Answer:
The moment of inertia will be increased by a factor of 1575
Explanation:
Let the moment of inertia of a wheel be given as;
[tex]I = \frac{1}{2} mr^2\\\\[/tex]
where;
I is the moment of inertia of the wheel
m is mass of the wheel
r is radius of the wheel
when the mass is increased by a factor of 7 and the radius is increased by a factor of 15
[tex]I_1 = \frac{1}{2}mr^2\\ \\I_2 = \frac{1}{2}(7m)(15r)^2\\\\I_2 = \frac{1}{2}(7m)(225r^2)\\\\I_2 = 7\times 225[\frac{1}{2}(mr^2)]\\\\I_2 = 1575[\frac{1}{2}(mr^2)]\\\\Recall, \frac{1}{2}(mr^2) = I_1 \\\\I_2 = 1575 I_1[/tex]
Thus, the moment of inertia will be increased by a factor of 1575
