Chapter 15, Problem 072A uniform circular disk whose radius R is 14.7 cm is suspended as a physical pendulum from a point on its rim. (a) What is its period? (b) At what radial distance r < R is there a pivot point thatgives the same period?(a) NumberUnits(b) NumberUnitsClick if you would like to Show Work for this question:Open Show Work

The moment of inertia of a uniform circular disc of radius R about an axis touching the disc at its diameter normal to the disc is I =32MR2a) For a pendulum with momentum inertia I, the time period will be T = 2πIMgd where, d = distance between center of mass and axis of rotation M = mass I = moment of inertia here d = R (radius of the disc) substituting the value of I we get, T = 2π32MR2MgRT = 2π 3R2g according to the question, R = 147 cm = 0.147 m acceleration due to gravity, g = 9.8 m/s2 T = 2π 3 ×0.1472 ×9.8 T = 0.942sb) If the disc is held at a distance r, such that r<R. then the moment of inertia of the disc will be I' = 12MR2 + Mr2 Now the time period will be, T' = 2πMR22+Mr2MgrT' = 2π R2 + 2r22gr according to the question we need to find 'r' such that T = T' 2π3R2g = 2π R2 + 2r22gr3R = R2+2r2rR2 +2r2 -3Rr = 0 above equation is a quadratic equation we can solve it for 'r' r = 3R ±3R2 - 8R24r = 3R ±R4 we get two solutions for 'r' r = 3R+R4 = R , this solution is not accepted as r must be less than R. r = 3R-R4 = R2, this solution for 'r' is accepted so, r = 14.72 = 7.35 cma) T = 0.942 sec b) r = 7.35 cm

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Chapter 15, Problem 072 A uniform circular disk whose radius R is 14.7 cm is suspended as a physical pendulum from a point on its rim. (a) What is its period? (b) At what radial distance r

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Question

Chapter 15, Problem 072
A uniform circular disk whose radius R is 14.7 cm is suspended as a physical pendulum from a point on its rim. (a) What is its period? (b) At what radial distance r < R is there a pivot point that
gives the same period?
(a) Number
Units
(b) Number
Units
Click if you would like to Show Work for this question:
Open Show Work

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