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A uniform rod of mass 1.2 kg and length 1.8 m is pivoted in the horizontal position as shown (black point).enter image description here The rod is at rest and then released. The acceleration due to gravity is $g = 9.8 m/s^2$. 1) Find the rotational inertia of the rod relative to the axis perpendicular to the screen and passing through the pivot point. 2) What is the angular speed (in rad/s, but do not include units) of the rod as it passes through the vertical position (when end marked B is at the bottom)?

My attempt of solution:

To find moment of inertia, I use the moment of inertia of a uniform rod respect to it's center of mass (middle in this case), which is $$I_{cm}=mL^{2}/12$$ where $m$ is mass of the rod and $L$ its length. Then I use the parallel axis theorem to find the moment of inertia in this case: $$mL^{2}/12+0.45^{2}m=0.567$$

For the 2º part, since the net torque on the rod is not constant I use potential energy to find first the speed of center of mass at the vertical position. The initial height of center of mass is 0.45 m taking as reference point position of the middle point of the rod at vertical position. So: $$mg0.45=1/2mv_{cm}^{2}+1/2I\omega ^{2}=1/2mv_{cm}^{2}+1/2I(v_{cm}^2/0.45^{2})$$

From here I get $v_{cm}$ and later the angular velocity.

Where am I wrong?

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  • $\begingroup$ What makes you think there is a mistake? And in which of the questions? $\endgroup$ – Enredanrestos Aug 8 '14 at 11:32
  • $\begingroup$ when I introduce an answer in the test I get it wrong. So I was wondering if there is something wrong with my reasoning or algebra @Enredanrestos $\endgroup$ – Mykolas Aug 8 '14 at 11:34
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I think it might be this $$mgh=\frac{1}{2}mv_{cm}^2+\frac{1}{2}I_{cm}\omega^2=\frac{1}{2}I\omega^2.$$ I mean, it seems that you added the kinetic energy due to center of mass velocity to the rotational kinetic energy respect to the rotational point. It should have been respect to the center of mass.

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