Finding the Speed of a Wedge and Rod System

[Satvik Pandey]

Looks good.

What to do next?

 

[TSny]

V_A=V_C+V_A/C

V_A={\frac{MV}{m}-\frac{-lωsinθ}{2}}\hat{i} -lωcosθ/2\hat{j}.
Is it right?
What to do next?

Looks like you have a sign error in the x component. For the y component of V_A did you include both the y component of V_C and the y component of V_A/C?

 

[Satvik Pandey]

Looks like you have a sign error in the x component. For the y component of V_A did you include both the y component of V_C and the y component of V_A/C?

{\frac{MV}{m}-\frac{lωsinθ}{2}}\hat{i} -lωcosθ\hat{j}.
How is it?
What to do next?
THANK YOU for guiding me.

 

[TSny]

{\frac{MV}{m}-\frac{lωsinθ}{2}}\hat{i} -lωcosθ\hat{j}.

That looks right.

Note that if you can find a relation between ω and V, then you will be able to express the energy equation in terms of just V. So, you will be able to solve for V.

To get a relation between ω and V, you need to use a constraint condition that you have not yet used.

To avoid messy expressions, you might want to go ahead and substitute the known values for m, M, l, and θ and simplify as much as possible your expressions for V_C and V_A.

 

[Satvik Pandey]

That looks right.

Note that if you can find a relation between ω and V, then you will be able to express the energy equation in terms of just V. So, you will be able to solve for V.

To get a relation between ω and V, you need to use a constraint condition that you have not yet used.

To avoid messy expressions, you might want to go ahead and substitute the known values for m, M, l, and θ and simplify as much as possible your expressions for V_C and V_A.

On putting M=5kg,m=1kg,l=1m, θ=30 and I=ml^{2}/12.
I got

[5V-\frac{ω}{48}]\hat{i}-\frac{√3 ω}{24}\hat{j}

According to wedge constraint the velocities of wedge and rod perpendicular to the to the contact surface should be same.

Velocity of wedge perpendicular to the contact surface is Vcos30
Velocity of A perpendicular to the contact surface=[5V-\frac{ω}{48}]cos150 \hat{i}-\frac{√3 ω}{24}cos60\hat{j}


Vcos30=-\frac{√3}{2}[5V-\frac{ω}{48}]\hat{i}-\frac{√3 ω}{48}\hat{j}

.

Is it right?

 

[Tanya Sharma]

@Satvik :Which standard/class are you in ?

 

[Satvik Pandey]

@Satvik :Which standard/class are you in ?

I am in class 10 now.
I think the foolish mistakes which I have made in this thread might have urged you to ask this.

 

[Tanya Sharma]

On putting M=5kg,m=1kg,l=1m, θ=30 and I=ml^{2}/12.
I got

[5V-\frac{ω}{48}]\hat{i}-\frac{√3 ω}{24}\hat{j}

According to wedge constraint the velocities of wedge and rod perpendicular to the to the contact surface should be same.

Velocity of wedge perpendicular to the contact surface is Vcos30
Velocity of A perpendicular to the contact surface=[5V-\frac{ω}{48}]cos150 \hat{i}-\frac{√3 ω}{24}cos60\hat{j}

Could you explain the above expressions ?

 

[Satvik Pandey]

Could you explain the above expressions ?

I tried to find the velocity of upper part(part A) of rod perpendicular to the contact surface.
I resolved the rectangular component of V_A in the direction perpendicular to the contact surface and then I wrote them together.
Is it right?

 

[Tanya Sharma]

On putting M=5kg,m=1kg,l=1m, θ=30 and I=ml^{2}/12.
I got

[5V-\frac{ω}{48}]\hat{i}-\frac{√3 ω}{24}\hat{j}

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

 

[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.