Taken from Tipler\'s Physics forScientists and Engineers 6th Edition chapter 9,

Question

   Taken from Tipler's Physics forScientists and Engineers 6th Edition chapter 9, problem #57. The problem states that there is a cone rotating "about anaxis parallel to its base and passing through itsapex." To me, this implies that the axis looks something likethis:       |< where | is the axis and < is the cone. However, with respects to a real-world application, this makesno sense. The cone is to be used in an advertisement. Why would the cone rotate in such a way? It seems likethe axis should be oriented like this:       -<- where the hyphens are the axis and < is the cone. Could someone help clarify this, as well as potentially offera solution to the problem as a whole? I have tried to workthe problem out in both situations and cannot seem to get theanswer shown in the back of the book. Thanks in advance!    Taken from Tipler's Physics forScientists and Engineers 6th Edition chapter 9, problem #57. The problem states that there is a cone rotating "about anaxis parallel to its base and passing through itsapex." To me, this implies that the axis looks something likethis:       |< where | is the axis and < is the cone. However, with respects to a real-world application, this makesno sense. The cone is to be used in an advertisement. Why would the cone rotate in such a way? It seems likethe axis should be oriented like this:       -<- where the hyphens are the axis and < is the cone. Could someone help clarify this, as well as potentially offera solution to the problem as a whole? I have tried to workthe problem out in both situations and cannot seem to get theanswer shown in the back of the book. Thanks in advance! where the hyphens are the axis and < is the cone. Could someone help clarify this, as well as potentially offera solution to the problem as a whole? I have tried to workthe problem out in both situations and cannot seem to get theanswer shown in the back of the book. Thanks in advance!

Explanation / Answer

Using an axis parallel to the base Moment of inertia of circular disk about a diameter is mR2 / 4 At distance h along axis of disk   I = mR2 / 4 + m h2 r = h tan       where isthe half angle at the vertex I = m h2 tan 2 + mh2    for a thin circular disk atdistance h from the vertex I = (tan2 + 1) mh2    dI = (tan2 + 1) h2 dm   at distance h from vertex dm = r2 dh = h2 tan2 dh Not sure here that using the variable r would be easier butwill see what happens dI = (tan2 +1) tan2 h4 dh I = (tan2 +1) tan2 H5 /5    integrating from 0 to H   V = R2 H / 3   for a rightcircular cone V = H3 tan2 / 3 M = V = H3 tan2 / 3 = 5 * H3 tan2 / (3 * 5) I = (5 / 3) M (tan2 + 1)H2     using M in our formula forI I = (5 / 3) M sec2 H2 Better check the math but I think this is what the problem islooking for It makes sense that the moment inertia would depend on theangle at the vertex because of the increase in the moment proportional to h2 as hincreases from the vertex. dI = (tan2 + 1) h2 dm   at distance h from vertex dm = r2 dh = h2 tan2 dh Not sure here that using the variable r would be easier butwill see what happens dI = (tan2 +1) tan2 h4 dh I = (tan2 +1) tan2 H5 /5    integrating from 0 to H   V = R2 H / 3   for a rightcircular cone V = H3 tan2 / 3 M = V = H3 tan2 / 3 = 5 * H3 tan2 / (3 * 5) I = (5 / 3) M (tan2 + 1)H2     using M in our formula forI I = (5 / 3) M sec2 H2 Better check the math but I think this is what the problem islooking for It makes sense that the moment inertia would depend on theangle at the vertex because of the increase in the moment proportional to h2 as hincreases from the vertex. Better check the math but I think this is what the problem islooking for It makes sense that the moment inertia would depend on theangle at the vertex because of the increase in the moment proportional to h2 as hincreases from the vertex.

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