A small object of mass m = 1.80 kg moves in a horizontal circle of radius r = 1.

Question

A small object of mass m = 1.80 kg moves in a horizontal circle of radius r = 1.95 m on a rough table. It is attached to a horizontal string fixed at the center of the circle. The speed of the object is initially 2.62 m/s. After completing one full trip around the circle, the speed of the object is 1.57 m/s. How many more revolutions will the object be making before coming to rest?

Explanation / Answer

friction force = f therefore by energy conservation 0.5*1.80*(2.62^2-1.57^2) = f*2*pi*1.95 =>f=0.323 N a= 0.323/1.80 = 0.17954 m/s2 0^2= 2.62^2-2*0.17954*s =>s= 19.116 m Number of revolutions =(19.116 m)/(2*pi*1.95) = 1.56

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