Finding the Speed of a Wedge and Rod System

[Satvik Pandey]

Yes. From that you should be able to show that the velocity of the upper end of the rod has a component perpendicular to the wedge hypotenuse equal to Vcos(30), where V is the velocity of the wedge.

So s=dcos(30)

$\frac{ds}{dt}$=$\frac{ddcos(30)}{dt}$
As θ is not going to vary with time.
So $\frac{ds}{dt}$=$\frac{dd}{dt}$cos(30)

$\frac{ds}{dt}$=Vcos(30).

Thank you TSny.

 

[Satvik Pandey]

$\hat{i}$ to the right and $\hat{j}$ upwards .

$\vec{V}_C = \frac{M}{m}V\hat{i} + \frac{l}{2}ωcosθ\hat{j}$

$\vec{V}_{A/C} = \frac{l}{2}ωsinθ\hat{i} + \frac{l}{2}ωcosθ\hat{j}$

I was unable to understand how you find V$_{C}$and V$_{A/C}$
I tried to understand your solution but it is a bit difficult.I tried this.
Let the rod rotate along CM
So,Vcosθ=lω/2
velocity of this component in y direction=lωcosθ/2
So V$_{A}$=-MV/m$\hat{i}$-lωcosθ$\hat{j}$
Is it right?I also have some sign error. I am unable to find V$_{A/C}$?










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[Satvik Pandey]

I was unable to understand how you find V$_{C}$and V$_{A/C}$
I tried to understand your solution but it is a bit difficult.I tried this.
Let the rod rotate along CM
So,Vcosθ=lω/2
velocity of this component in y direction=lωcosθ/2
So V$_{A}$=-MV/m$\hat{i}$-lωcosθ$\hat{j}$
Is it right?I also have some sign error. I am unable to find V$_{A/C}$?-

Here is the figure.

 

[TSny]

I was unable to understand how you find V$_{C}$and V$_{A/C}$
I tried to understand your solution but it is a bit difficult.I tried this.
Let the rod rotate along CM
So,Vcosθ=lω/2
velocity of this component in y direction=lωcosθ/2
So V$_{A}$=-MV/m$\hat{i}$-lωcosθ$\hat{j}$
Is it right?I also have some sign error. I am unable to find V$_{A/C}$?

Your y component of V$_{A}$ looks correct if ω represents the magnitude of the angular velocity (i.e., a positive number). However, the x-component is not correct. The equation Vcosθ=lω/2 is also not correct.

We have the relative velocity formulas $$\vec{V}_A = \vec{V}_C + \vec{V}_{A/C}$$ $$\vec{V}_B = \vec{V}_C + \vec{V}_{B/C}$$ where A is the upper end of the rod, B is the lower end of the rod, and C is the center of mass of the rod. See the attached figure.

Can you express the magnitudes of $\vec{V}_{A/C}$ and $\vec{V}_{B/C}$ in terms of $\omega$?

 

[Satvik Pandey]

Your y component of V$_{A}$ looks correct if ω represents the magnitude of the angular velocity (i.e., a positive number). However, the x-component is not correct. The equation Vcosθ=lω/2 is also not correct.

We have the relative velocity formulas $$\vec{V}_A = \vec{V}_C + \vec{V}_{A/C}$$ $$\vec{V}_B = \vec{V}_C + \vec{V}_{B/C}$$ where A is the upper end of the rod, B is the lower end of the rod, and C is the center of mass of the rod. See the attached figure.

Can you express the magnitudes of $\vec{V}_{A/C}$ and $\vec{V}_{B/C}$ in terms of $\omega$?

If the rod rotates around the CM of the rod then by v=rω
V$_{A/C}$=lω/2 (Considering CCW as +ve)
V$_{B/C}$=-lω/2
But how can we say that directions of V_A/c and V_B/C are perpendicular to the rod(from figure in#39) ?
It seems to be correct but are there any proofs.

 

[TSny]

Relative to C, all the points of the rod move in circles around C. The velocity of a point moving in a circle is tangent to the circle, and therefore perpendicular to the radius of the circle.

Yes, the magnitudes of $V_{A/C}$ and $V_{B/C}$ are each equal to $l\omega/2$.

Can you express the x and y components of $\vec{V}_{A/C}$ and $\vec{V}_{B/C}$ in terms of $l$, $\omega$ and $\theta$?

 

[Satvik Pandey]

Relative to C, all the points of the rod move in circles around C. The velocity of a point moving in a circle is tangent to the circle, and therefore perpendicular to the radius of the circle.

