Question

a,b,c please

(25%) Problem 4: A uniform rod of mass M and length L is free to swing back and forth by pivoting a distance x from its center. It undergoes harmonic oscillations by swinging back and forth under the influence of gravity. Randomized Variables M= 2.4 kg L=1.6 m x=0.43 m 33% Part (a) In terms of M, L, and x, what is the rod's moment of inertia / about the pivot point. A 33% Part (b) Calculate the rod's period T in seconds for small oscillations about its pivot point. A 33% Part (c) In terms of L, find an expression for the distance xm for which the period is a minimum. Grade Summary Deductions 0% 100% 0 ( 7 8 9 HOME ^ 4 5 6 * 1 2 3 + - 0 . END VO BACKSPACE DEL CLEAR Submissions Attempts remaining: 30 (2% per attempt) detailed view M. 0 Submit Hint Feedback I give up! Hints: 1% deduction per hint. Hints remaining: 3 Feedback: 0% deduction per feedback.

Answer 1

Here we apply concept of moment of inertia of solid body and parallel axis theorem as well as concept of simple harmonic motion of solid body.

Por May = Length = 2, M=2.4kg L=1-6m Position of pivot from center, x = 0.43m de Moment of Inertia of t e e red about center of mass tom 2 Ip, fem Ilainy Parallel cases theoxen fp = Tcm + Mx² - Ip = M22 Mx2 | tp = m ( +22 Ipo ) Let rod is making amall angle o with vertical Torque about pivat point ng I p = mg & Sino fox_elmall angle ', sino no Now, we know, T = Pp a Here a = angular acceleration _To wsimple harmonic mafione وم علوم تعليقي دور » Ip W²o = mga o عقاد قلد

Time Period, T = 24 - 25 le To max IT=an M (22/2+222 =21/(22 +1282) J Mga v 12g & s T=21 [cl•6m)?+1266.73m)2] 12 (9.8m/s2)60:43m). T= 1.932 sec » (es for T to be minimum, &I za JX

+ना +12x4 (1292) 27 dT = 4+2_d (12+12x2) ____dx (129X) 20 + 12 ---+12 =0