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en/2.3.28.md
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| + | ### Statement | ||
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| + | $2.3.28.$ [Insert the problem statement] | ||
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| + | ### Solution | ||
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| + | A helix is simply an inclined plane wrapped around a cylinder with an inclination angle of $\theta=\arctan(h/(2\pi R))$. A component of the normal force $N$ from the helix on the bead counters the component $mg\cos\theta$ of gravity perpendicular to the helix. At the depth $H$ below the point of release, the gravitational potential energy transforms into the kinetic energy, and the bead has velocity $v=\sqrt{2gH}$ along the helix, where its horizonal component is $v\cos\theta$. As the horizontal projection of the helix is a circle with radius $R$, the centripetal component of $N$ that keeps the bead moving around the cylinder is $m(v\cos\theta)^2/R$. Since the two components of $N$ are perpendicular to each other, we have | ||
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| + | \[N=\sqrt{(mg\cos\theta)^2+\left(\frac{m(v\cos\theta)^2}{R}\right)^2}=mg\cos\theta\sqrt{1+\left(\frac{2H\cos\theta}{R}\right)^2}.\] | ||
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| + | Using $\cos\theta=2\pi R/\sqrt{h^2+4\pi^2R^2}$, we get | ||
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| + | \[N=\frac{2\pi Rmg}{h^2+4\pi^2R^2}\cdot\sqrt{h^2+4\pi^2(R^2+4H^2)}.\] | ||
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| + | Note that we could use the velocity $v$ and the radius of curvature $(h^2+4\pi^2R^2)/(4\pi R)$ of an osculating circle to the helix to obtain the same expression for the centripetal component of $N$. | ||
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| + | #### Answer | ||
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| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $2.3.28.$ [Insert the problem statement] | |||
| ### Solution | |||
| A helix is simply an inclined plane wrapped around a cylinder with an inclination angle of $\theta=\arctan(h/(2\pi R))$. A component of the normal force $N$ from the helix on the bead counters the component $mg\cos\theta$ of gravity perpendicular to the helix. At the depth $H$ below the point of release, the gravitational potential energy transforms into the kinetic energy, and the bead has velocity $v=\sqrt{2gH}$ along the helix, where its horizonal component is $v\cos\theta$. As the horizontal projection of the helix is a circle with radius $R$, the centripetal component of $N$ that keeps the bead moving around the cylinder is $m(v\cos\theta)^2/R$. Since the two components of $N$ are perpendicular to each other, we have | |||
| \[N=\sqrt{(mg\cos\theta)^2+\left(\frac{m(v\cos\theta)^2}{R}\right)^2}=mg\cos\theta\sqrt{1+\left(\frac{2H\cos\theta}{R}\right)^2}.\] | |||
| Using $\cos\theta=2\pi R/\sqrt{h^2+4\pi^2R^2}$, we get | |||
| \[N=\frac{2\pi Rmg}{h^2+4\pi^2R^2}\cdot\sqrt{h^2+4\pi^2(R^2+4H^2)}.\] | |||
| Note that we could use the velocity $v$ and the radius of curvature $(h^2+4\pi^2R^2)/(4\pi R)$ of an osculating circle to the helix to obtain the same expression for the centripetal component of $N$. | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||