Classical mechanics problem

Note by Azimuddin Sheikh
8 months, 1 week ago

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Here is my solution

Initial assumptions:

1) The initial speed of the block is (3,0)(3,0) with respect to the ground
2) The initial position of the block is (0,0)(0,0) with respect to the ground
2) The speed of the platform is (0,4)(0,4) with respect to the ground
3) The friction force has a magnitude equal to the μmg\mu \, m \, g, and a direction opposite to the relative velocity of the block with respect to the platform. Both of these velocities are given with reference to ground. In other words:

Ffriction(vblockvplatform)\vec{F}_{friction} \propto \, -(\vec{v}_{block} - \vec{v}_{platform})

Running the simulation over time results in the block coming to rest relative to the platform after 1.6661.666 seconds, at which time the xx coordinate of the block is 2.52.5 with respect to ground, and the yy coordinate is 3.3333.333 with respect to ground. The block is a distance 4.16664.1666 from the origin, and it has moved a total arc length of 4.7534.753 with respect to ground.

The (x,y)(x,y) trajectory of the block relative to ground is shown below:

Steven Chase - 8 months, 1 week ago

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Sir initial speed of block should be wrt platform . As such in ground frame it has initial speed to be net 5

Azimuddin Sheikh - 8 months, 1 week ago

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If that is the case, my last paragraph changes as follows:

Running the simulation over time results in the block coming to rest relative to the platform after 11 second, at which time the xx coordinate of the block is 1.51.5 with respect to ground, and the yy coordinate is 44 with respect to ground. The block is a distance 4.2724.272 from the origin, and it has moved a total arc length of 4.3484.348 with respect to ground.

Steven Chase - 8 months, 1 week ago

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@Steven Chase Sir but my energy method I am getting 7/6m which is actually the answer given sir.

Azimuddin Sheikh - 8 months, 1 week ago

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@Azimuddin Sheikh Perhaps you can post your solution

Steven Chase - 8 months, 1 week ago

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@Steven Chase This is by energy method sir . Sir I have posted above with the question.

Azimuddin Sheikh - 8 months, 1 week ago

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@Azimuddin Sheikh @Steven Chase Sir I have posted my energy method above with the question.

Azimuddin Sheikh - 8 months, 1 week ago

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@Azimuddin Sheikh @Steven Chase Sir can u pls tell.is there anything wrong I did ?

Azimuddin Sheikh - 8 months, 1 week ago

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@Azimuddin Sheikh I wanted to check my original solution, since it was a bit involved. So then I remembered the first postulate of special relativity: "The laws of physics are the same in all inertial frames of reference". Since the platform is not accelerating, an observer sitting on the platform and watching should observe "ordinary" physics. And we can apply simple kinematics to that case. Let v0v_0 be the initial velocity of the block relative to the platform. There is a constant acceleration relative to the platform equal to μg \mu \, g .

v0μg=tftf=1\frac{v_0}{\mu \, g} = t_f \\ t_f = 1

Then the distance the block slides relative to the platform is:

D=12μgtf2=32D = \frac{1}{2} \, \mu \, g \, t_f^2 = \frac{3}{2}

And over the one second, the platform has moved 4 meters. So the final position of the block with respect to the ground is (32,4) (\frac{3}{2}, 4). This matches the more complicated approach I took earlier.

Then, to find the total arc length traveled with respect to ground, begin with the expressions for the velocity relative to ground.

(vx,vy)=(33t,4)v=2518t+9t2(v_x,v_y) = (3 - 3 t, 4) \\ v = \sqrt{25 - 18t + 9 t^2}

The total arc length is:

L=012518t+9t2dt=4.3484L = \int_0^1 \sqrt{25 - 18t + 9 t^2} \, dt = 4.3484

This also matches the result of my previous solution

Steven Chase - 8 months, 1 week ago

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@Steven Chase sir please help with this problem discussion on a David Morin Question.

Harsh Poonia - 1 month, 2 weeks ago

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Can it be solved without energy method ? @Steven Chase sir @Aaghaz Mahajan bro ??

