Answer: a) 0.315 (V/L)
Explanation:
From Conservation of angular momentum, we know that
L1 = L2 ,
Therefore MV L/2 = ( Irod + Ib) x W
M/4 x V x L/2 = (M (L/2)^2 + 1/3xMxL^2) x W
M/8 X VL = (ML^2/16 + ML^2 /3 )
After elimination we have,
V/8 = 19/48 x L x W
W = 48/8 x V/19L = 6/19 x V/L
Therefore W = (0.136)X V/L
The value of final angular speed of the uniform rod which rests on the frictionless horizontal surface is,
[tex]\omega=\dfrac{6v}{19L}[/tex]
The angular speed of a body is the rate by which the body changed its angle with respect to the time. It can be given as,
[tex]\omega= \dfrac{\Delta \theta}{\Delta t}[/tex]
A uniform rod of length L rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end.
The rod is initially at rest. A bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed v strikes the rod at its center and becomes embedded in it.
The mass of the bullet is one-fourth the mass of the rod. The diagram for the above condition is attached below.
In the attached image the angular momentum about the point A is constant just before and after the collision. Thus,
[tex]L_i=L_f\\L_i=\dfrac{m}{4}\times\dfrac{vL}{2}=I\omega\\\dfrac{m}{4}\times\dfrac{vL}{2}=I\omega[/tex]
Put the value of inertia as,
[tex]\dfrac{m}{4}\times\dfrac{vL}{2}=\left(\dfrac{mL^2}{3}+\dfrac{mL^2}{4\times4}\right)\\\dfrac{mvL}{8}\times\dfrac{vL}{2}=\left(\dfrac{19mL^2}{48}\right)[/tex]
Solving it further we get,
[tex]\dfrac{v}{8}=\dfrac{19}{48}L\omega\\\omega=\dfrac{6v}{19L}[/tex]
Hence, the value of final angular speed of the uniform rod which rests on the frictionless horizontal surface is,
[tex]\omega=\dfrac{6v}{19L}[/tex]
Learn more about the angular speed here;
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