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

from Density D = mass/Volume

mass m = DV

m = 2.64 *10^3 * (0.23* 0.19 * 0.012)

mass of the Block m = 1.384 kgs

Moment of inertia about its centre Icm = M/12*( a^2 + b^2)

MOI abt any given axis I = Icm + Md^2

here d^2 = (a/2)^2 + (b/2)^2

d^2 = (0.23/2)^2 + (0.19/2)^2

d^2 = 0.0225 m^2

I = (1.384/12) *((0.23^2 + 0.19^2)) + (1.384* 0.0225)

I = 0.0414 kgm^2

angular momentum L = I W

W = L/I

W = 0.104/0.0414

W = 2.512 rad/s

KE rotational = 0. I W^2

KEr = 0.5 * 0.0414 * 2.512^2

KEr = 0.1306 Joules