I am not understanding the set up of this problem from your on-line example. Her

Question

I am not understanding the set up of this problem from your on-line example. Here is the problem: "A this uniform rod (mass 4.3 kg, length 4.7 m) rotates freely about a horizontal axis A that is perpendicular to the rod and passes through a point at a distance d - 1.3 m from the end of the rod. The kinetic energy of the rod as it passes through the vertical position is 24 J (a) What is the rotational inertia of the rod about axis A? (b) What is the linear speed of the end B of the rod as the rod passes through the vertical position? (c) At what angle will the rod momentarily stop in its upward swing?

In part a, I used I = 1/12 mL^2 + md^2 to get I = 15.2 kg m^2 Answer appears to be wrong.

In part b I have kinetic energy (final) = initial PE, or 24J = mgy. I'm not sure where the 1*(1-cos theta) comes from in your solution.

Thank you.

Explanation / Answer

The mass of the rod m = 4.3kg the length of the rod L = 4.7m the position from end d = 1.3m (a) The moment of inertia from parallel axis theorem        I = Icm^2 + Id ^2          = 1/12 mL^2 + m[ (L/2) -d]^2           I = 1/12 mL^2 + md^2           = (1/12) (4.3)(4.7)^2 + (4.3)(1.05)^2           = 12.65 kgm^2 (b) The rotational kinetic energy         K = 1/2 I^2          24J = (0.5)(12.65) ^2 then ^2 =  3.79             =  1.95 rad/s now the linear speed v = r = (3.4)(1.95)                                              = 6.63 m/s (c) The maximum angle is obtained when all the kinetic energy is transformed into potential energy             the center of mass rises to h = d(1-cos) then            K = mgd(1-cos)             25 = (4.3)(9.8)(1.3)(1-cos)             1-cos = 0.45635 therefore the angle         = 57 deg therefore the angle         = 57 deg

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