Question

what is the angular velocity of a shaft if the power is 2.5kw and the torque is 7 nm

Answer 1

Explanation:

Instantaneous power ‘P' of a shaft is defined as the rate of change in it's kinetic energy which is dE/dt , now, if the saft has moment of inertia ‘I' and it is rotating with an angular velocity ‘w’, then it's kinetic energy ‘E' can be given as

E = (Iw² )/2 ….. (1)

Now, as we're interested to find the rate of change of kinetic energy, we need to differentiate this energy equation (eq. 1) with respect to time to get the equation as below

dE/dt =Iw(dw/dt) or,

P=dE/dt =Iw(dw/dt) ……(2)

dw/dt is nothing but the rate at which angular velocity is changing, also known as angular acceleration, and nett torque ‘T’ is ‘ moment of inertia times angular acceleration', hence, equation (2) can be written as follows

P = Tw or w=P/T

Now, using the given values of P and T we get the angular velocity w of the shaft as under

w = 2500W/7Nm = 357.14 rad/sec