Question 4 of 7 view policies current attempt in progress...

Question 4 of 7 View Policies Current Attempt in Progress The mass center G of the slender bar of mass 0.60 kg and length 0.46 m is falling vertically with a velocity $v = 2.4 m/s$ at the instant depicted. Calculate the angular momentum of the bar about point O if the angular velocity of the bar is (a) $\omega_o = 8.4 rad/s$ clockwise and (b) $\omega_b = 8.4 rad/s$ counterclockwise. The angular momentum is positive if counterclockwise, negative if clockwise. 0.30 m 0.46 m Answers: (a) $H_o = $ kg-m²/s (b) $H_o = $ kg-m²/s eTextbook and Media

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00:01 Hello everyone so let's start with the question so in a question as you can see in the diagram there is a rod thin rod of a uniform mass distribution which is pivoted about one of its end by a pane passing through that point and the mass of the rod which is given that is m equal to what 0 .440 kg also the length we have given that is 1 .53 meter now when the rod is released from horizontal position suppose it is the horizontal position and it is released from the horizontal position it swings down to a vertical position.

00:36 So it will be going this means in this direction and it swings down to the, it comes to this point.

00:43 Okay, verticals.

00:44 So we have to find, we have to find what? what is the angular momentum? angular momentum of the center of mass, of the center of mass of the rod.

01:07 About the pain, about the pain.

01:14 And also we have to, we have to state the relationship between, relationship between angular velocity, angular and tangential velocity.

01:37 So, now, so let's start with solution.

01:41 So first we have to know that is the moment of inertia of rod.

01:47 Moment of inertia of rod about pinpoint so it will be that is 1 upon 3 1 divided by 3 m l square now using the equation of motion using the equation of motion that is angular motions we get the formula that is omega square of final point minus omega square initial point equal to 2 alpha theta this is a similar equation of linear equation that is v square minus u square equal to 2 as okay and here alpha is our angular acceleration or omega is our linear acceleration and theta is our angular displacement now so here initially initial angular velocity is zero radiant per seconds therefore i can write that is omega square f equal to 2 alpha theta now as the rod is swinging means it is moving so there will be some torque and acting.

02:48 So i can write, i can write that is talk acting on rod is about mg l by 2...

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