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<meta property="og:description" content="The rod rests its ends on the sides of a right angle. The upper end of the rod is lifted at a speed of v. Find how the speed of its lower end depends on time. Take the moment when the upper end is at the top of the corner as the beginning of the time reference. Rod length L.">
<title>The rod rests its ends on the sides of a right angle. The upper end of the rod is lifted at a speed of v. Find how the speed of its lower end depends on time. Take the moment when the upper end is at the top of the corner as the beginning of the time reference. Rod length L.</title>
$1.5.16.$ The rod rests its ends on the sides of a right angle. The upper end of the rod is lifted at a speed of $v$. Find how the speed of its lower end depends on time. Take the moment when the upper end is at the top of the corner as the beginning of the time reference. Rod length $L$.
<p>NO: Before viewing the solution to this problem, I recommend that you familiarize yourself with the solution to <a href="../1.5.14">1.5.14</a> and <a href="../1.5.15">1.5.15</a></p>
<p>At time $t$, the height to which the upper end has risen</p>
<p>
$$ y = v t\;(1) $$
</p>
<p>Consider the change in coordinate $x = \sqrt{L^2-y^2}$ over a small period of time $dt$</p>
<p>
$$ dx = \frac{ydy}{\sqrt{L^2 - y^2}}$$
</p>
<p>Considering that $v = \frac{dy}{dt}$ and $v_A = \frac{dx}{dt}$</p>
<p>If you connect the center of the rod with the vertex of the right angle, then this segment will be the median of the right triangle and will be equal to half the length of the hypotenuse. It turns out that the center of the rod moves along a circle with the center at point $O$.</p>
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
<meta name="keywords" content="Savchenko Problems in Physics, Savchenko solutions, physics problems, physics olympiad preparation, IPhO, Jaan Kalda">
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<meta name="description" content="The rod rests its ends on the sides of a right angle. The upper end of the rod is lifted at a speed of v. Find how the speed of its lower end depends on time. Take the moment when the upper end is at the top of the corner as the beginning of the time reference. Rod length L.">
<meta name="description" content="The rod rests its ends on the sides of a right angle. The upper end of the rod is lifted at a speed of v. Find how the speed of its lower end depends on time. Take the moment when the upper end is at the top of the corner as the beginning of the time reference. Rod length L.">
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<meta property="og:title" content="The rod rests its ends on the sides of a right angle. The upper end of the rod is lifted at a speed of v. Find how the speed of its lower end depends on time. Take the moment when the upper end is at the top of the corner as the beginning of the time reference. Rod length L.">
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<meta property="og:description" content="The rod rests its ends on the sides of a right angle. The upper end of the rod is lifted at a speed of v. Find how the speed of its lower end depends on time. Take the moment when the upper end is at the top of the corner as the beginning of the time reference. Rod length L.">
<meta property="og:description" content="The rod rests its ends on the sides of a right angle. The upper end of the rod is lifted at a speed of v. Find how the speed of its lower end depends on time. Take the moment when the upper end is at the top of the corner as the beginning of the time reference. Rod length L.">
<title>The rod rests its ends on the sides of a right angle. The upper end of the rod is lifted at a speed of v. Find how the speed of its lower end depends on time. Take the moment when the upper end is at the top of the corner as the beginning of the time reference. Rod length L.</title>
<title>The rod rests its ends on the sides of a right angle. The upper end of the rod is lifted at a speed of v. Find how the speed of its lower end depends on time. Take the moment when the upper end is at the top of the corner as the beginning of the time reference. Rod length L.</title>
$1.5.16.$ The rod rests its ends on the sides of a right angle. The upper end of the rod is lifted at a speed of $v$. Find how the speed of its lower end depends on time. Take the moment when the upper end is at the top of the corner as the beginning of the time reference. Rod length $L$.
$1.5.16.$ The rod rests its ends on the sides of a right angle. The upper end of the rod is lifted at a speed of $v$. Find how the speed of its lower end depends on time. Take the moment when the upper end is at the top of the corner as the beginning of the time reference. Rod length $L$.
<p>NO: Before viewing the solution to this problem, I recommend that you familiarize yourself with the solution to <a href="../1.5.14">1.5.14</a> and <a href="../1.5.15">1.5.15</a></p>
<p>NO: Before viewing the solution to this problem, I recommend that you familiarize yourself with the solution to <a href="../1.5.14">1.5.14</a> and <a href="../1.5.15">1.5.15</a></p>
<p>At time $t$, the height to which the upper end has risen</p>
<p>At time $t$, the height to which the upper end has risen</p>
<p>
<p>
$$ y = v t\;(1) $$
$$ y = v t\;(1) $$
</p>
</p>
<p>Consider the change in coordinate $x = \sqrt{L^2-y^2}$ over a small period of time $dt$</p>
<p>Consider the change in coordinate $x = \sqrt{L^2-y^2}$ over a small period of time $dt$</p>
<p>
<p>
$$ dx = \frac{ydy}{\sqrt{L^2 - y^2}}$$
$$ dx = \frac{ydy}{\sqrt{L^2 - y^2}}$$
</p>
</p>
<p>Considering that $v = \frac{dy}{dt}$ and $v_A = \frac{dx}{dt}$</p>
<p>Considering that $v = \frac{dy}{dt}$ and $v_A = \frac{dx}{dt}$</p>
<p>If you connect the center of the rod with the vertex of the right angle, then this segment will be the median of the right triangle and will be equal to half the length of the hypotenuse. It turns out that the center of the rod moves along a circle with the center at point $O$.</p>
<p>If you connect the center of the rod with the vertex of the right angle, then this segment will be the median of the right triangle and will be equal to half the length of the hypotenuse. It turns out that the center of the rod moves along a circle with the center at point $O$.</p>
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
