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CBSE Class 12-science Answered

A cubical block of side a moving with velocity v on a smooth horizontal plane. It hits a rigid point. What will be the angular speed of the block after it hits the rigid point.
Asked by Today's | 02 Feb, 2016, 02:59: PM
answered-by-expert Expert Answer
 
The cubical box is moving towards the rigid support (red spot) with velocity v.
When it hits the support it tumbles and acquires an angular speed ω.
Now, the initial angular momentum just before it hits the support is

begin mathsize 14px style straight L subscript initial equals straight r cross times straight p equals AO cross times Mv equals Mvd over 2 end style
Now, the moment of inertia of the cube about central axis passing through O is I = Md2/6
 
Therefore, the moment of inertia about an axis passing through B and perpendicular to it is
 
begin mathsize 14px style straight I subscript straight B equals straight I subscript straight O plus straight M open parentheses BO close parentheses squared equals Md squared over 6 plus straight M open parentheses fraction numerator straight d over denominator square root of 2 end fraction close parentheses squared straight I subscript straight B equals Md squared over 6 plus Md squared over 2 equals fraction numerator 4 Md squared over denominator 6 end fraction end style
 
Therefore, the angular momentum after hitting is
 
begin mathsize 14px style straight L subscript final equals straight I subscript straight B straight omega equals fraction numerator 4 Md squared over denominator 6 end fraction straight omega Therefore comma space applying space the space law space of space conservation space of space angular space momentum space we space get Mvd over 2 equals fraction numerator 4 Md squared over denominator 6 end fraction straight omega therefore straight omega equals Mvd over 2 cross times fraction numerator 6 over denominator 4 Md squared end fraction equals 3 over 4 straight v over straight d end style
Answered by Romal Bhansali | 02 Feb, 2016, 06:16: PM

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