A large iron cylinder of mass 360 kg rests so that its circular face rolls along

Question

A large iron cylinder of mass 360 kg rests so that its circular face rolls along the edge of a table as shown in the figure below. A smaller disk (radius 16 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 4.0 kg is attached to the free end of the cable and allowed to descend through a distance h = 37 cm. Calculate the moment of inertia and the total mass of the entire assembly. moment of inertia kg m2 total mass kg What is the total kinetic energy of the system after the attached mass descends through a height h? J If the entire assembly rolls without slipping, what is its velocity at this time? cm/s How far has the system moved along the horizontal? cm

Explanation / Answer

small R1 = 16 cm = 0.16 m.....m1 = m2/4 = 360/4 =90 kg

large R2 = 2*R1 = 32 cm = 0.32m..........m2 = 360kg


a) I = 0.5*m1*R1^2 + 0.5*m2*R2^2

= 0.5*90*0.16^2 + 0.5*360*0.32^2

I =19.584 kg m^2

M = 160+90+4= 254 kg

b) torque = mg*R1 = I*a/R1

a = m*g*R1^2/I = 4*9.8*0.16^2/19.584= 0.05124183006m/s^2

KE = m g h = 4*9.8*37 = 1450.4 J

c) KE = 0.5*M*V^2 + 0.5*M*W^2


W = V/R2

KE = 0.5*M*V^2+0.5*M*V^2/R2^2


KE = V^2 *M*0.5*(1+1/R2^2)


1450.41450.4 = 0.5*(360+90)*0.5*(1+(1/0.32)^2)*V^2


V = 1.09433453869 m/s

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