Rotating Bar?. A thin, uniform 12.0-kg bar that is 2.0 m long rotates uniformly about a pivot at one end, making 5.0 complete revolutions every 3.00 seconds. What is the kinetic energy of this bar? (?Hint?: Different points in the bar have different speeds Break the bar up into infinitesimal segments of mass ?dm? and integrate to add up the kinetic energies of all these segments.)
Solution 61P Kinetic energy is calculated from the equation K = mv , where m is the mass of an 2 object and v is its speed. We are asked to consider a segment in the bar that has a mass of dm. Therefore, if the segment has a speed of v , its kinetic energy is given by dK = dm × v …..(1) 2 Let the length of the segment be dx. 12.0 kg is the mass of the rod and it has a length of 2.0 m. Therefore, the mass per unit 12 kg kg length of the rod is = 2 m = 6 m Therefore, the mass of the segment dx is = 6dx kg Or, dm = 6dx…..(2) Let the segment is located at a distance of x from the pivot. So, its speed is given by v = x , where is angular speed of the segment. Now, 5 revolutions = 10 rad Therefore, = 10/3 rad/s Therefore, v = x × 10/3 m/s v = 10.47x m/s Now, substituting the values of dm and v in equation (1), dK = 1× 6dx × (10.47x) 2 2 2 dK = 329x dx…..(3) Now, integrating this equation from x = 0to x = 2.0 m 2 dK = 329 x d 2 0 K = 329 × ( ) x3 2 3 0 8 K = 329 × 3 J K = 877 J Therefore, the approximate kinetic energy of the bar is 877 J.