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5. 40 Pendulum and Peg * A pendulum of length L is held with its string horizontal, and then released. The string runs into a peg a distance d below the pivot, as shown in Fig. 5.29. What is the smallest value of d for which the string remains stretched at all times? L Fig. 5.29 You may expound on the answer below: 5.40. Pendulum and peg The radius of the circle is L – d, so as in Exercise 5.39 we need v? > g(L – d) at the top of the circle. But the top of the circle is a distance L - below the starting point, so conservation of energy gives v² = 2g(2d – L) at the top. Therefore, 2g(2d – L) > g(L – d) = d> 3L/5. – 2(L – d) = 2d – L |

5. 40 Pendulum and Peg * A pendulum of length L is held with its string horizontal, and then released. The string runs into a peg a distance d below the pivot, as shown in Fig. 5.29. What is the smallest value of d for which the string remains stretched at all times? L Fig. 5.29 You may expound on the answer below: 5.40. Pendulum and peg The radius of the circle is L – d, so as in Exercise 5.39 we need v? > g(L – d) at the top of the circle. But the top of the circle is a distance L - below the starting point, so conservation of energy gives v² = 2g(2d – L) at the top. Therefore, 2g(2d – L) > g(L – d) = d> 3L/5. – 2(L – d) = 2d – L |

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5. 40 Pendulum and Peg * A pendulum of length L is held with its string horizontal, and then released. The string runs into a peg a distance d below the pivot, as shown in Fig. 5.29. What is the smallest value of d for which the string remains stretched at all times? Fig. 5.29 You may expound on the answer below: 5.40. Pendulum and peg The radius of the circle is L – d, so as in Exercise 5.39 we need v? > g(L – d) at the top of the circle. But the top of the circle is a distance L – 2(L – d) = 2d – L below the starting point, so conservation of energy gives v? = 2g(2d – L) at the top. Therefore, 2g(2d – L) > g(L – d) = d> 3L/5.
Transcribed Image Text:5. 40 Pendulum and Peg * A pendulum of length L is held with its string horizontal, and then released. The string runs into a peg a distance d below the pivot, as shown in Fig. 5.29. What is the smallest value of d for which the string remains stretched at all times? Fig. 5.29 You may expound on the answer below: 5.40. Pendulum and peg The radius of the circle is L – d, so as in Exercise 5.39 we need v? > g(L – d) at the top of the circle. But the top of the circle is a distance L – 2(L – d) = 2d – L below the starting point, so conservation of energy gives v? = 2g(2d – L) at the top. Therefore, 2g(2d – L) > g(L – d) = d> 3L/5.
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