Azimuddin Sheikh - 8 months, 1 week ago

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Yes........I mean, I solved it using relative velocity.......View the motion of the block with respect to the horizontal platform........The time after which it stops is 1.66666 seconds.....I think you can proceed from here onwards....:)

Aaghaz Mahajan - 8 months, 1 week ago

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@Aaghaz Mahajan bro but then from that frame answer I am getting is wrong , can u tell what r u getting ?? Its difficult for me to get answer using relative frame I used that. .

Azimuddin Sheikh - 8 months, 1 week ago

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@Azimuddin Sheikh Well, what did you try??? What is your answer??

Aaghaz Mahajan - 8 months, 1 week ago

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@Aaghaz Mahajan I am getting answer as 7/6 m from energy method , but not from this . Can u show by relative frame @Aaghaz Mahajan bro

Azimuddin Sheikh - 8 months, 1 week ago

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@Azimuddin Sheikh Well, what is the correct answer??? I am getting the answer to be 5 m.........Here is my method:- Taking the velocity of block wrt to the platform, we get it to be <3,-4> i.e. v = 5 m/s. Also, the acceleration is -3 m/s^2. So, the time taken is 1.66666666 seconds. Now, the displacement of the block in that time is <5,-20/3> and the displacement of the platform is <0,4>. Using relative velocity again, we get the displacement of the block wrt ground to be <5,0>......

Aaghaz Mahajan - 8 months, 1 week ago

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@Aaghaz Mahajan Answer is 7/6 m only @Aaghaz Mahajan bro. We need total distance not displacement

Azimuddin Sheikh - 8 months, 1 week ago

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@Azimuddin Sheikh @Steven Chase Sir u r method is correct , but what's incorrect in my method (energy)??

Azimuddin Sheikh - 8 months ago

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@Azimuddin Sheikh I have several comments:

1) I interpret the "4" in your energy expression as being the final velocity with respect to ground, which is the same velocity as the platform. It's important to note that it is relative to ground.

2) The "3" in your energy expression is relative to the platform (per your previous clarification). So you have two different references in your energy equation, which seems problematic.

3) The other concern is that the work calculation has the following general form:

W=Fd W = \int \vec{F} \cdot \vec{d \ell}

If the force magnitude is constant AND the force is everywhere parallel to the motion, this reduces to:

W=FΔ W = |\vec{F}| \, \Delta \ell

How do you know that these simplifications should occur?

Steven Chase - 8 months ago

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@Steven Chase Seems like 1 is causing trouble to my energy method , will try again

Azimuddin Sheikh - 8 months ago

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Which dimension is gravity in (i, j, or k)?

Steven Chase - 8 months, 1 week ago

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Sir, it is written that x-y plane is a horizontal plane, so I think g should be in k.....

Aaghaz Mahajan - 8 months, 1 week ago

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@Aaghaz Mahajan Ok, thanks

Steven Chase - 8 months, 1 week ago

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@Steven Chase @Steven Chase sir u got the answer? By other method?

Azimuddin Sheikh - 8 months, 1 week ago

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@Azimuddin Sheikh I have posted my solution

Steven Chase - 8 months, 1 week ago

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vb=3i,a=3(3i+4j)5a=3,vb/p=5    t0=5/3s.\vec v_b=3 \vec{i}, \vec{a}= \dfrac{3(-3 \vec i +4 \vec j)}{5}\, |\vec a|=3, |\vec v_{b/p}|=5 \implies t_0=5/3 s. (Using formula 05=3t00-5=-3t_0) For block, sb=3t0i+(1/2)at02\vec{s}_b=3t_0 \vec i + (1/2) \vec{a} {t_0}^2 Use this to get s=5i+103j\vec{s}=5 \vec i + \dfrac{10}{3} \vec j Note: vbv_b is velocity of block in ground frame, vb/pv_{b/p} is velocity of block in platform frame.

Harsh Poonia - 1 month, 3 weeks ago

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