12 Points) Pivot Point 2) A uniform thin rod of length L and mass M is allowed T

Question

12 Points) Pivot Point 2) A uniform thin rod of length L and mass M is allowed The moment of inertia for this configuration is I-(ML 1/3 to pivot about its end, as shown in the figure above. (a) The rod is allowed to fall from the horizontal position A through the vertical position B (shown above). Starting with conservation of energy, derive an expression (create an equation) for the velocity of the free end of the rod at Position B. The equation you derive can only contain the following symbols: M, L and g (numbers, squares, and square roots, and other algebraic necessities are acceptable as well). mva n experiment is designed to test the expression found in part (a). A student uses rods of various lengths that all ave a uniform distribution. The student releases each of the rods from the horizontal position A and uses otogates to measure the velocity of the free end at position B. The data are recorded below. ngth (m) city (m/s)2.7 0.5 3.8 0.751.00 1.25 1.50 4.6 0.25 5.2 5.8 6.3

Explanation / Answer

from conservation of energy

initial total energy = final total energy

mghi = mghf+1/2Iw^2

here I = 1/3mL^2

mg(L-L/2) =1/3* 1/2*mL^2*w^2

mgL/2 = 1/6*mL^2w^2

g = 1/3Lw^2

w^2 = 3g/L

w = sqrt(3g/L)

w is the angular velocity

from the relation v = rw

here r = L

v = Lw = sqrt(3gL)

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.