Question

A grindstone of mass 12 kg and radius 0.3 m is initially rotating freely at 48 rad/sec. An axe is brought into contact with the grindstone, which brings it to a stop in 6.6 seconds. Assume the grindstone is a uniform solid cylinder. 1) What is the moment of inertia of the grindstone around its rotation axis

Answer 1

Answer:

$I = 0.54\,kg\cdot m^{2}$

Explanation:

1) The moment of inertia of the grindstone is:

$I = \frac{1}{2}\cdot m \cdot r^{2}$

$I = \frac{1}{2}\cdot (12\,kg)\cdot (0.3\,m)^{2}$

$I = 0.54\,kg\cdot m^{2}$

Answer 2

Answer:

Moment of inertia of the grindstone around its rotation axis = 0.54 Kg.m²

Explanation:

We are given that;

Mass; m = 12kg

Radius; r = 0.3m

Now,moment of inertia of a cylinder is given as;

I = ½mr²

Plugging in the relevant values, we obtain

= ½ x 12kg x (0.3m)²

= 6kg x 0.09m²

= 0.54 kg·m²