Question

Consider a thin 46 m rodpivoted at one end. A uniform density spherical object (whose massis 2 kg and radius is 12 m) is attached to the free end of the rodand the moment of inertia of the rod about anend is Irod= 1 / 3 mL^2

and themoment of inertia of the sphere about its center of mass isIsphere = 2 / 5 mr^2 .

What is the angular acceleration of therod immediately after it is released from its initial position of63? from thevertical? The acceleration of gravity g = 9.8 m/s2 .Answer in units ofrad/s2.

Answer 1

   [Newton's Second Law forrotational motion]

   [Solving for α; the nettorque is the weight of the center of mass at a distance of 'r'from our origin (chosen

                       to be the rotation axis]

   EQI. [Rotational inertia is the sum of the inertias forthe sphere and rod (respectively)]

You have all of the variables except "r" and θ. Here, "r" isthe distance of the center of mass of this object to the rotationalaxis.

    [Definition of centerof mass; Lrod = length of the rod] EQII

Plug in your givens to this equation to find the center ofmass. This is given as "R" in Equation 1.

Because this is falling at an angle, you need θ. (e.g. - thetorque is not perpendicular to the axis). Here's a picture of whatI mean:

As you can see, the vertical axis makes a 90 degree angle with thehorizontal. So the angle we need in equation 1 that is relative toFg is the complementary angle to θ. Thus, our θis 27 degrees.

This is a challenging problem so let me know if you havequestions. Look over this solution and follow the steps firstthough.