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anonymous
 5 years ago
A cubical block of side "a" is moving with velocity v on a horizontal smooth plane. It hits a ridge at point O. Find the angular speed of the block after it hits the point O. Please view the attached image for more details
anonymous
 5 years ago
A cubical block of side "a" is moving with velocity v on a horizontal smooth plane. It hits a ridge at point O. Find the angular speed of the block after it hits the point O. Please view the attached image for more details

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This is the attached diagram

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So this thing is just heading towards O which then acts as a point of rotation?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay...I just to do something, then I'll take a look.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0*just go to do something*

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, I will be waiting

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm thinking you can attack the problem by considering conservation of energy, assuming only internal, conservative forces are acting.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You have two regimes  translational (before the ridge), then rotational.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The kinetic energy of the block before hand is \[K_i=\frac{1}{2}Mv^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The kinetic energy of the block just after impact, as it enters the rotational regime, is\[K_f=\frac{1}{2}I \omega^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is this making sense so far?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If I am not mistaken then the angular speed is coming to be (v/a)(sqrt6)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now, your problem is now about finding the moment of inertia for a block whose axis of rotation is about one of its edges. A solid, rectangular parallelpiped is given by...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I get something close, but different.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh I am sorry, I calculated the I around the Cm

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So how do you calculate it around that corner?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Parallel Axis Theorem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh yes, you are right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The new moment of inertia will be\[I=I_{cm}+Mh^2\] where h is the distance from the CM axis to the new, parallel axis.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[I_{cm}=\frac{Ma^2}{6}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[Mh^2=M \left( \frac{a}{2} \right)^2=M \frac{a^2}{4}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think it is \[1/6(Ma^2)+2M\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So \[I=\frac{5Ma^2}{12}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You see the corner wise distance will be sqrt 2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I mean sqrt 2 times a

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[w^2=\frac{12v^2}{5a^2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, you're right there.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So change h to a.sqrt(2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[h=\frac{a}{\sqrt{2}}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So I am supposed to solve the equation \[0.5 Mv^2=0.5[(1/6)Ma^2+2M]\omega\] Am I right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think the moment of inertia should be\[I=\frac{Ma^2}{6}+Mh^2=\frac{Ma^2}{6}+M \left( \frac{a}{\sqrt{2}} \right)^2=\frac{Ma^2}{6}+\frac{Ma^2}{2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[=\frac{4Ma^2}{6}=\frac{2Ma^2}{3}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Then\[\frac{1}{2}Mv^2=\frac{1}{2}\frac{2Ma^2}{3} \omega ^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i.e.\[v^2=\frac{2a^2}{3} \omega ^2 \rightarrow \omega ^2 = \frac{3v^2}{2a^2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Are you agreeing? I'm doing this onthefly, so if you pick up anything, interrupt.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Any how I have got the thing

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Did your answer agree?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Lokisan, I have got what I expected, I can handle the rest

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There was some problem with the site, and some of our converstation got split up

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol, thanks for the genius comment. we'll just see how this question pans out.
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