Question

A uniform block of granite in the shape of a book has face dimensions of 23 cm and 19 cm and a thickness of 1.2 cm. The density (mass per unit volume) of granite is 2.64 g/cm3. The block rotates around an axis that is perpendicular to its face and half way between its center and a corner. Its angular momentum about that axis is 0.104 kg·m2/s. What is its rotational kinetic energy about that axis?

Answer 1

KInteic energy is given by:

KE = 0.5*I*w^2

Angular momentum is given by:

L = I*w = 0.104 kg-m^2/sec

KE = 0.5*L^2/I

I = moment of inertia about center of mass

a = length = 23 cm = 0.23 m

b = width = 19 cm = 0.19 m

h = height = 1.2 cm = 0.012 m

rho = 2.64 gm/cm^3 = 2640 kg/m^3

Book is spining about the point = sqrt ((a/4)^2 + (b/4)^2)

Moment of inertia about this point will be

I = Icenter + Iz

I = M*(a^2 + b^2)/12 + M*(a/4)^2 + M*(b/4)^2

Mass = density*Volume = rho*a*b*h

So,

KE = 0.5*L^2/[rho*a*b*h*((a^2 + b^2)/12 + (a/4)^2 + (b/4)^2)

KE = 0.5*0.104^2/[2640*0.23*0.19*0.012*[(0.23^2 + 0.19^2)/12 + (0.23/4)^2 + (0.19/4)^2]]

KE = 0.3009 J