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Old Jul 2nd 2017, 06:10 PM   #1
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Rotational Motion Question!


So this is the question:

A uniform rigid rod with mass Mr = 2.7 kg, length L = 3.1 m rotates in the vertical xy plane about a frictionless pivot through its center. Two point-like particles m1 and m2, with masses m1 = 6.7 kg and m2 = 1.6 kg, are attached at the ends of the rod. What is the magnitude of the angular acceleration of the system when the rod makes an angle of 51.1 degrees with the horizontal? (m2 is 51.1 degrees above the horizontal, and m1 is 51.1 degrees below the horizontal).

My efforts:

I began by calculating inertia, like this:

I = mL2/12 = (2.7kg)(3.1m)2/12=2.16225 kgm2

Then I calculated the gravitational forces from m1 and m2:


Then I determined the component of these forces which act perpendicular to the lever arm:


Then I calculated the net torque/moment (clockwise negative, counterclockwise positive):

Radius = 1/2(3.1m)=1.55m
Torque = (1.55m41.23N)-(1.55m9.84N) = 48.65Nm

Then I calculated the angular acceleration:

Angular acceleration = Torque/Inertia
Angular acceleration = 48.65Nm/2.16kg*m2
Angular acceleration = 22.52 rad/s2

I have been told my answer is incorrect through an online system where I am able to check my answers, however I'm not sure where I went wrong. If anyone would be able to spot an error, or guide me in the right direction, it would be greatly appreciated!!
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Old Jul 3rd 2017, 09:03 AM   #2
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Join Date: Aug 2008
Posts: 80
Looks like you omitted the rotational inertia of the endpoint masses

$\alpha = \dfrac{\tau_{net}}{I}$

$\tau_{net} = m_1g\cos{\theta} \cdot \frac{L}{2} - m_2g\cos{\theta} \cdot \frac{L}{2} \approx 48.65 \, N~m$

$I_{sys} = m_1 \left(\frac{L}{2}\right)^2 + m_2 \left(\frac{L}{2}\right)^2 + \frac{m_r L^2}{12} \approx 22.1 \, kg~m^2$

$\alpha \approx 2.2 \, rad/s^2$
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