| Grade Summary Submissions A uniform rod of mass M and length L is pivoted at d
ID: 1402923 • Letter: #
Question
|
Grade Summary
Submissions
A uniform rod of mass M and length L is pivoted at distance x from its center and undergoes harmonic oscillations.Randomized VariablesM = 3.4 kg
L = 1.7 m
x = 0.43 m 33% Part (a) In terms of M, L, and x, what is the rod’s moment of inertia I about the pivot point. I =
|
a d g h j k L M n P t x ( ) 7 8 9 HOME ^ ^ 4 5 6 / * 1 2 3 + - 0 . END () BACKSPACE DEL CLEARGrade Summary
Deductions 0% Potential 100%Submissions
Attempts remaining: 5(0% per attempt) detailed view Hints: 1% deduction per hint. Hints remaining: 2 Feedback: 1% deduction per feedback. 33% Part (b) Calculate the rod’s period T in seconds for small oscillations about its pivot point.33% Part (c) In terms of L, find an expression for the distance xm for which the period is a minimum.
Explanation / Answer
Part a)
we know that from centre of rod :
I cm = (1/12) ML^2
Apply parallel axis theorem, to find moment of inerta at a distance x from centre
I = (1/12) ML^2 + Mx^2 <----Answer
part b)
T= 2pi * sqrt (I/MgL)
I cm = (1/12) ML^2 = (1/12)*3.4*(1.7)^2 = 0.82 Kgm^2
Apply parallel axis theorem, to find moment of inerta at a distance x from centre
I = Icm + Mx^2
= 0.82 + 3.4*(0.43)^2
= 1.45 Kgm^2
Putting values:
T= 2pi * sqrt (I/MgL)
= 2pi* sqrt(1.45/3.4*9.8*1.7)
= 1 second
Answer: 1 sec
part c)
T= 2pi * sqrt (I/MgL)
clearly period is minimum if I is minimum.
I will be minimum at centre.
so, x = 0 m
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.