imported>Patrick at 16:46, 12 August 2011

2011-08-12T16:46:03Z

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'''(a)''' The speed of the particle in uniform circular motion can be found from the period (T = 2 s) and the radius (r = 0.5 m):

<math>v = {2 \pi r \over T}</math>

<math>v = 1.57 m/s</math> 

'''(b)''' <math>K = \frac{1}{2} m v^2 = \frac{1}{2}(0.5)(1.57)^2 = 0.616 J</math> 

'''(c)''' The force acting on the particle acts towards the center (i.e.: is centripetal) and is given by:

<math>F_C = {m v^2 \over r} = 2 \times {\frac{1}{2} m v^2 \over r} = 2 \times {K \over r} = 2 \times {(0.616) \over (0.5)} = 2.46 N</math> 

'''(d)''' There is no work done by the force. The displacement <math>\Delta r</math> is in the direction of the velocity. The centripetal force is perpendicular to the velocity and hence to <math>\Delta r</math>, so the dot product is zero: 
[[image:Helena_Work_Ex_5_Soln.png|TOP]] 

<math>W = F \cdot \Delta r = F r \cos(90^{\circ}) = 0 J</math>
 
Since there is no work done on the particle, there is no change in kinetic energy and its speed stays constant even though there is a force acting on it.

Work EX 5 - Revision history

imported>Patrick at 16:46, 12 August 2011