Lagrange equation for block and incline

[member 731016]

Homework Statement

Please see below

Relevant Equations

Please see below

For this problem,

Does someone please know where the term highlighted in blue came from?

Thanks!

 

[erobz]

Homework Statement: Please see below
Relevant Equations: Please see below

For this problem,
View attachment 344125
Does someone please know where the term highlighted in blue came from?

Thanks!

Write the kinetic energy of the block in the inertial frame. $\dot y$ is w.r.t. the ( accelerating) wedge.

 

[member 731016]

Write the kinetic energy of the block in the inertial frame. $\dot y$ is w.r.t. the ( accelerating) wedge.

Thank you for your reply @erobz!

$T = \frac{1m}{2}(\dot x)^2 + \frac{1m}{2}(\dot y)^2$

Is the please correct?

THanks!

 

[Orodruin]

Thank you for your reply @erobz!

$T = 1/2m\dotx + 1/2m\doty$

Is the please correct?

THanks!

No, it is incorrect (even including if you add the missing squares). Note that the x and y directions of motion for the block are not orthogonal.

 

[erobz]

Thank you for your reply @erobz!

$T = \frac{1m}{2}(\dot x)^2 + \frac{1m}{2}(\dot y)^2$

Is the please correct?

THanks!

No. The wedge is moving with velocity $\dot x$, the block is riding on the wedge. $\dot y$ is with respect to the wedge, the wedge is not an inertial frame.

 

[member 731016]

No, it is incorrect (even including if you add the missing squares). Note that the x and y directions of motion for the block are not orthogonal.

Thank you for your reply @Orodruin!

Sorry I am confused. I know the axes are not orthongal, put that does not make a difference does it (since the velocity is both along the axes)?

Thanks!

 

[erobz]

Thank you for your reply @Orodruin!

Sorry I am confused. I know the axes are not orthongal, put that does not make a difference does it (since the velocity is both along the axes)?

Thanks!

Pretend you are standing in the ground watching this unfold. What velocity do you measure for the block in the horizontal and vertical directions? Remember $\dot y$ is being measured with respect to the wedge.

 

[Orodruin]

put that does not make a difference does it

It most certainly does. Imagine for example that they were parallel. What would the speed be?

 

[kuruman]

Does someone please know where the term highlighted in blue came from?

The drawing is misleading. It shows two arrows labeled $\dot x$, one on the wedge and another to the right of the sliding block. The first arrow clearly represents the horizontal velocity of the wedge relative to the ground. The second arrow is meant to be the horizontal velocity of the sliding block relative to the ground. It cannot be given the same symbol because it is the horizontal velocity of the sliding block relative to the wedge plus the horizontal velocity of the wedge relative to the ground $\dot x$. The highlighted term is the cross term that arises when you square the correct expression for the horizontal velocity of the block relative to the ground.

 

[member 731016]

Pretend you are standing in the ground watching this unfold. What velocity do you measure for the block in the horizontal and vertical directions? Remember $\dot y$ is being measured with respect to the wedge.

It most certainly does. Imagine for example that they were parallel. What would the speed be?

The drawing is misleading. It shows two arrows labeled $\dot x$, one on the wedge and another to the right of the sliding block. The first arrow clearly represents the horizontal velocity of the wedge relative to the ground. The second arrow is meant to be the horizontal velocity of the sliding block relative to the ground. It cannot be given the same symbol because it is the horizontal velocity of the sliding block relative to the wedge plus the horizontal velocity of the wedge relative to the ground $\dot x$. The highlighted term is the cross term that arises when you square the correct expression for the horizontal velocity of the block relative to the ground.

Thank you for your replies @erobz , @Orodruin and @kuruman!


I had another think about this and since the time derivative vectors of x and y, they form a non-right angle triangle, so I must use the cosine rule I think to find the correct magnitude of the velocity where I think the extra term comes from. Thanks!

 

[erobz]

Thank you for your replies @erobz , @Orodruin and @kuruman!


I had another think about this and since the time derivative vectors of x and y, they form a non-right angle triangle, so I must use the cosine rule I think to find the correct magnitude of the velocity where I think the extra term comes from. Thanks!

I doesn’t sound to me like you understand the crux of the problem yet.

Let’s start with the straightforward part. What is the vertical component of the blocks velocity in the ground frame?

 

[Orodruin]

I doesn’t sound to me like you understand the crux of the problem yet.

Let’s start with the straightforward part. What is the vertical component of the blocks velocity in the ground frame?

The cosine rule is perfectly fine to use for addition of velocities here.

 

[Steve4Physics]

To add to what @Orodruin just said (and since I’d already done a diagram!), vector-addition of the 2 velocities using the cosine rule is a simple approach:

Remembering that $\cos(\alpha) = -\cos(180^o - \alpha)$.

 

[erobz]

The cosine rule is perfectly fine to use for addition of velocities here.

Ok, I wasn’t familiar with it by name.