You are designing an athletics exhibit containing a baseball bat that will be su

Question

You are designing an athletics exhibit containing a baseball bat that will be suspended horizontally by a single thin wire attached to it midway between its ends. Although shapes, sizes, and weights vary a bit, a typical baseball bat is 34.0 . long, weighs 8.60 , and has its center of mass three-fourths of the way from the end of the handle. In your exhibit, you glue a 150 baseball to the thick end of the bat.


A.)In order that the bat hang horizontally, where should you attach a small 15.0 weight so that the center of mass of this system will be at the point where the support wire is attached to the bat?

Explanation / Answer

i apparently interpreted the question incorrectly the wire is going to be attached at the mid point of the bat, 17". you need to set the equation up so that the center of mass equals 17". center of mass without the weights = 25.5in (3/4 * 34in) 15.0N weight = 1.53kg 8.6N bat = .877 kg baseball = .150 kg 17in = (mass of weight * X in) + (mass bat * 25.5in) + (mass of baseball * 34.0in)/ (mass of weight + mass of bat + mass of baseball) solve for X. X = 10.5in from the handle.

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