Finding the Speed of a Wedge and Rod System

[Satvik Pandey]

How did you use I=ml^{2}/12 in the above expression ?

That was my mistake .It should be l(length of rod).Sorry:shy:

 

[Satvik Pandey]

Velocity of A perpendicular to the contact surface=[5V-\frac{ω}{4}]cos150 \hat{i}-\frac{√3 ω}{2}cos60\hat{j}
According to wedge constraint the velocities of wedge and rod perpendicular to the to the contact surface should be same.
So,
Vcos30=-\frac{√3}{2}[5V-\frac{ω}{4}]\hat{i}-\frac{√3 ω}{4}\hat{j}
Is it right?

 

[Tanya Sharma]

Velocity of A perpendicular to the contact surface=[5V-\frac{ω}{4}]cos150 \hat{i}-\frac{√3 ω}{2}cos60\hat{j}
According to wedge constraint the velocities of wedge and rod perpendicular to the to the contact surface should be same.
So,
Vcos30=-\frac{√3}{2}[5V-\frac{ω}{4}]\hat{i}-\frac{√3 ω}{4}\hat{j}
Is it right?

This is incorrect .The last equation doesn't even make any sense.

 

[Satvik Pandey]

I tried to use dot product approach in finding the component of V_{A} perpendicular to the surface.
Unit vector perpendicular to the surface \vec{U}=-\frac{√3}{2}\hat{i}-\frac{1}{2}\hat{j}.


\vec{A}=[5V-\frac{ω}{4}]\hat{i}-\frac{√3 ω}{2}\hat{j}

Projection of A on U =\vec{A}.\vec{U}/U

={[5V-\frac{ω}{4}]\hat{i}-\frac{√3 ω}{2}\hat{j}}.{-\frac{√3}{2}\hat{i}-\frac{1}{2}\hat{j}}/1


=\frac{-5√3V}{2}+\frac{3√3}{8}
But this is equal to Vcos(30)
So -20V+3ω=4V

Hence ω=8V.
Now
V_{C}=\frac{MV}{m}\hat{i}-\frac{lωcosθ}{2} \hat{j}

V_{c}=√37 V

As
(mgl/2)(sin60° – sin30°) = (1/2)MV²+ (1/2)mv²+ (1/2)Iω²
On putting values-
\frac{3(√3-1)g}{2}=142V^2

V=√3(√3-1)/142

Now I think it is right. At last I found it.

 

[Satvik Pandey]

Thank you TSny for guiding me.
And thank you Tanya Sharma for posting this question.

 

[TSny]

\frac{3(√3-1)g}{2}=142V^2

GOT IT!

V=√3(√3-1)/142

What happened to g and the factor of 2 in the denominator?

 

[Satvik Pandey]

GOT IT!
What happened to g and the factor of 2 in the denominator?

Sorry I missed that.My foolish mistakes continue.
V=√15(√3-1)/142.
Are my other steps correct?
THANK YOU TSny for guiding me.You helped me a lot.

 

[Satvik Pandey]

Tanya mam I forgot to wish you "HAPPY INDEPENDENCE DAY".
Thank you.

 

[TSny]

V=√15(√3-1)/142.
Are my other steps correct?

Yes, everything looks good to me.

Good work!

 

[Satvik Pandey]

Yes, everything looks good to me.

Good work!

It couldn't have been possible without your help.

 

[FeynmanIsCool]

Hey Guys,
I'm a little confused. Why would some of the P.E from the center of mass of the rod be converted to Rotational Kinetic energy? The the rod is not rotating. I see the idea that as it slips down its "rotating" about the center of mass, but is it correct to imply that in our reference frame its actually rotating. If this was the case, it seems like all problems of this sort would involve a rotational kinetic energy. If a ladder was perched up on a wall and fell over, you wouldn't assume that some of the P.E is converted into rotational K.E would you (assuming the end on the ground was stationary)? Or a rod was sliding down a parabola shaped ramp? I just have a weird feeling about it. Would any of you guys care to explain?

 

[Orodruin]

You would have to take rotation into account in all of the examples you described. Also if you consider a ball rolling down a slope you would have to take rotation into account. The point is that you cannot just describe things based on how the CoM moves, this will in general not give you the full kinetic energy.

Of course you could consider the movement of every single mass point in an object and how it moves (including any motion from rotation), but you can just as well split the energy cleverly into rotation and translation and end up with a simpler problem.