Nonuniform cylindrical object. In Figure 11-39, a cylindrical object of mass M a

Question

Nonuniform cylindrical object. In Figure 11-39, a cylindrical object of mass M and radius R rolls smoothly from rest down a ramp and onto a horizontal section. From there it rolls off the ramp and onto the floor, landing a horizontal distance d = 0.453 m from the end of the ramp. The initial height of the object is H = 0.89 m; the end of the ramp is at height h = 0.15 m. The object consists of an outer cylindrical shell (of a certain uniform density) that is glued to a central cylinder (of a different uniform density). The rotational inertia of the object can be expressed in the general form I =

Explanation / Answer

consider the horizontal surface
let the velocity of cylinder be u when it falls on the horizontal surface
we have v2 - u2 = 2*a*s
v= final velocity =0 ; a= *g where - coefficient of friction and g-accelaration due to gravity;s=d=0.453

v=sqrt(0.906**g) --------- eq1

now consider the ramp

we have *M*g*cos = I* where -angle of the ramp ; -angular accelaration ; I-moment of inertia

since the cylinder rolls we have =a/R2

*M*g*cos = I*(a/R2) ------------- eq2

we have M*g*sin - *M*g*cos = M*a --------- eq3

now divide eq2 by eq3

*cos/(sin-*cos) = I/(M*R2)

= M*R2*tan/(M*R2+I)

substitute in eq1

we have work done by all forces = change in kinetic energy
work done by gravity wg = M*g*(0.89-0.15)
change in kinetic energy = 1/2*I*w2 --------- eq4

w=v/R

substituing v from eq1 and I=*M*R2 in eq4 we get

change in kinetic energy = (0.453*tan*g)/(1+(1/))

so, wg = change in kinetic energy

M*g*(0.89-0.15) = (0.453*tan*g)/(1+(1/))

that gives = 1/[(0.612*tan)-1]

** i think should be given in the problem to find value numerically **

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