Question

what is the grinding wheels kinetic energy when it is rotating at 1200 rev/min?

a 1.5-kg grinding wheel is in the form of a solid cylinder of radius 0 10 m that rotates about an axis through its center..

Answer 1

Rotational kinetic energy is given by:

KErot = 0.5*I*w^2

w = angular speed = 1200 rev/min

w = (1200 rev/min)*(2*pi rad/1 rev)*(1 min/60 sec) = 125.66 rad/sec

I = moment of inertia of Solid Cylinder = M*R^2/2

M = mass of wheel = 1.5 kg

R = radius of cylinder = 0.10 m

I = 1.5*0.10^2/2 = 7.5*10^-3 kg-m^2

So, Using these values

KErot = 0.5*(7.5*10^-3 kg-m^2)*(125.66 rad/sec)^2

KErot = 59.2 J

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