphyschem
06.12.2021
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Table of Contents

  • Vertical Circular Motion
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Vertical Circular Motion

 

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      TT =  N    TB =  N  m =  kg  vave =  m/s
    r =  m    a =   m/s2   T =  s      ω =  rev/s

For vertical circular motion, the motion is not uniform, as the object increases speed on the downward swing and
decreases on the way up. When analyzing vertical circular motion, one finds that there are two forces acting on the
object. These two forces are T, the tension in the string, and FW, the weight of the object, as shown below.


The weight, FW, can be resolved into a tangential component expressed as FWsinθ and a radial component expressed
as FWcosθ. Applying Newton’s second law to express the tangential acceleration gives:
 

aT = FT/m = FWsinθ/m = mgsinθ/m = gsinθ
 
Similarly, the radial acceleration can be expressed as:
 

aR = FR/m = (T – FWcosθ)/m = (T – mgcosθ)/m
 
Substituting v2/r for aR and solving for T yields:
 

T = m(v2/r + gcosθ)
 
Two interesting points to consider are the top and the bottom of the circle. First the bottom:
 
At the lowest point of the circular path, θ = 0° and cos 0° = 1. Substituting into the equation for T yields:
 

T = m(v2/r + g)
 
At the highest point of the circular path, θ = 180° and cos 180° = -1. Substituting into the equation for T yields:
 

T = m(v2/r – g)
 
At both the highest and lowest point, the tangential acceleration is zero because sin 0° = sin 180° = 0 and the    
acceleration is strictly radial. Physically at these two points, could there be similarities between being on an elevator      
and being on a ferris wheel?
 
At the top of a vertical circle, there are two possible forces that can exist: FW – the weight of the object, and FT – the
tension in the string. The simulator solves problems assuming the only force at the top is FW. It assumes that the weight
is the only agent providing the centripetal force which represents the minimum speed the mass must have. At any smaller
velocity, the string does not remain taught and the mass falls out of its circular path.
 
The x and y velocity components are displayed to further show that circular motion is as sinusoidal as a pendulum. Just
as with the Horizontal Circular Motion simulator, after entering the input values, press Play and the remaining values will
be computed. The simulator accepts simple factors for entries. For example, if a problem gives the weight of an object
(685 N) rather than its mass, simply enter 685/9.80 for its mass.
 
 
1) A man with a mass of 75 kg rides on a ferris wheel which is 21.5 m in diameter and rotates once every 14.0 s.
    (a) How much would the man appear to weigh at the top of the ferris wheel?
    (b) How much would the man appear to weigh at the bottom of the ferris wheel?
 
2) A motorcyclist rides down a ramp and around the loop-the-loop as shown below. If the loop has a radius of 17 m,
    what is the slowest speed the rider can have at the top of the loop to avoid falling?
 


 
3) A 80.0 kg person is spinning around the equator of a planet which is rotating at 3.2 x 103 km/h. The centripetal force
    present is 8.0 N. What is the radius of the planet?
 
4) An object with a mass of 8.0 kg is swung in a vertical circle of radius 2.4 m with a speed of 6.0 m/s.
    (a) Determine the maximum and minimum tension in the string.
    (b) The string breaks when the tension exceeds 340 N. Determine the maximum speed of the mass and where the
          mass will be when the string breaks. Justify your answer as to where the mass will be.

 

Answers

Physlets

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PowerPoint

CF Solver

 

 

 

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