# Answer to Question #23234 in Optics for Anitha

Question #23234

A circular disc of mass 5kg and diameter 0.2m initially at rest attains a angular velocity of 600rpm in 10 sec. Calculate the torque acting on it and the rate at which work is being done at the end of 10 sec

Expert's answer

Moment of inertia I = 0.5*m*r^2

torque T = I*acc

angular acceleration acc = w / t

torque T = 0.5*m*r^2 * w / t = 0.5* 5 kg *(0.2 m)^2 * (600 1/min) / (10 sec) =

0.1 N*m

rate at which work is being done is power P = T * w = 0.1 N*m * (600 1/min) = 1

W

Answer: torque 0.1 N*m, power 1 W

torque T = I*acc

angular acceleration acc = w / t

torque T = 0.5*m*r^2 * w / t = 0.5* 5 kg *(0.2 m)^2 * (600 1/min) / (10 sec) =

0.1 N*m

rate at which work is being done is power P = T * w = 0.1 N*m * (600 1/min) = 1

W

Answer: torque 0.1 N*m, power 1 W

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