Falling beads

7635 /www/nrich/html/content/id/7635/ 1 0000-00-00T00:00:00 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Suppose that you have a large number of knitting needles and thousands of beads which can slide along needles without friction. All beads are released from rest at the same time from a point A to slide down the needles. <ul> <li style="text-align: justify;">Find a shape which is formed by the beads after time $t$.</li> <li style="text-align: justify;">Describe how this shape is changing with respect to time.</li> </ul> <hr></hr> <mdo:image alt="" src="knitting_needles1.bmp" style="width: 400px; height: 270px;"></mdo:image><mdo:image alt="" src="800px-Hama_beads.jpg" style="width: 200px; height: 260px;"></mdo:image> <hr></hr> Extension: In a real situation there will a friction force. The coefficient of friction is $\mu$. Suppose it is not dependent on the speed of a bead. Find the shape of beads after time $t$ and describe how it is changing. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> First of all let's choose a convenient Cartesian coordinate system as shown in the picture and suppose that the bead which is sliding through the needle inclined at an angle $\alpha$ to x-axis is at distance $r(t)$ from the origin. <mdo:image alt="" src="knitting_needles123.bmp" style="width: 700px; height: 414px;"></mdo:image> The gravity force is acting on the bead and it makes the bead accelerate along the needle. Write II-Newton's law in the direction along the needle and in the direction perpendicular to the needle. $ma = mg\sin(\alpha)$ $0 = N - mg\cos(\alpha)$ where $N$ is a reaction force. Beads start from rest, so $r(t) = \frac{at^2}{2} = \frac{g\sin(\alpha) t^2}{2}$. Change the polar coordinates to Cartesian coordinates by substituting $r = \sqrt{x^2 + y^2}$ and $\sin(\alpha) = \frac{y}{\sqrt{x^2 + y^2}}$. $$\sqrt{x^2 + y^2} = \frac{y}{\sqrt{x^2 + y^2}}\frac{gt^2}{2}$$ Multiply both sides by $\sqrt{x^2 + y^2}$ and complete the square. $$x^2 + (y -\frac{gt^2}{4})^2 = (\frac{gt^2}{4})^2$$ <ul> <li style="text-align: justify;">The shaped formed by beads after time $t$ is a circle of radius $\frac{gt^2}{4}$ with centre coordinates $(0, \frac{gt^2}{4})$.</li> <li style="text-align: left;">From the previous answer we deduce that there is a circle whose centre is falling with the acceleration $\frac{g}{2}$ and this circle is expanding with the same acceleration. A picture shows the shape of beads after time 0.5 s, 0.75 s, 1 s, 1.25 s, 1.5 s, 1.75 s, 2 s.<mdo:image alt="" src="circles1.bmp" style="width: 700px; height: 435px;"></mdo:image> <hr></hr> Extension: Now we have a friction force due to sliding. We modify our previous equations:</li> </ul> $ma = mg\sin(\alpha) -\mu N$ $0 = N - mg\cos(\alpha)$ where $N$ is a reaction force. $\therefore$ $$a = g(\sin(\alpha) - \mu\cos(\alpha))$$ It is important to notice that if $a < 0$ then the bead is not even starting to slide. Thus, the beads will slide for $\alpha$ from $\tan^{-1}(\mu)$ to $(\pi - \tan^{-1}(\mu))$. To simplify the situation, notice that the shape will be symmetrical in the y-axis. Thus, we need to analyse what happens for $\alpha$ from $\tan^{-1}(\mu)$ to $\frac{\pi}{2}$. $$r(t) = \frac{at^2}{2} = (\sin(\alpha) - \mu \cos(\alpha)) \frac{gt^2}{2}$$ Change to the Cartesian coordinate system $x^2 + y^2 = \frac{gt^2}{2} (y - \mu x)$ for $\frac{\pi}{2} > \tan^{-1} (\frac{y}{x}) > \tan^{-1} (\mu)$. Complete the squares to get $$(x + \frac{\mu gt^2}{4})^2 + (y - \frac{gt^2}{4})^2 = (1+ \mu^2)(\frac{gt^2}{4})^2$$ It is easier to plot a whole circle and then to take a suitable arc. Similarly, to reflect a circle in y-axis change $x$ to $-x$ then we have an equation $$(-x + \frac{\mu gt^2}{4})^2 + (y - \frac{gt^2}{4})^2 = (1+ \mu^2)(\frac{gt^2}{4})^2$$ and we take a suitable arc again. Let's take $\mu = \frac{\sqrt3}{3}$ then $\tan^{-1}(\frac{\sqrt3}{3}) = \frac{\pi}{6}$. We plot circles which help to show how the shape of beads in red changes. These graphs represent the shape of beads after time 0.5 s, 0.75 s, 1 s, 1.25 s, 1.5 s, 1.75 s, 2 s. Circles are traveling along green lines which equations can be found knowing that the coordinates of center is $(\frac{\mu gt^2}{4}, \frac{gt^2}{4}) = \frac{gt^2}{4}(\mu, 1)$ and $\frac{gt^2}{4}(-\mu, 1)$ <mdo:image alt="" src="circlefriction.bmp" style="width: 800px; height: 441px;"></mdo:image> <hr></hr> To sum up: <ul> <li style="text-align: justify;">The shape of beads after time $t$ is an union of two arcs.