Phys21(W09)MT2 Solutions

Phys21(W09)MT2 Solutions - 5‘0 L 3770 m Physics 21...

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Unformatted text preview: 5‘0 L 3770 m Physics 21 Midterm # 2 Winter 2009 1. (30 points) A wheel consists of a circular rim of radius R and mass M, together with 6 spokes, each of mass M12, as shown. This wheel is placed on an incline of angle 6, upon which it rolls without slipping. With what acceleration does it move down the incline? (Treat each pair of spokes as a thin rod, and recall that the moment of inertia of a thin rod of mass M and length L about its center is ML2 /12.) 0‘1 4 7L Hui m Mme par-cf m far, So we fan/field] Jim {LAW/W J, Icw’“{C/M)K(; éL/Wfl \2 WALSMKQ CM/io/ 0/ fl/ =9 9Mjltsmc02— Mt): “151,41 J/ZL/f/naz/wé, {”1” 6/4?" ‘73?”1‘ aéwf 6/2, (h‘f’r (”A {/42 ou/W/ mamm- M»: m é \E f’ffl‘f flfv/f, A?" a“ All“ Mara work. 2. A plank of length L width W and mass M, is lying at rest on a horizontal frictionless surface, as shown. A second identical plank is sliding across the surface moving in the x-direction with speed V0, as shown. Just as the moving plank passes the resting plank, their ends are fused together (crazy glue?) to make one long plank. The moment of inertia of such a plank about its center ofmass is M(L2 +W2)/12.‘ (a. 10 points) With what velocity (magnitude and direction) does the center of mass of the combined planks move?.. (b. 15 points) With what angular velocity (magnitude and direction) do the combined planks rotate? (c. 15 points) What fraction of the initial kinetic energy is lost in this collision? tws t—ws ~ L L N //w+2 O {0 L575 CIA-MM?" C‘Alwég/ 4% XML} éH-L our" orrjm‘ 47L 6 TUJFF AJJI‘L ffickffla LWJ 2.? __.J Lym-zMW—[Z 5 {AW [’Hof, 7 Arm. ST'IChwa . Lr Ruse-r 2 {MVdL/?)II W g/MHLUZ/L} HA Ml/ol/l 1L cm glam {7? mfim (Lay/L _M'—'$ Va % ~+ 3 U: {[l' $331 3. A thin rod of length 2a has a non—uniform density, which is lower near the center of the rod and increases toward the ends of the rod. The linear density fix), is given by 20:): 20(x/a)2, where 10 is a constant with dimensions of kg/m, and xis the distance from the center of the rod. (This means that the mass 5m , of a short piece of the rod of length 6x , located a distance x from the center of the rod is given by 6m = l(x)5x .) (a. 15 points) Calculate the mass M, of the rod. (b. 15 points) Calculate the moment of inertia of the rod about an axis passing through its center, and express your result in terms of M and a. ...
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