Angular momentum is conserved, just before the clay hits and just after;
<span>mv(L/2) = Iw </span>
<span>I is the combined moment of inertia of the rod, (1/12)ML^2 , and the clay at the tip, m(L/2)^2 ; </span>
<span>I = [(1/12)ML^2 + m(L/2)^2] </span>
<span>Immediately after the collision the kinetic energy of rod + clay swings the rod up so the clay rises to a height "h" above its lowest point, giving it potential energy, mgh. From energy conservation in this phase of the problem; </span>
<span>(1/2)Iw^2 = mgh </span>
<span>Use the "w" found in the conservation of momentum above; and solve for "h" </span>
<span>h = mv^2L^2/8gI </span>
<span>Next, get the angle by noting it is related to "h" as; </span>
<span>h = (L/2) - (L/2)Cos() </span>
<span>So finally </span>
<span>Cos() = 1- 2h/L = 1 - mv^2L/4gI </span>
<span>m=mass of clay </span>
<span>M=mass of rod </span>
<span>L=length of rod </span>
<span>v=velocity of clay</span>
weeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee true eeeeeeeeeeeeeeeeeeeee
Answer:
3.605551275463989
Explanation:
solve using Pythagorean theorem
Answer:
um how about no.. this is not the site for what you're looking for...
Explanation:
