Finding the Speed of a Wedge and Rod System

[Satvik Pandey]

The axis of rotation is perpendicular to the page.


If you mean exactly where is the axis located? I think there are several correct ways to look at it.

For example, you could look at it as rotation about the (rod's) center of mass, (although the center of mass moving vertically and horizontally)

Another way is to look at it as rotation about the end of the rod that is touching the floor
(I think this way is more simple, because this way the rotation axis is only moving horizontally, not vertically)


There are other ways too, and if I'm not mistaken, all of them are valid (although some are more simple than others)

Thank you Nathanael.
This is the case in which rod is going through translation and rotational motion simultaneously.
Is it so?

 

[Satvik Pandey]

Thanks for the response ,TSny .

Component of velocity of the upper end of the rod perpendicular to the hypotenuse face = Vcos30° .
.

I am unable to understand why velocity of the upper end of the rod perpendicular to the hypotenuse face = Vcos30° .It would be nice if you could help.
From this figure

h^{2}= p^{2}+ b^{2}
differentiating the above equation with respect to time we get
0=2p\frac{dp}{dt}+ 2b \frac{db}{dt}
0=2pv'+2bv
-\frac{b}{p}v=v'
-cot θv=v'
Component of this velocity perpendicular to h=v'cosθ
=-cotθvcosθ but this is not what you have found.

 

[TSny]

I am unable to understand why velocity of the upper end of the rod perpendicular to the hypotenuse face = Vcos30°.

See the attached figure. The points A and A’ represent the initial and final locations of the upper end of the rod as the wedge moves horizontally to the left by a small distance d. The distance s is the distance that the upper end of the rod moves perpendicularly to the hypotenuse of the wedge. Using trig on triangle abc, what is the relationship between s and d?

 

[Satvik Pandey]

See the attached figure. The points A and A’ represent the initial and final locations of the upper end of the rod as the wedge moves horizontally to the left by a small distance d. The distance s is the distance that the upper end of the rod moves perpendicularly to the hypotenuse of the wedge. Using trig on triangle abc, what is the relationship between s and d?

Is it \frac{s}{d}=cos(30)

 

[TSny]

Is it \frac{s}{d}=cos(30)

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.

 

[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}?










-

 

[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.

 

[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 ?