Edits to “Statement”, “Solution”, “Answer”

Valter edited Aug 31, 2026, 7:02 PM

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revision #20474 parent #20473 ← older

en/2.1.64.md +7 −10

		@@ -1,15 +1,11 @@
			### Statement

		−	$2.1.64.$ [Insert the problem statement]
		−
		−	### Solution
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		−	### Statement
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			$2.1.64.$ What is the maximum speed a motorcyclist can ride on a horizontal plane, describing a circle of radius $R$, if the coefficient of friction is $\mu$? At what angle from the vertical must they lean? By what factor does the maximum permissible speed of the motorcyclist increase when moving on an inclined track with an angle of inclination $\alpha$ to the horizontal compared to the maximum permissible speed when moving on a horizontal track with the same turning radius and the same friction coefficient?

			### Solution

		+	![Forces acting on motorcyclist\|723x290, 50%](../../img/2.1.64/hdsgasgdf.png)
		+
			<b>1. Motion on a horizontal track:</b>
			The forces acting on the motorcyclist are gravity $m\vec{g}$, the normal reaction force $\vec{N}$, and the static friction force $\vec{F}_{\text{fr}}$.
			The centripetal acceleration can only be provided by the static friction force directed radially towards the center $O$.

			Newton's second law in projections:
			$$Ox: F_{\text{fr}} = m \frac{v_1^2}{R}$$
			$$Oy: N - mg = 0 \implies N = mg$$

		@@ -21,10 +17,15 @@Solution
			Given that the maximum static friction force is $F_{\text{fr}} = \mu N = \mu mg$, we find the maximum speed $v_1$ on the horizontal track:
			$$m \frac{v_1^2}{R} = \mu mg \implies v_1 = \sqrt{\mu g R}$$

		+	![Forces acting on motorcyclist during a turn\|727x343, 50%](../../img/2.1.64/sjafkhslkfjhs.png)
		+
			To prevent a tipping moment (torque), the resultant of the normal force and the friction force must pass through the center of gravity of the motorcyclist. Therefore, the motorcyclist must lean towards the center of the circle by an angle $\beta$ from the vertical:
			$$\tan\beta = \frac{F_{\text{fr}}}{N} = \frac{\mu mg}{mg} = \mu \implies \beta = \arctan\mu$$

			<b>2. Motion on an inclined track:</b>
		+
		+	![\|214x247, 50%](../../img/2.1.64/uryeuiwhfdjs.png)
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			Now consider the motion on a track inclined at an angle $\alpha$ to the horizontal. To achieve the maximum speed $v_2$, the static friction force $\vec{F}_{\text{fr}}$ must act downwards along the incline (preventing the motorcycle from sliding outward).

			Newton's second law in projections onto the vertical ($Y$) and horizontal ($X$) axes:
			$$Y: N \cos\alpha - F_{\text{fr}} \sin\alpha - mg = 0 \quad (1)$$
			$$X: N \sin\alpha + F_{\text{fr}} \cos\alpha = m \frac{v_2^2}{R} \quad (2)$$

			To solve for $N$ and $F_{\text{fr}}$, we manipulate the system.
			Multiply (1) by $\cos\alpha$ and (2) by $\sin\alpha$, then add them:
			$$N(\cos^2\alpha + \sin^2\alpha) = mg \cos\alpha + m \frac{v_2^2}{R} \sin\alpha \implies N = m\left(g \cos\alpha + \frac{v_2^2}{R} \sin\alpha\right)$$

			Multiply (1) by $-\sin\alpha$ and (2) by $\cos\alpha$, then add them:
			$$F_{\text{fr}}(\sin^2\alpha + \cos^2\alpha) = m \frac{v_2^2}{R} \cos\alpha - mg \sin\alpha \implies F_{\text{fr}} = m\left(\frac{v_2^2}{R} \cos\alpha - g \sin\alpha\right)$$

			Since $F_{\text{fr}}$ is the static friction force, its maximum value must satisfy $F_{\text{fr}} \le \mu N$:
			$$m\left(\frac{v_2^2}{R} \cos\alpha - g \sin\alpha\right) \le \mu m\left(g \cos\alpha + \frac{v_2^2}{R} \sin\alpha\right)$$

			Divide both sides by $m$ and rearrange to isolate terms with $v_2$:
			$$\frac{v_2^2}{R} (\cos\alpha - \mu \sin\alpha) \le g(\mu \cos\alpha + \sin\alpha)$$

			Divide both sides by $\cos\alpha$ (assuming $\cos\alpha > 0$):
			$$\frac{v_2^2}{R} (1 - \mu \tan\alpha) \le g(\mu + \tan\alpha)$$

			Assuming $(1 - \mu \tan\alpha) > 0$, the maximum speed $v_2$ on the inclined track is:
			$$v_2 = \sqrt{\frac{gR(\mu + \tan\alpha)}{1 - \mu \tan\alpha}}$$

			<b>3. Ratio of maximum speeds:</b>
			The ratio of the maximum speed on the inclined track to the maximum speed on the horizontal track is:
			$$\frac{v_2}{v_1} = \frac{\sqrt{\frac{gR(\mu + \tan\alpha)}{1 - \mu \tan\alpha}}}{\sqrt{\mu g R}} = \sqrt{\frac{\mu + \tan\alpha}{\mu(1 - \mu \tan\alpha)}}$$

			#### Answer
		@@ -58,7 +59,3 @@Answer
			$$v = \sqrt{\mu g R}$$
			$$\beta = \arctan\mu$$
			$$\frac{v_2}{v_1} = \sqrt{\frac{\mu + \tan\alpha}{\mu(1 - \mu \tan\alpha)}}$$
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		−	#### Answer
		−
		−	[Insert a concise answer or boxed result]