A 0.84-lb collar with center of mass at G and a uniform cylindrical horizontal a

Question

A 0.84-lb collar with center of mass at G and a uniform cylindrical horizontal arm A of length L = 1 ft, radius r i = 0.021 ft, and weight W A = 1.30 lb arc rotating as shown with omega0 = 1.42 rad/s while the collar's mass center is at a distance d = 0.47 ft from the z axis. The vertical shaft has radius e = 0.03 ft and negligible mass. After the cord restraining the collar is cut, the collar slides with no friction relative to the arm. Assuming that no external forces or moments arc applied to the system, determine the collar's impact speed with the end of A if the collar is modeled as a particle coinciding with its own mass center (in this case, neglect the collar's dimensions), and the collar is modeled as a uniform hollow cylinder with length l = 0.13 ft, inner radius ri, and outer radius r0= 0.044 ft. There are three separate principles that are helpful to address the current problem. First, if no external moments are applied, the angular momentum of the system is conserved about the z -axis: Hz = HO = (IO + mCrO2)omega O = (IO + mCr2)omega Second, if no external forces are applied, the collar's radial equation of motion takes the form Combining the two equations, Finally, using a variation of one of our earlier kinematics relationships, Integrate the LHS of the equation over r and the right over to find the collar's impact speed at A .

Explanation / Answer

(a) Moment of inertia of arm I = WaL2/3 = 1.3*12/3 = 0.433 lb-ft2

Initial moment of inertia of system H0 = (I+mcd2)0 = (0.433+0.84*0.472)*1.42 = 0.879 lb-ft2/s

We have d2r/dt2 = H02r/(I+mcr2)2

and (d2r/dt2) dr = (dr/dt)(d/dr(dr/dt))

So, H02r/(I+mcr2)2 dr = (dr/dt)(d(dr/dt))

Integrating left hand side over r and RHS over dr/dt we have,

-H02/(2mc(I+mcr2)) + C1 = (dr/dt)2/2 + C2

Taking the integral from r = d to r = L,

-H02/(2mc(I+mcL2))+H02/(2mc(I+mcd2)) = 1/2*((dr/dt)L2 - (dr/dt)d2)

Putting (dr/dt)d = 0

We have the impact speed at A, (dr/dt)L = H0[1/mc(I+mcd2) - 1/mc(I+mcL2)]

= 0.879 * [1/0.84*(0.433+0.84*0.472) - 1/0.84*(0.433+0.84*12)]

= 0.879*(1.9246 - 0.935)

= 0.874 ft/s

(b) Now, at radius r, Moment of inertia of cylindrical collar about z-axis is, Icr = 1/12*mc[3*(rin2 + rout2) + l2] + mcr2

Moment of inertia of the system = I + Icr = 0.433 + 1/12*mc[3*(rin2 + rout2) + l2] + mcr2

This expression is similar to expression in part (a) except for the term 1/12*mc[3*(rin2 + rout2) + l2] = 0.001682

Hence, replacing I by I+0.001682 (=0.4347) in part (a) we get,

So, H0 = (0.4347+0.84*0.472)*1.42 = 0.8807 lb-ft2/s

We have the impact speed at A, (dr/dt)L = H0[1/mc(I+mcd2) - 1/mc(I+mcL2)]

= 0.8807 * [1/0.84*(0.4347+0.84*0.472) - 1/0.84*(0.4347+0.84*12)]

= 0.8807*(1.9195 - 0.9339)

= 0.8012 ft/s

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