You shove a cube of inertia m and side length d so that it slides along a smooth

Question

You shove a cube of inertia m and side length d so that it slides
along a smooth table with speed v_i   (Figure a). The
cube then hits a raised lip at the end of the table. After it hits
the lip, the block begins to rotate about it (Figure b).
(a) Show that the magnitude of the block%u2019s angular mo-
mentum about the lip before the collision is L = mdv_i/2.
(b) Explain why the angular momentum still has that value
at the instant of collision, before the block has had time to
rotate much. (c) What is the rotational acceleration of the
block the instant after it hits the lip? (d) What maximum
initial speed can the block have so that it does not topple
over the lip?

Explanation / Answer

Let me help you

a)

In these types of problem , we do not find angular momentum from anywhere we want to ,

we calculate angular momentum only at the point where the block collides because when the block collides, there will be unknown inpulses will which will pass through the point of collision..

SOn if we calculate angulat momentum from that point ( i.e. the lip here ) then the moment due to collision forces would be zero and we will be able conserve angular momentum

SO calculating angular momentum about the lip

The velocity vector of block = v_i i^

location vector of center of block from the lip = ( - d /2 ) i^ + d/2 j^

angular momentum = m ( v vector ) x ( r vector )

= m ( v_i i^ ) x ( ( - d /2 ) i^ + d/2 j^ )

= m v_i * d /2 k^

b) The explaination is in part a itself .. because there wont be any affect of impact forces if we calculate angular momentum about the point of rotaion

c) This question is wrongly asked .. you mean rotational velocity right ...

So here we go .. .

Let the motational velocity be w

Inertia of the body about the corner = inertia about the center + m * ( d / sqrt ( 2 ) ) ^2

= (1/ 12 ) * [ md^2 + md^2 ] + md2 / 2

= md^2 / 6 + md^2 / 2

= (2/3 ) m*d^2

conserving angular momentum just before and after the collision

m*v_i ( d/2 ) = (2/3) * md^2 * w

so, w = (3/4) v_i / d

d)

Energy just after collision = 0.5 * Inertia * w^2

= 0.5 * (2/3) * m*d^2 * (9/16) * v_i^2 / d^2

= (3/16) * m*v_i^2

Just before toppling , the center of cube will be at height = d / sqrt ( 2 )

so, (3/16) * m*v_i^2 = m * g *d / sqrt ( 2 )

so, v_i = sqrt ( (16 / 3 root (2) ) * g * d )

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