Question based on physical mech

Диск радиусом R вращается против часовой стрелки. А — точка на диске. B — неподвижный наблюдатель в плоскости диска. Если скорость А с точки зрения В равна 4 м/с, то какова скорость В относительно А. OB = 3m.

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I am getting answer as -4 m/s in downward direction… is it correct?

I don’t think so, because you didn’t use the radius of the disc and the distance from the center to B. I think that you also take into account the motion of the center of the disc.

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How will we do it then? Could you explain please.

What answer are you getting?

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You can divide the motion of any point on the disk into the motion of that point relative to the center of the disk and the motion of the center:

\overrightarrow {u_{rel}} - velocity vector of point P relative to the center
\overrightarrow {v_c} - velocity vector of the center
\overrightarrow {v_{abs}} - velocity vector of the point P for the observer on the ground
These velocities are connected with vector equation:

\overrightarrow {v_{abs}}=\overrightarrow {u_{rel}}+\overrightarrow {v_{abs}}

In this problem

u_{rel}=\omega R=v_c, v_{abs}=v_A
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I did the same thing but my answer is coming out to be 4. What did you get?

The answer is not 4, should i give the solution or you want to do solve it by yourself?

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please give the solution

The angular frequency:


(which is obvious)
In this problem, we probably must find the velocity of point B with respect to A, because we don’t know the speed of the disk’s center (in the above example I considered that the disk rolls on the ground, but in this problem we do not know anything about the disk’s center - its just rotates)
This is how B see A:

And this is how A see B:

I think that (I’m not quite sure, so @Alisher must check this) B rotates in the opposite direction (clockwise) and with the same frequency as A. It simply rotates on a bigger circle with radius x.

v_B=\omega x=v_a\frac{x}{R}
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I don’t think this is correct… You have found out velocity of B with respect to centre of disc… Now, I know that with respect to all points on disc the velocity will be same that is 6 m/s… but this method is not right… the question has asked with respect to A… @Alisher please have a look !

No, i meant the answer is 6 m/s… but the method is wrong. This version doesnt yield the correct answer. We will get 7.2 m/s which is wrong. Considering radius as 2m.

Previous method yield the correct answer, so it means that also the method is correct (I don’t think its just randomness)

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Actually we might get the same answer using wrong methods as well… the previous method we found the answer with respect to the centre, not point A. But I think Alisher needs to help us out!

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In this problem we actually needn’t consider velocity of point O (because OB is given as 3\space \text{m}, so apparently this length is fixed). But the Damir’s solution (without considering this velocity) is correct. Just go into frame of reference where both O and A are motionless. It turns out that the whole Universe except the disc rotates at angular velocity -\vec \omega with respect to O, so

\vec v_B ' = -\vec\omega \times \vec x \quad \Rightarrow \quad|\vec v_B'| = v_A \frac{x}{R}.

p.s. i spent much time trying to give more rigour and general solution but then realized an easier one.

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Thank you for clarifying, master Alisher. :pray::clap:

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By the way, if B moved, it would be solved exactly the same way? The velocity of point B would be the absolute velocity, and by Galileo’s relativity, it would be possible to obtain the value of the relative velocity of the point. The absolute speed and the portable one could be combined according to Pythagoras. Right?

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it doesn’t really complicate the problem, the answer would be simply \dot{\vec x} + \vec\omega\times\vec x

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