Question

A 32 cm -diameter circular saw blade has a mass of 0.84 kg , distributed uniformly as in a disk. a) What is its rotational kinetic energy at 2700 rpm ? b) What average power must be applied to bring the blade from rest to 2700 rpm in 3.5 s ?

Answer 1

(a) Here the rotational inertia of a uniform solid disk is I = 1/2MR^2.

put the values -

I = 1/2 x 0.84 x (0.16)^2 = 0.010752 kg.m^2

Again, rotational kinetic energy, K = 1/2 x I x (w)^2

where (w) is the angular speed.

given that the saw is rotating at 2700 rpm = 2700 / 60 rps

so w = (2700/60) x 2 x PI radians / s (as one revolution is equal to 2 x PI radians or 360 degrees)

K = 1/2 x I x (w)^2
= 1/2 x 0.010752 x ((2700 / 60) x 2 x PI)^2
= 429.6 J

(b) For the average power to bring the blade from rest.

P = delta(E) / delta(t)
= 429.6 / 3.5
= 122.7 W