Physics Help. Answer please?

[匿名]

2015-12-10 4:55 pm

A long thin uniform rod of length 1.50 m is to be suspended from a frictionless pivot located at some point along the rod so that its pendulum motion takes 3.00 s. How far from the center of the rod should the pivot be located?
8.73 cm
8.40 cm
23.4 cm
7.98 cm
7.52 cm

回答 (1)

NCS

2015-12-10 10:39 pm

✔ 最佳答案

The equivalent length Leq = I / md
where I is the MoI about the pivot point
and m is the mass
and d is the distance from the CoM to the pivot point

period T = 2π√(Leq / g)
so for a period of 3.00 s, we need
3.00 s = 2π√(Leq / 9.8m/s²)
Leq = 2.234 m

By the parallel axis theorem,
I = Icm + md² = mL²/12 + md², so
2.234 m = (mL²/12 + md²) / md = L²/12d + d → multiply by 12d
and drop units for ease:
26.8d = L² + 12d²
Substitute L = 1.50 and rearrange
12d² - 26.8d + 2.25 = 0
quadratic with solutions at
d = 0.0874 m = 8.74 cm ◄ solution
and d = 2.15 m ← does not fall on the rod

Hope this helps!

參考: http://www.wolframalpha.com/input/?i=12d%C2%B2+-+26.8d+%2B+2.25+%3D+0&a=*MC.d^2-_*Variable-&a=UnitClash_*d.*Days.dflt--

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