Edits to “Statement”, “Solution”, “Answer”
en/2.3.28.md
+6 −4
| @@ -1,19 +1,21 @@ | |||
| ### Statement | |||
| − | $2.3.28.$ [Insert the problem statement] | ||
| + | $2.3.28.$ A bead of mass $m$ slides along a vertical smooth and solid helix. The radius of the helical loop is $R$, and the helical pitch (vertical distance between adjacent turns) is $h$. With what force does the bead act on the helix at the moment when it has descended vertically to the distance $H$? The initial velocity of the bead is zero. | ||
| ### 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 | |||
| − | \[N=\sqrt{(mg\cos\theta)^2+\left(\frac{m(v\cos\theta)^2}{R}\right)^2}=mg\cos\theta\sqrt{1+\left(\frac{2H\cos\theta}{R}\right)^2}.\] | ||
| + | \[N=\sqrt{(mg\cos\theta)^2+\left(\frac{m(v\cos\theta)^2}{R}\right)^2}=mg\cos\theta\cdot\sqrt{1+\left(\frac{2H\cos\theta}{R}\right)^2}.\] | ||
| 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)} | ||
| + | \[N=\frac{2\pi Rmg}{h^2+4\pi^2R^2}\cdot\sqrt{h^2+4\pi^2(R^2+4H^2)},\] | ||
| + | which has the same magnitude as the force that the bead acts on the helix. | ||
| + | |||
| 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] | ||
| + | $\frac{2\pi Rmg}{h^2+4\pi^2R^2}\cdot\sqrt{h^2+4\pi^2(R^2+4H^2)}$ | ||
| @@ -1,19 +1,21 @@ | |||
| ### Statement | ### Statement | ||
| $2.3.28.$ [Insert the problem statement] | $2.3.28.$ A bead of mass $m$ slides along a vertical smooth and solid helix. The radius of the helical loop is $R$, and the helical pitch (vertical distance between adjacent turns) is $h$. With what force does the bead act on the helix at the moment when it has descended vertically to the distance $H$? The initial velocity of the bead is zero. | ||
| ### Solution | ### 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 | 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 | ||
| \[N=\sqrt{(mg\cos\theta)^2+\left(\frac{m(v\cos\theta)^2}{R}\right)^2}=mg\cos\theta\sqrt{1+\left(\frac{2H\cos\theta}{R}\right)^2}.\] | \[N=\sqrt{(mg\cos\theta)^2+\left(\frac{m(v\cos\theta)^2}{R}\right)^2}=mg\cos\theta\cdot\sqrt{1+\left(\frac{2H\cos\theta}{R}\right)^2}.\] | ||
| Using $\cos\theta=2\pi R/\sqrt{h^2+4\pi^2R^2}$, we get | 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)} |
\[N=\frac{2\pi Rmg}{h^2+4\pi^2R^2}\cdot\sqrt{h^2+4\pi^2(R^2+4H^2)},\] | ||
| which has the same magnitude as the force that the bead acts on the helix. | |||
| 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 $N$. | 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 | #### Answer | ||
| [Insert a concise answer or boxed result] | $\frac{2\pi Rmg}{h^2+4\pi^2R^2}\cdot\sqrt{h^2+4\pi^2(R^2+4H^2)}$ | ||