Правка разделов «Statement», «Solution», «Answer»
en/2.7.26.md
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| ### Statement | |||
| − | $2.7.26.$ [Insert the problem statement] | ||
| + | $2.7.26.$ The center of the thin ring is just above the edge of the table. The ring starts rolling off the table without slipping out of its resting state. At what angle will the ring turn before it breaks away from the edge of the table? Is this angle greater or less if a ball rolls off the table? | ||
| ### Solution | |||
| − |  | ||
| The center of the ring will rotate about point $P$ at the edge of the table until it leaves the table. Suppose the moment of inertia of the ring about its center is $I$. Then the moment of inertia of the ring about $P$ is $I+mr^2$, where $m$ and $r$ are the mass and the radius of the ring, respectively. Assume that the ring leaves the table after its center has rotated through an angle $\alpha$. Then, its height drops by $r-r\cos\alpha$, and gravitational potential energy becomes rotational kinetic energy; that is, | |||
| \[mgr(1-\cos\alpha)=\frac{1}{2}(I+mr^2)\omega^2\qquad\Rightarrow\qquad\omega^2=\frac{2mgr(1-\cos\alpha)}{I+mr^2}.\] | |||
| When the ring is about to lose contact with the table, the centripetal force that keeps the center of the ring rotating about $P$ is a component of gravity. At this point, | |||
| \[mg\cos\alpha=m\omega^2r\qquad\Rightarrow\qquad\cos\alpha=\frac{2mr^2(1-\cos\alpha)}{I+mr^2}.\] | |||
| Using $I=mr^2$ for the ring, we have $\cos\alpha=1/2$, and so $\alpha=60^\circ.$ For a ball with the same mass and radius, $I<mr^2$, and hence | |||
| \[\cos\alpha>1-\cos\alpha\qquad\Rightarrow\qquad\cos\alpha>\frac{1}{2}\qquad\Rightarrow\qquad\alpha<60^\circ.\] | |||
| @@ -20,4 +20,4 @@Solution | |||
| #### Answer | |||
| − | |||
| + | $60^\circ$ for the ring and smaller angle for the ball. | ||
| @@ -1,10 +1,10 @@ | |||
| ### Statement | ### Statement | ||
| $2.7.26.$ [Insert the problem statement] | $2.7.26.$ The center of the thin ring is just above the edge of the table. The ring starts rolling off the table without slipping out of its resting state. At what angle will the ring turn before it breaks away from the edge of the table? Is this angle greater or less if a ball rolls off the table? | ||
| ### Solution | ### Solution | ||
|  | ||
| The center of the ring will rotate about point $P$ at the edge of the table until it leaves the table. Suppose the moment of inertia of the ring about its center is $I$. Then the moment of inertia of the ring about $P$ is $I+mr^2$, where $m$ and $r$ are the mass and the radius of the ring, respectively. Assume that the ring leaves the table after its center has rotated through an angle $\alpha$. Then, its height drops by $r-r\cos\alpha$, and gravitational potential energy becomes rotational kinetic energy; that is, | The center of the ring will rotate about point $P$ at the edge of the table until it leaves the table. Suppose the moment of inertia of the ring about its center is $I$. Then the moment of inertia of the ring about $P$ is $I+mr^2$, where $m$ and $r$ are the mass and the radius of the ring, respectively. Assume that the ring leaves the table after its center has rotated through an angle $\alpha$. Then, its height drops by $r-r\cos\alpha$, and gravitational potential energy becomes rotational kinetic energy; that is, | ||
| \[mgr(1-\cos\alpha)=\frac{1}{2}(I+mr^2)\omega^2\qquad\Rightarrow\qquad\omega^2=\frac{2mgr(1-\cos\alpha)}{I+mr^2}.\] | \[mgr(1-\cos\alpha)=\frac{1}{2}(I+mr^2)\omega^2\qquad\Rightarrow\qquad\omega^2=\frac{2mgr(1-\cos\alpha)}{I+mr^2}.\] | ||
| When the ring is about to lose contact with the table, the centripetal force that keeps the center of the ring rotating about $P$ is a component of gravity. At this point, | When the ring is about to lose contact with the table, the centripetal force that keeps the center of the ring rotating about $P$ is a component of gravity. At this point, | ||
| \[mg\cos\alpha=m\omega^2r\qquad\Rightarrow\qquad\cos\alpha=\frac{2mr^2(1-\cos\alpha)}{I+mr^2}.\] | \[mg\cos\alpha=m\omega^2r\qquad\Rightarrow\qquad\cos\alpha=\frac{2mr^2(1-\cos\alpha)}{I+mr^2}.\] | ||
| Using $I=mr^2$ for the ring, we have $\cos\alpha=1/2$, and so $\alpha=60^\circ.$ For a ball with the same mass and radius, $I<mr^2$, and hence | Using $I=mr^2$ for the ring, we have $\cos\alpha=1/2$, and so $\alpha=60^\circ.$ For a ball with the same mass and radius, $I<mr^2$, and hence | ||
| \[\cos\alpha>1-\cos\alpha\qquad\Rightarrow\qquad\cos\alpha>\frac{1}{2}\qquad\Rightarrow\qquad\alpha<60^\circ.\] | \[\cos\alpha>1-\cos\alpha\qquad\Rightarrow\qquad\cos\alpha>\frac{1}{2}\qquad\Rightarrow\qquad\alpha<60^\circ.\] | ||
| @@ -20,4 +20,4 @@Solution | |||
| #### Answer | #### Answer | ||
| $60^\circ$ for the ring and smaller angle for the ball. | |||