anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
This is the attached diagram
1 Attachment
anonymous
  • anonymous
So this thing is just heading towards O which then acts as a point of rotation?
anonymous
  • anonymous
Yes

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anonymous
  • anonymous
Okay...I just to do something, then I'll take a look.
anonymous
  • anonymous
*just go to do something*
anonymous
  • anonymous
Okay, I will be waiting
anonymous
  • anonymous
Still waiting?
anonymous
  • anonymous
Yes
anonymous
  • anonymous
I'm thinking you can attack the problem by considering conservation of energy, assuming only internal, conservative forces are acting.
anonymous
  • anonymous
You have two regimes - translational (before the ridge), then rotational.
anonymous
  • anonymous
The kinetic energy of the block before hand is \[K_i=\frac{1}{2}Mv^2\]
anonymous
  • anonymous
The kinetic energy of the block just after impact, as it enters the rotational regime, is\[K_f=\frac{1}{2}I \omega^2\]
anonymous
  • anonymous
Is this making sense so far?
anonymous
  • anonymous
Yes
anonymous
  • anonymous
If I am not mistaken then the angular speed is coming to be (v/a)(sqrt6)
anonymous
  • anonymous
Now, 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
  • anonymous
I get something close, but different.
anonymous
  • anonymous
Oh I am sorry, I calculated the I around the Cm
anonymous
  • anonymous
Which is wrong
anonymous
  • anonymous
Yes
anonymous
  • anonymous
So how do you calculate it around that corner?
anonymous
  • anonymous
Parallel Axis Theorem
anonymous
  • anonymous
Oh yes, you are right
anonymous
  • anonymous
The 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
  • anonymous
\[I_{cm}=\frac{Ma^2}{6}\]
anonymous
  • anonymous
\[Mh^2=M \left( \frac{a}{2} \right)^2=M \frac{a^2}{4}\]
anonymous
  • anonymous
I think it is \[1/6(Ma^2)+2M\]
anonymous
  • anonymous
So \[I=\frac{5Ma^2}{12}\]
anonymous
  • anonymous
You see the corner wise distance will be sqrt 2
anonymous
  • anonymous
I mean sqrt 2 times a
anonymous
  • anonymous
\[w^2=\frac{12v^2}{5a^2}\]
anonymous
  • anonymous
Yes, you're right there.
anonymous
  • anonymous
So change h to a.sqrt(2)
anonymous
  • anonymous
a/sqrt(2)
anonymous
  • anonymous
\[h=\frac{a}{\sqrt{2}}\]
anonymous
  • anonymous
So I am supposed to solve the equation \[0.5 Mv^2=0.5[(1/6)Ma^2+2M]\omega\] Am I right
anonymous
  • anonymous
am i right?
anonymous
  • anonymous
I 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
  • anonymous
\[=\frac{4Ma^2}{6}=\frac{2Ma^2}{3}\]
anonymous
  • anonymous
Then\[\frac{1}{2}Mv^2=\frac{1}{2}\frac{2Ma^2}{3} \omega ^2\]
anonymous
  • anonymous
i.e.\[v^2=\frac{2a^2}{3} \omega ^2 \rightarrow \omega ^2 = \frac{3v^2}{2a^2}\]
anonymous
  • anonymous
Ok I understand that
anonymous
  • anonymous
Are you agreeing? I'm doing this on-the-fly, so if you pick up anything, interrupt.
anonymous
  • anonymous
Any how I have got the thing
anonymous
  • anonymous
Did your answer agree?
anonymous
  • anonymous
Lokisan, I have got what I expected, I can handle the rest
anonymous
  • anonymous
There was some problem with the site, and some of our converstation got split up
anonymous
  • anonymous
Okay. Happy physics.
anonymous
  • anonymous
lol, thanks for the genius comment. we'll just see how this question pans out.

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