s m attached to the end of a thin. y about the fixed axis uniform rod of length

Question

s m attached to the end of a thin. y about the fixed axis uniform rod of length & and mass Sm. The object can rotate freel that is perpendicular to the rod and pass through the center of the rod as shown in the figure. The object is held in its horizontal position and then released from rest. swings downward due to the torque produced by gravity. Ignore friction and air resistance Rod has mass 5m and length Particle mass L/2 Determine the distance between the axis O and the center of mass of the object (i.e. find center-of-mass for the rod-particle system). Given: Land physisal a. L/12 b. What is the rotational inertia of the object (both rod and particle combined)? and physical constants 2

Explanation / Answer

a)

x1 = center of mass of the rod = 0

m1 = mass of rod = 5 m

x2 = center of mass of the particle = L/2

m2 = mass of particle = m

Xcm = center of mass of rod-particle system

Xcm = (m1 x1 + m2 x2)/(m1 + m2) = ((5m) (0) + (m) (L/2))/(5m + m) = L/12

b)

I = Irod + Iparticle = (5m)L2/12 + m(L/2)2 = (5m)L2/12 + mL2/4 = (0.67) mL2

c)

using conservation of energy

rotational KE = Potential energy

KE= (6m) g h

KE= (6m) g (L/12) = mgL/2

d)

(0.5) I w2 = mgL/2

(0.5) (2/3)mL2 w2 = mgL/2

w = sqrt(3g/2L)

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