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It is generally a good idea to gain an understanding of the \"size\" of units. E

ID: 1441223 • Letter: I

Question

It is generally a good idea to gain an understanding of the "size" of units. Either using an Internet search, or referring to the hint panel below, find the official size and masses of the sports balls listed below and calculate their moments of inertia about their center of mass. Rounding your results to the nearest power of 10, categorize each of the sports balls according to the order of magnitude of the moment of inertia. (Note that some of the balls are hollow.) 10_5kgm2 10-4kgm2 10_3kg-m2 10-2kgm2 10-1kgm2 1 kgm2As a physics demonstration, you want a special bowling ball made to demonstrate exactly 1 kg - m2, so that your students can rotate the ball about its center of mass to get a "feel" for how "big" 1 kg- m^2 is. The bowling balls most familiar to your students has a weight of 15.6 pounds and have a circumference of 26.6 inches, but do not have a moment-of-inertia equal to 1 kg - m2. Since the sporting goods manufacturer has no understanding of how "big" 1 kg - m2 is, calculate the diameter of the demo bowling ball (in inches) it will need to manufacture. Assume that bowling balls are solid, with a constant density.

Explanation / Answer

The equation for moment of inertia for a sphere is I=2/5mR2

where m = mass and R = radius

Using the data given for a standard bowling ball (15.6 lb = 26.6 in circumference) lets do some unit conversions to find the density in kg/m^3

15.6lb x 1 kg/2.2lb = 7.1kg
26.6 in x 2.54 cm/in x 1 m/ 100 cm = 0.67564 m in circum. = 2R => R = 0.107531 m

V=4/3R3 = 4/3(0.107531 m)3 = 0.005208 m3

Density = 7.1 kg/0.005208 m3 = 1363.23 kg/m3

So, Density = = m/V

rearranging that m=*V

plugging into moment of inertia equation: I = 2/5(*V*R2) = 2/5(*4/3*R3*R2)

Now to solve for R when I = 1 kg m2

1 kg m2 = 2/5(1363.23 kg/m3*4/3*R3*R2)

Radius = R = 0.2129 m = 8.38 in

Diameter = 2R = 16.76 inches

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