A sheet of metal in the shape of a triangle massing(adsbygoogle = window.adsbygoogle || []).push({}); 10 kgper square meter is to be spun at an angular velocity of4 radiansper second about some axis perpendicular to the plane of the sheet. The triangle is a right triangle with both short sides of length1 meter.

(a)The axis of rotation is the line through the right angle of the triangle and perpendicular to the plane of the sheet. What is the resulting kinetic energy?

(b)The axis of rotation is instead through the centroid of the triangle, or equivalently, through the center of mass of the sheet. That center of mass is two thirds of the way from any vertex to the midpoint of the side opposite it. Set up, but do not evaluate, a double integral which if evaluated would give the kinetic energy of the spinning plate. (The formula for kinetic energy is (1/2)mv^{2}where m denotes mass and v denotes regular speed, which is not the same thing as angular velocity.)

My solution so far:

a) Moment of inertia

I = 1/2*mrw^{2}= 1/2 * 10 * 0.5 * 4^{2}=40kgm2

w^{2}= v^{2}/r

v^{2}= wr

v^{2}= 4*0.5

v = sqrt 2

KE = 1/2 * (2/3 * m) * v^{2}

KE = 1/2 * 2/3 * 10 * (sqrt 2)^{2}

KE = 6.67J

How do I go about part b?

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# Setting up the double integral

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