Calculate the moment of inertia of the disk. What percent is this of the speed o

Question

Calculate the moment of inertia of the disk.



What percent is this of the speed of a particle sliding down africtionless ramp?
A disk of mass M and radius R has a hole of radius r centered on the axis. Calculate the moment of inertia of the disk. Express your answer in terms of the variables M, R, And r. A 5.0- cm-diameter disk with a 3.0- cm-diameter hole rolls down a 54- cm-long, 25 degree ramp. What is its speed at the bottom? What percent is this of the speed of a particle sliding down a friction less ramp?

Explanation / Answer

mass surface density = M/[(R2 -r2)] mass of the disk with radius r is m =r2whose moment of inertia =mr2/2 = r4/2 mass of the disk with radius R (without hole) is M' =R2 whose moment of inertia = M'R2/2 =R4/2 so the moment of inertia in this question =R4/2 - r4/2= (R4 - r4)/2 =M/[(R2 - r2)] *(R4 -r4)/2 I = M(R2 + r2)/2 R = 2.5 cm, r = 1.5 cm, L = 54 cm, = 25o, findv Mg(Lsin) = Mv2/2 + I2/2 =Mv2/2 + [M(R2 +r2)/2)*(v/R)2/2 = Mv2*[3 +(r/R)2]/4 v = {4gLsin/[3 + (r/R)2]} = 1.63m/s for a particle: vp = (2gLsin) = 2.11m/s v/vp = 77%

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