Правка разделов «Statement», «Solution»
en/2.1.65.md
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| ### Statement | |||
| − | $2.1.65.$ [Insert the problem statement] | ||
| + | $2.1.65^*.$ A speed skater on an ice track tries to negotiate a turn as close to the inner edge as possible. A cyclist on a velodrome, conversely, takes a turn as far from the inner edge as possible. How can this difference in cornering tactics be explained? The profile of the velodrome becomes steeper with distance from its inner edge. | ||
| − | ### Solution | ||
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| − | ### Statement | ||
| − | |||
| − | $2.1.65.$ A speed skater on an ice track tries to negotiate a turn as close to the inner edge as possible. A cyclist on a velodrome, conversely, takes a turn as far from the inner edge as possible. How can this difference in cornering tactics be explained? The profile of the velodrome becomes steeper with distance from its inner edge. | ||
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| ### Solution | |||
| + |  | ||
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| Let's first consider the <b>speed skater</b>. The centripetal acceleration of the skater is provided by the static friction force against the ice: | |||
| $$F_{\text{fr}} = \mu N$$ | |||
| Since the skater does not move in the vertical direction, the normal reaction force $N$ is equal in magnitude to the gravity force $Mg$. Therefore, the maximum friction force is: | |||
| $$F_{\text{fr}} = \mu Mg$$ | |||
| According to Newton's second law for circular motion, the maximum speed is determined by: | |||
| $$\frac{Mv^2}{R} = \mu Mg \implies v = \sqrt{\mu g R}$$ | |||
| When making a turn, the skater covers a distance proportional to the radius, $S \sim R$ (for example, for a semicircle $S = \pi R$). The time taken to complete the turn is: | |||
| $$t = \frac{S}{v} = \frac{\pi R}{\sqrt{\mu g R}} = \pi\sqrt{\frac{R}{\mu g}}$$ | |||
| @@ -21,7 +19,3 @@Solution | |||
| The larger the radius $R$ of the circle along which the skater moves, the longer this time. Although the maximum speed increases with the turning radius ($\sim \sqrt{R}$), the distance traveled increases even more rapidly ($\sim R$). This is why the speed skater tries to take the turn as close to the inner edge as possible to minimize their time. | |||
| For the <b>cyclist</b> on the velodrome, the mechanics change due to the banked track. The angle of inclination increases as one moves further from the inner edge. On a steeply banked track, the horizontal component of the normal reaction force provides a massive centripetal force, allowing for much higher speeds ($v \sim \sqrt{g R \tan\alpha}$). Because the steepness ($\tan\alpha$) of the velodrome increases towards the outer edge, the maximum permissible speed increases much faster than $\sqrt{R}$, which more than compensates for the longer path distance $S$. Thus, the cyclist minimizes their time by riding higher up on the steeper, outer part of the track. | |||
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| − | #### Answer | ||
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| − | [Insert a concise answer or boxed result] | ||
| @@ -1,15 +1,13 @@ | |||
| ### Statement | ### Statement | ||
| $2.1.65.$ [Insert the problem statement] | $2.1.65^*.$ A speed skater on an ice track tries to negotiate a turn as close to the inner edge as possible. A cyclist on a velodrome, conversely, takes a turn as far from the inner edge as possible. How can this difference in cornering tactics be explained? The profile of the velodrome becomes steeper with distance from its inner edge. | ||
| ### Solution |  | ||
| ### Statement | |||
| $2.1.65.$ A speed skater on an ice track tries to negotiate a turn as close to the inner edge as possible. A cyclist on a velodrome, conversely, takes a turn as far from the inner edge as possible. How can this difference in cornering tactics be explained? The profile of the velodrome becomes steeper with distance from its inner edge. | |||
| ### Solution | ### Solution | ||
|  | |||
| Let's first consider the <b>speed skater</b>. The centripetal acceleration of the skater is provided by the static friction force against the ice: | Let's first consider the <b>speed skater</b>. The centripetal acceleration of the skater is provided by the static friction force against the ice: | ||
| $$F_{\text{fr}} = \mu N$$ | $$F_{\text{fr}} = \mu N$$ | ||
| Since the skater does not move in the vertical direction, the normal reaction force $N$ is equal in magnitude to the gravity force $Mg$. Therefore, the maximum friction force is: | Since the skater does not move in the vertical direction, the normal reaction force $N$ is equal in magnitude to the gravity force $Mg$. Therefore, the maximum friction force is: | ||
| $$F_{\text{fr}} = \mu Mg$$ | $$F_{\text{fr}} = \mu Mg$$ | ||
| According to Newton's second law for circular motion, the maximum speed is determined by: | According to Newton's second law for circular motion, the maximum speed is determined by: | ||
| $$\frac{Mv^2}{R} = \mu Mg \implies v = \sqrt{\mu g R}$$ | $$\frac{Mv^2}{R} = \mu Mg \implies v = \sqrt{\mu g R}$$ | ||
| When making a turn, the skater covers a distance proportional to the radius, $S \sim R$ (for example, for a semicircle $S = \pi R$). The time taken to complete the turn is: | When making a turn, the skater covers a distance proportional to the radius, $S \sim R$ (for example, for a semicircle $S = \pi R$). The time taken to complete the turn is: | ||
| $$t = \frac{S}{v} = \frac{\pi R}{\sqrt{\mu g R}} = \pi\sqrt{\frac{R}{\mu g}}$$ | $$t = \frac{S}{v} = \frac{\pi R}{\sqrt{\mu g R}} = \pi\sqrt{\frac{R}{\mu g}}$$ | ||
| @@ -21,7 +19,3 @@Solution | |||
| The larger the radius $R$ of the circle along which the skater moves, the longer this time. Although the maximum speed increases with the turning radius ($\sim \sqrt{R}$), the distance traveled increases even more rapidly ($\sim R$). This is why the speed skater tries to take the turn as close to the inner edge as possible to minimize their time. | The larger the radius $R$ of the circle along which the skater moves, the longer this time. Although the maximum speed increases with the turning radius ($\sim \sqrt{R}$), the distance traveled increases even more rapidly ($\sim R$). This is why the speed skater tries to take the turn as close to the inner edge as possible to minimize their time. | ||
| For the <b>cyclist</b> on the velodrome, the mechanics change due to the banked track. The angle of inclination increases as one moves further from the inner edge. On a steeply banked track, the horizontal component of the normal reaction force provides a massive centripetal force, allowing for much higher speeds ($v \sim \sqrt{g R \tan\alpha}$). Because the steepness ($\tan\alpha$) of the velodrome increases towards the outer edge, the maximum permissible speed increases much faster than $\sqrt{R}$, which more than compensates for the longer path distance $S$. Thus, the cyclist minimizes their time by riding higher up on the steeper, outer part of the track. | For the <b>cyclist</b> on the velodrome, the mechanics change due to the banked track. The angle of inclination increases as one moves further from the inner edge. On a steeply banked track, the horizontal component of the normal reaction force provides a massive centripetal force, allowing for much higher speeds ($v \sim \sqrt{g R \tan\alpha}$). Because the steepness ($\tan\alpha$) of the velodrome increases towards the outer edge, the maximum permissible speed increases much faster than $\sqrt{R}$, which more than compensates for the longer path distance $S$. Thus, the cyclist minimizes their time by riding higher up on the steeper, outer part of the track. | ||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||