An electric motor running at 300rev/min does 920J of work per second. (1)what is the torque of the motor spindle? (2)what force is being exerted,if the radius of the spindle is0.3m?
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Asked by Kennymore on 24th October, 2013
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6 years ago
= 184/0.3 = 613.3N/M
Force exerted= mv^2/r = 18.4*5^2/0.3= 1500N