A 2 kg mass and a 6 kg mass are attached to either end of a 3 m long massless ro

Question

A 2 kg mass and a 6 kg mass are attached to either end of a 3 m long massless rod.

a.) Find the center of mass of the system.
m, from the 2 kg mass.

Find the rotational inertia (I) of the system when rotated about:
b.) the end with the 2 kg mass.
kg m2
c.) the end with the 6 kg mass.
kg m2
d.) the center of the rod.
kg m2
e.) the center of mass of the system.
kg m2
(Compare this to parts b-d. Is this what you expect?)

f.) If the system is rotated about the center of mass by a force of 6 N acting on the 6 kg mass, (perpendicular to the rod), what will the size of the angular acceleration of the system be?
rad/s2

Explanation / Answer

The one mass, m1 = 2 kg The second mass, m2 = 6 kg The distance between the two masses, d = 3 m a) Let the distance of the centre of mass from mass m1 is x and is at a distance of (d-x) from the mass m2. Then we have m1x = m2(d-x) x(m1 + m2) = m2d From the above we have, x = m2d/(m1+m2) = 0.75 m So the distance of centre of mass from the mass 2kg is x1 = 0.75 m The distance of centre of mass from the mass 6 kg is x2 = 3 - 0.75 = 2.25 m b) I = m2d^2 = 5 * 3^2 = 45 kgm^2 c) I = m1d^2 = 2 * 3^2 = 18 kgm^2 d) I = m1(d/2)^2 + m2(d/2)^2 = 18 kg m^2 e) I = m1x1^2 + m2x2^2 = 31.5 kg m^2 f) The moment of inertia about centre of mass, I = 18 kg m^2 The distance of centre of mass from 6kg, r = 2.25 m Applied force, F = 6 N We have a formula for the torque as = I = Fr From the above we have        = Fr/I = 6*2.25/18 = 0.75 rad/s^2 f) The moment of inertia about centre of mass, I = 18 kg m^2 The distance of centre of mass from 6kg, r = 2.25 m Applied force, F = 6 N We have a formula for the torque as = I = Fr From the above we have        = Fr/I = 6*2.25/18 = 0.75 rad/s^2

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