38. Advance Illustration | Rigid Body Dynamics | A Block toppling over a Ridge | by Ashish Arora
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38. Advance Illustration | Rigid Body Dynamics | A Block toppling over a Ridge | by Ashish Arora

January 4, 2020


In this illustration we’ll analyze a block
toppling over a ridge. we are given that a cubical block of mass m, moving at velocity
u on a smooth ground. and it hits a ridge o on ground, we are required to find minimum
speed u, of the block for which it will topple over the ridge. now, in this situation. we
can see. due to the momentum in the block. when it hits the ridge. it starts rotating
about the ridge, say this is the ridge o. when the block strikes this ridge, it has
a momentum in forward direction due to which, it starts rotating about the ridge and it
may topple over here. and it will topple only if its center of mass will cross this vertical
line like this. so if its center of mass will cross over this line, then it’ll topple otherwise
it’ll fall back. so in this situation first we can calculate the angular speed with which
it starts. rotating about the ridge. here we can write. at the point of. hitting. we
use. conservation of angular momentum. as. here we can write m, u multiplied by ay by
2 was initial. angular momentum of. the block, about this ridge, and after collision this
starts rotating about point o. so its moment of inertia we can write. for, a cubical block
about a diagonal its moment of inertia we can write as. 2 by 3 m ay square. multiplied
by the angular speed omega. so this is giving us the value angular speed just after collision
that is 3 u by, 4 a, and, we can write, for, toppling. over
the ridge. the omega is such that. center of mass of, block. crosses over. the ridge.
so that the weight of gravity will support in toppling then it’ll fall over the ridge.
so, here we can write. thus. by, work energy theorem. we use. here the initial kinetic
energy of, this block when it started rotation is, half. its i is to 2 by 3, m ay square.
multiplied by omega square is 3 u by. 4 ay whole square. minus. the, rise in center of
mass we can see initially its center of mass was at a height ay by 2. and in this final
situation its center of mass is at a height, ay by, root 2. so the work done again gravity
is m g multiplied by. ay by root 2 minus, ay by 2. should be greater then equals to
zero. here, m gets cancelled out. here ay square also gets cancelled out and simplifying.
this relation. here, we can see we’ll be getting, this 2 we can also cancel out. so this will
give us the value. 3 u square, by, 16. should be greater then or equal to, this g a multiplied
by. root 2 minus 1 divided by 2. and this giving us the value of u should be greater
then equal to root of. 8 g a by 3 multiplied by, root 2 minus 1 that is a result of this
problem.

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