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![[Post New]](/templates/default/images/icon_minipost_new.gif) 31 Mar 2008 20:15:16 IST
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A cubical block of side a is moving with a velocity v on a horizontal smooth plane as shown in the figure. It hits a ridge at O. The angular speed of the block after it hits the ridge at O., is
(a) 3v/4a
(b) 3v/2a
(c)  3v/ 2a
(d) 0
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The deepest secret is that life is not a process of discovery, but a process of creation. You are not discovering yourself, but creating yourself anew. Seek therefore, not to find out who you are; seek to determine who you want to be."
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31 Mar 2008 20:23:55 IST
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come on guys fast fast...
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The deepest secret is that life is not a process of discovery, but a process of creation. You are not discovering yourself, but creating yourself anew. Seek therefore, not to find out who you are; seek to determine who you want to be."
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31 Mar 2008 20:33:14 IST
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The ridge is gonna exert some force on the bloc
but we can conserve angular momentum abt the ridge, as that force doesnt give any torque abt the ridge
Just before impact angl momentum abt the ridge = M*v*a/2
Final angular momentum = I*W
were I is the moment of inertia abt the ridge and W the angular velocity as seen from the frame of the ridge.
=[Ma^2/6+Ma^2/2 ]*W
equate both
n u get
W=3v/4a
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