A uniform rod, supported and pivoted at its midpoint, but initially at rest, has

Question

A uniform rod, supported and pivoted at
its midpoint, but initially at rest, has a mass
of 86 g and a length 5 cm. A piece of clay with
mass 21 g and velocity 1.7 m/s hits the very
top of the rod, gets stuck and causes the clay-
rod system to spin about the pivot point O at
the center of the rod in a horizontal plane.
Viewed from above the scheme is
!
!
1.7 m/s

(a) before (b) during (c) after
5 cm
After the collisions the clay-rod system has an
diby (ad33488) – HW8 – mackie – (10612) 3
angular velocity ! about the pivot.
With respect to the pivot point O, what is
the magnitude of the initial angular momen-
tum Li of the clay-rod system?
Answer in units of kgm2/s
014 (part 2 of 3) 10.0 points
With respect to the pivot point O, what is
the final moment of inertia If of the clay-rod
system?
Answer in units of kgm2
015 (part 3 of 3) 10.0 points
The final angular speed !f of the clay-rod
system is
Answer in units of rad/s

Explanation / Answer

      The mass of the rod, M = 86 g                                            = (86 g)(10-3 kg/1 g)                                            = 0.086 kg       The length of the rod, L = 5 cm                                             = (5 cm)(10-2 m/ 1 cm)                                             = 0.05 m       The mass of the clay, m = 21 g                                             = (21 g)(10-3 kg/1 g)                                             = 0.021 kg       The velocity of the clay, v = 1.7 m/s       The distance between clay and pivot point of the rod, r = L/2                                                                                               = 0.05 m/2                                                                                                   = 0.025 m _______________________________________________________________       The moment of inertia of the clay, Ic= mr2       The moment of inertia of the rod, Ir= (1/12)mL2 ________________________________________________________________ a)       Before:       Initially the rod is in rest position, so the initial angular       momentum of the rod is zero.       Initially the rod is in rest position, so the initial angular       momentum of the rod is zero.       So, the initial angular moment of the clay-rod system is                    Li = mvr                       = (0.021 kg)(1.7 m/s)(0.025 m)                        = 0.0008925 kg.m2/s _________________________________________________________________ _________________________________________________________________ b)       During:       The final moment of inertia If of the clay-rod system is                       If = (1/12)ML2+mr2                          = (1/12)(0.086 kg)(0.05 m)2+(0.021 kg)(0.025 m)2                           = 0.0000310416 kg.m2 __________________________________________________________________ __________________________________________________________________ c)       The angular momentum related with angular speed can be expressed as                                 L = I       After:       The final angular speed of of the clay-rod system is                       f = Lf/If
                            = (0.0008925 kg.m2/s)/(0.0000310416 kg.m2)                             = 28.75 rad/s              The final angular speed of of the clay-rod system is                       f = Lf/If
                            = (0.0008925 kg.m2/s)/(0.0000310416 kg.m2)                             = 28.75 rad/s       

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