Question

Suppose the engine has a mass of 1010 kg and is shifted in location 63.0 cm toward the inside of the curve. Assume that the engine is relatively small in comparison to the radius of the curvature of the track and that the radius of the track turn is R = 28.5 m. Calculate the absolute value of the percent change in the moment of inertia of the engine about the center of curvature of the track caused by the shift in location.

Answer 1

Answer:

4.37%

Explanation:

M = Mass of engine = 1010 kg

R = Track turn radius = 28.5 m

r = Shift in location = 28.5-0.63 = 27.87 m

Percent change in the moment of inertia is given by

$\Delta I=\dfrac{\dfrac{1}{2}MR^2-\dfrac{1}{2}Mr^2}{\dfrac{1}{2}MR^2}\times 100\\\Rightarrow \Delta I=\dfrac{R^2-r^2}{R^2}\times 100\\\Rightarrow \Delta I=\dfrac{28.5^2-27.87^2}{28.5^2}\times 100\\\Rightarrow \Delta I=4.37\%$

The percent change in the moment of inertia is 4.37%