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A uniform rod of length $L$ and mass $m$ is supported as shown. If the cable attached at end $B$ suddenly breaks, determine ( $a$ ) the acceleration of end $B,(b)$ the reaction at the pin support.
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Physics 101 Mechanics
Plane Motion of Rigid Bodies: Forces and Accelerations
Constrained Plane Motion
Motion Along a Straight Line
University of Washington
Simon Fraser University
University of Winnipeg
In physics, kinematics is the description of the motion of objects and systems in the frame of reference defined by the observer. An observer has to be specified, otherwise the term is meaningless.
In mathematics, a position is a point in space. The concept is abstracted from physical space, in which a position is a location given by the coordinates of a point. In physics, the term is used to describe a family of quantities which describe the configuration of a physical system in a given state. The term is also used to describe the set of possible configurations of a system.
End $A$ of the 8 -kg unifo…
The uniform rod $A B$ with…
A rod $A B,$ hinged at $A$…
So here we have a free body diagram for the beam. We're going to solve clockwise to be positive self. For the moment, about a This is equal, of course, the effective moment about any. And we then have the weight multiplied by half of the length. This will be then equal to the moment of inertia multiplied by the angular acceleration plus the mass times the linear acceleration multiplied by L over to. And so we can then say that m g l over to this would be equal to, um al Squared Alfa over 12 plugging in for the moment of inertia of a rod plus then and oh, squared Alfa over four. And so this is equal. And then, um, I'll squared also over three. And then we can cancel out EMS. Cancel her one of the els, and we have then that the angular acceleration is equaling three g over to l. So at this point, um, we can first, when we have this, we can actually start on part B and say that if upwards is positive that some of the forces in the Y direction would be equal to the sum of the effective forces in the Y direction and we have then a minus MGI Equalling negative mass times acceleration, linear acceleration. This is an Equalling negative. Um oh, off over two. And so we can then say that a minus and G equals the negative m multiplied by L over two multiplied by three g over to our the Els cancel out and we have then but a is equaling MGI over four upwards. So this would be your answer for part B and then for part A you want to find the acceleration of B This would be equal to based on ace of Be over beat a Equalling than zero plus the full length multiplied by, uh, the angular acceleration. And so we can say that Then the acceleration of be equaling l multiplied by three G over to l. And this is giving us three g over to and this would be downwards. So this would be our answer for party that is thean of the solution. Thank you for watching
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