Table of Contents
Vertical Circular Motion
|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 =
|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.