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A small 4.0 kg brick is released from rest 2.5 $\mathrm{m}$ above a horizontal seesaw on a fulcrum at its center, as shown in Figure 10.52 . Find the angular momentum of this brick about a horizontal axis through the fulcrum and perpendicular to the plane of the figure (a) the instant the brick is released and (b) the instant before it strikes the seesaw.

a) 0 $\mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s}$b) $ 44.8 \mathrm{kg} . \mathrm{m}^{2} / \mathrm{s}$

Physics 101 Mechanics

Chapter 10

Dynamics of Rotational Motion

Newton's Laws of Motion

Rotation of Rigid Bodies

Equilibrium and Elasticity

University of Michigan - Ann Arbor

University of Washington

University of Winnipeg

Lectures

02:34

In physics, a rigid body i…

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Brett (mass 70 $\mathrm{kg…

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Well, we're working Problem number 25 what we have going on here is we have a small brick that has a massive four kilograms, and it is released from rest, which means its velocity initially is zero and it hits a seesaw. And there's two questions. First is what is it angular momentum before or as the brick is released here and when it hits just at the instant it hits the seesaw. And I want the angular Momenta MME about being middle of the seesaw, the pivot point there in the middle, we know angular momentum is defined is I Omega. And because this is a particle with respect to this, see, so its angular momentum would be m r squared times the makeup. Now we know that if the velocity is zero, the angular momentum initially would be zero. So that makes initial angular momentum. Zero now be says, What about when it hits the seesaw? Well, we will have angular momentum at that point. Are angular velocity at that point because we'll have velocity. At that point, we can use our kin O Matics formulas to find the final velocity. We know it fell 2.5 meters. So we want to find that we know that the square is the initial square plus two and it's free. Falling seller acceleration is G in the displacement of why solving for V in Subbing in No, it fell down 2.5. That gives us a velocity of seven meters per second at this point. So how do we make that angular where we remember that our Trent 10 gentle variables are related by the radius. So we know this is seven. We know the distance apart 1.6 or and that would help us get our angular spirit of the home. Yeah, and I have 4.375 radiance for a second. So substituting in now, massive. Our block is four kilograms. R lever arm here is 1.6 and are angular. Speed is 4.375 It gives me 44 point a kilograms. *** squared per second. Thank you for

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