Question

A. (10 points) A physical pendulum consists of a uniform rod of mass m and length L pivoting by its end as shown. If the rod makes 14 complete oscillations in 17 seconds, what is the length of the rod? Solve completely symbolically before inserting values. B. (10 points) Now consider a uniform rod of mass m and length L pivoting the rod about a point L/5 from its end. What is the period of the rod? Solve completely symbolically. (This is a new rod, do not use L from part A.) L/5 - C. (10 points) Explain in your own words how you could calculate where to place the pivot point for a physical pendulum consisting of a uniform rod of m and length L so that it would complete an oscillation in the least amount of time. Do not do the calculation.

Answer 1

L= 3x9.8x0-0373 L = 0.549om at the place where x will be maximum and also I must be man imum So I will be less. L = 0.55m L=0.55m I = ML? 12 mx² B) I = my m -) T= 25 m2 ² mx? 12 mgo = mc²9ML? 12 100 for physical pendulum T= 20 1 mga In moment of inentia about fixed point X= distance of com from fixed point. A T= 20 m12/3 T= 20 1² - 13ML? 75 legx + x= 1/3 - 1/5 for maximum Tod I = 0 1 + O 2 129x T= 20 mg. (1/2) 13m2² 75 mg (3/10) Teg 하 L2 2L T: शा 2 3g T- 2 m 130 ML 2 28 225 mgh 129 - x=512 L 26 L T 17 2L 14 39 0.0373 = 2L 39 45 9 x= 0.28864 -0.29h from centere of rod. T= 25 264 usg

for least time distance of fixed point must be at 0.29 L from mid point of rod.

or at 0.21L from end of rod.

Thanks for asking.