Two Particles on Thin Rods Two particles, each with mass m = 5.6 g , are fastene

Question

Two Particles on Thin Rods

Two particles, each with mass m = 5.6 g, are fastened to each other and to a rotation axis at P, by two thin rods, each with length L = 0.90 m and a mass of 7.4 g, as shown. The combination rotates around the rotation axis with an angular velocity of 15.0 rad/s.

A) Find the rotational inertia of the combination about P?

B)What is the kinetic energy associated with the rotation about P?

I am stuck on this one. For part A) I have tried 0.0307, 2.74, 20.412, 0.0706, 0.01755 (units kg*m^2) None of these are correct. I know I need to find part A) before I can solve part B). Been working on this for a few hours now I need help with it.Thank you in advance.

Explanation / Answer

m = 5.6g L = 0.9m M = 7.4g = 15 rad/s I = ?

The moment of inertia of the system about the axis P is the sum of the moments of inertia for each rod and mass. Since the rods are equal length and mass and since they are attached in a line, they can be treated as a single rod of mass 2M and length 2L.

i.e I = 1/3 (2M) (2L)2 + mr12 + mr22

Since the masses are given in grams, first convert them to kilograms by dividing by 1000. The radius to the first mass is the length of the first rod. The radius of the second mass is the length of both rods.

i.e I = 8/3 ML2 + mL2 + m(2L)2 = (8/3)(.0074)(0.9)2+(.0056)(0.9)2+(.0056)4(0.9)2

=> I = 0.03866 Kg.m2

B) Kinetic energy associated = (1/2)Iw2 = (1/2)(.03866)(15)2 = 4.35J

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