Правка раздела «Solution»
en/1.3.2.md
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| ### Statement | |||
| $1.3.2.$ a. From the top point of the circle, a ball begins to slide along a smooth chute at an angle $\varphi$ to the vertical. How long will it take for it to reach the circle, if its diameter is $D$? | |||
| b. From point $A$, small beads begin to slide along the spokes with different slopes at the same time without friction. What curve will the beads be on at time $t$? | |||
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| ### Solution | |||
| @@ -9,7 +9,7 @@Solution | |||
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| − | a) A ball will move along a smooth chute with acceleration equal to the projection of the free-fall acceleration in the direction of motion, i.e. $$a = g \cdot\cos{\varphi }$$ The displacement of the ball is the chord of a circle of diameter $D$, the magnitude of which is related to the diameter, by the following relation $$r = D \cdot\cos{\varphi }$$ Let's write further the equation of accelerated motion of the ball and from it find the time of motion $${r=\frac{at^{2}}{2},\quad D\cos\varphi=\frac{g\cos\varphi}{2}t^{2},\quad t=\sqrt{\frac{2D}{g}} .}$$ $$\fbox{ $t=\sqrt{\frac{2D}{g}}$ } \ | ||
| + | a) A ball will move along a smooth chute with acceleration equal to the projection of the free-fall acceleration in the direction of motion, i.e. $$a = g \cdot\cos{\varphi }$$ The displacement of the ball is the chord of a circle of diameter $D$, the magnitude of which is related to the diameter, by the following relation $$r = D \cdot\cos{\varphi }$$ Let's write further the equation of accelerated motion of the ball and from it find the time of motion $${r=\frac{at^{2}}{2},\quad D\cos\varphi=\frac{g\cos\varphi}{2}t^{2},\quad t=\sqrt{\frac{2D}{g}} .}$$ $$\fbox{ $t=\sqrt{\frac{2D}{g}}$ } \tag{1}$$ | ||
| b) Note that the expression $(1)$ does not include the value of the angle, so all balls will be dropped simultaneously. They will lie on a circle of radius $r = g t^2/2$, as shown in the animation | |||
| #### Answer | |||
| a. $t = \sqrt{2D/g}$ | |||
| b. On a circle of radius $\frac{gt^{2}}{2}$ with top point $A$. | |||
| ещё строк без изменений 3 | |||
| ### Statement | ### Statement | ||
| $1.3.2.$ a. From the top point of the circle, a ball begins to slide along a smooth chute at an angle $\varphi$ to the vertical. How long will it take for it to reach the circle, if its diameter is $D$? | $1.3.2.$ a. From the top point of the circle, a ball begins to slide along a smooth chute at an angle $\varphi$ to the vertical. How long will it take for it to reach the circle, if its diameter is $D$? | ||
| b. From point $A$, small beads begin to slide along the spokes with different slopes at the same time without friction. What curve will the beads be on at time $t$? | b. From point $A$, small beads begin to slide along the spokes with different slopes at the same time without friction. What curve will the beads be on at time $t$? | ||
|  |  | ||
| ### Solution | ### Solution | ||
| @@ -9,7 +9,7 @@Solution | |||
|  |  | ||
| a) A ball will move along a smooth chute with acceleration equal to the projection of the free-fall acceleration in the direction of motion, i.e. $$a = g \cdot\cos{\varphi }$$ The displacement of the ball is the chord of a circle of diameter $D$, the magnitude of which is related to the diameter, by the following relation $$r = D \cdot\cos{\varphi }$$ Let's write further the equation of accelerated motion of the ball and from it find the time of motion $${r=\frac{at^{2}}{2},\quad D\cos\varphi=\frac{g\cos\varphi}{2}t^{2},\quad t=\sqrt{\frac{2D}{g}} .}$$ $$\fbox{ $t=\sqrt{\frac{2D}{g}}$ } \ |
a) A ball will move along a smooth chute with acceleration equal to the projection of the free-fall acceleration in the direction of motion, i.e. $$a = g \cdot\cos{\varphi }$$ The displacement of the ball is the chord of a circle of diameter $D$, the magnitude of which is related to the diameter, by the following relation $$r = D \cdot\cos{\varphi }$$ Let's write further the equation of accelerated motion of the ball and from it find the time of motion $${r=\frac{at^{2}}{2},\quad D\cos\varphi=\frac{g\cos\varphi}{2}t^{2},\quad t=\sqrt{\frac{2D}{g}} .}$$ $$\fbox{ $t=\sqrt{\frac{2D}{g}}$ } \tag{1}$$ | ||
| b) Note that the expression $(1)$ does not include the value of the angle, so all balls will be dropped simultaneously. They will lie on a circle of radius $r = g t^2/2$, as shown in the animation | b) Note that the expression $(1)$ does not include the value of the angle, so all balls will be dropped simultaneously. They will lie on a circle of radius $r = g t^2/2$, as shown in the animation | ||
| #### Answer | #### Answer | ||
| a. $t = \sqrt{2D/g}$ | a. $t = \sqrt{2D/g}$ | ||
| b. On a circle of radius $\frac{gt^{2}}{2}$ with top point $A$. | b. On a circle of radius $\frac{gt^{2}}{2}$ with top point $A$. | ||
| ещё строк без изменений 3 | |||