A 1 kg mass and a 6 kg mass are attached to either end of a 3 m long massless ro

Question

A 1 kg mass and a 6 kg mass are attached to either end of a 3 m long massless rod.

Find the rotational inertia (I) of the system when rotated about:
b.) the end with the 1 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

f.) Suppose a frictionless pivot is attached at the center of mass, such that the system is free to rotate about this point in the horizontal plane. A force of 6 N is exerted perpendicular to the rod, pushing directly on the 6 kg mass. What is the size of the system's angular acceleration that would result?
rad/s2

Explanation / Answer

b) inertia = 6*3^2 = 54 Kg m^2

c) inertia = 1 * 3^2 = 9 kg m^2

d) inertia = (6+1)*1.5^2 = 15.75 kg m^2

e) center of mass = 1*3 / 7 = 0.42857 m from 6 kg mass
= 1.5 - 0.42857 = 1.07142857 m from the center..

I_cm + 7*1.07142857^2 = 15.75
so.. I_cm = 7.7142857 kg m^2

f) torque applied = force * distance = 6 * 0.42857 = 2.57142 N-m
inertia = 7.7142857 kg m^2

so acceleraion = torque / inertia = 2.57142 / 7.7142857 = 0.33333 rad/sec2

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