A large iron cylinder of mass 150 kg rests so that its circular face rolls along
ID: 2144398 • Letter: A
Question
A large iron cylinder of mass 150 kg rests so that its circular face rolls along the edge of a table as shown in the figure below. A smaller disk (radius 15 cm) is fastened to the circular face with its center axis aligned with the large cylinder such that the small disk extends beyond the edge of the table. A cable is wrapped around the smaller disk, which has half the radius and one-fourth the mass of the cylinder. With the entire assembly at rest, a mass of 4.5 kg is attached to the free end of the cable and allowed to descend through a distance h = 29 cm.
(a) Calculate the moment of inertia and the total mass of the entire assembly.
Explanation / Answer
small R1 = 15 cm = 0.15 m.....m1 = m2/4 = 150/4 = 37.5 kg
large R2 = 2*R1 = 30 cm = 0.30m..........m2 = 150 kg
a) I = 0.5*m1*R1^2 + 0.5*m2*R2^2 = 0.5*637.5*0.15^2 + 0.5*150*0.30^2
I = 13.922 kg m^2
M = 150+37.5+4.5 = 192 kg
b) torque = mg*R1 = I*a/R1
a = m*g*R1^2/I = 4.5*9.8*0.15^2/13.922 = 0.07127 m/s^2
KE = m g h = 4.5*9.8*0.29 = 12.789 J
c) m g h = KE of (cylinder + disk)+ KE rot of (cylinder +disk) + KE of m
m g h = 0.5*M*Vcm^2 + 0.5*I*Vcm^2/R^2 + 0.5*m*V^2
12.789 = 0.5*(150+37.5)*V^2 + 0.5*13.922*V^2/ R2^2 + 0.5*m*(Vcm*R1/R2)^2
12.789 = 0.5*Vcm^2*((150+37.5) + (13.922/0.09) + (4.5*0.25))
Vcm = 27.3 cm/s
d)
d + (R1/R2)*d = h
3d/2 = h
d = 2h/3 = 2*29/3 = 19.33 cm
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