OK got it.
THANK YOU FOR HELP.

 

[Satvik Pandey]

Relative to C, all the points of the rod move in circles around C. The velocity of a point moving in a circle is tangent to the circle, and therefore perpendicular to the radius of the circle.

Yes, the magnitudes of $V_{A/C}$ and $V_{B/C}$ are each equal to $l\omega/2$.

Can you express the x and y components of $\vec{V}_{A/C}$ and $\vec{V}_{B/C}$ in terms of $l$, $\omega$ and $\theta$?

V_A/C=-lωsinθ/2 $\hat{i}$-lωcosθ/2 $\hat{j}$
V_B/C=-lωsinθ/2 $\hat{i}$-lωcosθ/2 $\hat{j}$

i hat +ve towards right and j hat +ve in upward direction.

 

[TSny]

OK for $\vec{V}_{A/C}$, but the signs are wrong for $\vec{V}_{B/C}$. (I'm assuming $\omega$ is a positive number.)

 

[Satvik Pandey]

OK for $\vec{V}_{A/C}$, but the signs are wrong for $\vec{V}_{B/C}$. (I'm assuming $\omega$ is a positive number.)

Yes there should be +ve sign instead of -ve sign.

V$_{B/C}$=lωsinθ/2 $\hat{i}+lωcosθ/2\hat{j}$

 

[Satvik Pandey]

I tried to find V_C.
As V_C/B=V_C-V_B
V$_{C}$=V$_{C/B}$+V$_{B}$

V$_{C/B}$=-lωsinθ/2$\hat{i}$ -lωcosθ/2$\hat{j}$.V$_{B}$=MV/m $\hat{i}$
Comb. these

V$_{C}$=-lωsinθ/2$\hat{i}$ -lωcosθ/2$\hat{j}$+MV/m $\hat{i}$
Is it right?

I have considered ω here w.r.t axis passing through B.
Will it be same as ω w.r.t axis passing through C which I have used earlier?

 

[TSny]

V$_{B}$=MV/m $\hat{i}$

This equation isn't correct.

The idea is to use the energy equation to solve for the speed of the wedge, V. The energy equation contains V as well as the speed of the center of mass of the rod, V_C, and the angular speed of the rod, ω. So, we need to express V_C and ω in terms of V.

Using conservation of momentum in the x-direction, what can you deduce about the x component of $\vec{V}_C$?

To get some information about the y-component of $\vec{V}_C$, what do you get if you take the y-component of the equation $\vec{V}_B = \vec{V}_C + \vec{V}_{B/C}$?

 

[Satvik Pandey]

This equation isn't correct.

The idea is to use the energy equation to solve for the speed of the wedge, V. The energy equation contains V as well as the speed of the center of mass of the rod, V_C, and the angular speed of the rod, ω. So, we need to express V_C and ω in terms of V.

Using conservation of momentum in the x-direction, what can you deduce about the x component of $\vec{V}_C$?

To get some information about the y-component of $\vec{V}_C$, what do you get if you take the y-component of the equation $\vec{V}_B = \vec{V}_C + \vec{V}_{B/C}$?

THANK YOU TSny.
By conservation of momentum where v$_{x}$ is the velocity of CM (x-comp.)
MV=mv$_{x}$

v$_{x}$=MV/m

$\vec{V}_B = \vec{V}_C + \vec{V}_{B/C}$
0=V$_{C}$+$\frac{lωcosθ}{2}$ $\hat{j}$

V$_{C}$=-$\frac{lωcosθ}{2}$ $\hat{j}$

V$_{C}$=$\frac{MV}{m}$$\hat{i}$-$\frac{lωcosθ}{2}$ $\hat{j}$
Is it right?

 

[Satvik Pandey]

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?

 

[TSny]

THANK YOU TSny.
By conservation of momentum where v$_{x}$ is the velocity of CM (x-comp.)
MV=mv$_{x}$

v$_{x}$=MV/m

$\vec{V}_B = \vec{V}_C + \vec{V}_{B/C}$
0=V$_{C}$+$\frac{lωcosθ}{2}$ $\hat{j}$

V$_{C}$=-$\frac{lωcosθ}{2}$ $\hat{j}$

V$_{C}$=$\frac{MV}{m}$$\hat{i}$-$\frac{lωcosθ}{2}$ $\hat{j}$
Is it right?

Looks good.

 

[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.:rofl:

 

[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.:rofl:

 

[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.