</li> <li style="text-align: justify;">Centre of circles are traveling with $\frac{g}{2} \sqrt{1+\mu^2}$ acceleration and expanding with the same acceleration.</li> </ul> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> It might help to express the distance traveled by a particular bead in the Cartesian coordinates in order to describe what's going on. Extension: Think about the symmetry and the possible shape of beads. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> First of all let's choose a convenient Cartesian coordinate system as shown in the picture and suppose that the bead which is sliding through the needle inclined at an angle $\alpha$ to x-axis is at distance $r(t)$ from the origin. <mdo:image alt="" src="knitting_needles123.bmp" style="width: 700px; height: 414px;"></mdo:image> The gravity force is acting on the bead and it makes the bead accelerate along the needle. Write II-Newton's law in the direction along the needle and in the direction perpendicular to the needle. $ma = mg\sin(\alpha)$ $0 = N - mg\cos(\alpha)$ where $N$ is a reaction force. Beads start from rest, so $r(t) = \frac{at^2}{2} = \frac{g\sin(\alpha) t^2}{2}$. Change the polar coordinates to Cartesian coordinates by substituting $r = \sqrt{x^2 + y^2}$ and $\sin(\alpha) = \frac{y}{\sqrt{x^2 + y^2}}$. $$\sqrt{x^2 + y^2} = \frac{y}{\sqrt{x^2 + y^2}}\frac{gt^2}{2}$$ Multiply both sides by $\sqrt{x^2 + y^2}$ and complete the square. $$x^2 + (y -\frac{gt^2}{4})^2 = (\frac{gt^2}{4})^2$$ <ul> <li style="text-align: justify;">The shaped formed by beads after time $t$ is a circle of radius $\frac{gt^2}{4}$ with centre coordinates $(0, \frac{gt^2}{4})$.</li> <li style="text-align: left;">From the previous answer we deduce that there is a circle whose centre is falling with the acceleration $\frac{g}{2}$ and this circle is expanding with the same acceleration. A picture shows the shape of beads after time 0.5 s, 0.75 s, 1 s, 1.25 s, 1.5 s, 1.75 s, 2 s.<mdo:image alt="" src="circles1.bmp" style="width: 700px; height: 435px;"></mdo:image> <hr></hr> Extension: Now we have a friction force due to sliding. We modify our previous equations:</li> </ul> $ma = mg\sin(\alpha) -\mu N$ $0 = N - mg\cos(\alpha)$ where $N$ is a reaction force. $\therefore$ $$a = g(\sin(\alpha) - \mu\cos(\alpha))$$ It is important to notice that if $a < 0$ then the bead is not even starting to slide. Thus, the beads will slide for $\alpha$ from $\tan^{-1}(\mu)$ to $(\pi - \tan^{-1}(\mu))$. To simplify the situation, notice that the shape will be symmetrical in the y-axis. Thus, we need to analyse what happens for $\alpha$ from $\tan^{-1}(\mu)$ to $\frac{\pi}{2}$. $$r(t) = \frac{at^2}{2} = (\sin(\alpha) - \mu \cos(\alpha)) \frac{gt^2}{2}$$ Change to the Cartesian coordinate system $x^2 + y^2 = \frac{gt^2}{2} (y - \mu x)$ for $\frac{\pi}{2} > \tan^{-1} (\frac{y}{x}) > \tan^{-1} (\mu)$. Complete the squares to get $$(x + \frac{\mu gt^2}{4})^2 + (y - \frac{gt^2}{4})^2 = (1+ \mu^2)(\frac{gt^2}{4})^2$$ It is easier to plot a whole circle and then to take a suitable arc. Similarly, to reflect a circle in y-axis change $x$ to $-x$ then we have an equation $$(-x + \frac{\mu gt^2}{4})^2 + (y - \frac{gt^2}{4})^2 = (1+ \mu^2)(\frac{gt^2}{4})^2$$ and we take a suitable arc again. Let's take $\mu = \frac{\sqrt3}{3}$ then $\tan^{-1}(\frac{\sqrt3}{3}) = \frac{\pi}{6}$. We plot circles which help to show how the shape of beads in red changes. These graphs represent the shape of beads after time 0.5 s, 0.75 s, 1 s, 1.25 s, 1.5 s, 1.75 s, 2 s. Circles are traveling along green lines which equations can be found knowing that the coordinates of center is $(\frac{\mu gt^2}{4}, \frac{gt^2}{4}) = \frac{gt^2}{4}(\mu, 1)$ and $\frac{gt^2}{4}(-\mu, 1)$ <mdo:image alt="" src="circlefriction.bmp" style="width: 800px; height: 441px;"></mdo:image> <hr></hr> To sum up: <ul> <li style="text-align: justify;">The shape of beads after time $t$ is an union of two arcs.</li> <li style="text-align: justify;">Centre of circles are traveling with $\frac{g}{2} \sqrt{1+\mu^2}$ acceleration and expanding with the same acceleration.</li> </ul> </mdoxml> 2 4 0 0 0 1 1 Falling beads