The question is based on Rotational Mechanics. Find the correct option (s)?

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The question is based on Rotational Mechanics. Find the correct option (s)?

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Answer 1 · 2017-04-06T12:34:33

See below.

Explanation:

We have

$l = \sqrt{24} a$ $l=sqrt(24)a$
$l_{g} = l + \frac{1}{5} l = \frac{6}{5} l$ $l_g=l+1/5 l = 6/5l$
${\to r}_{g} = (0, \frac{l_{g}}{cos α}, 0)$ $vec r_g = (0,l_g/cosalpha,0)$

where $(u, v, w)$ $(u,v,w)$ must be understood as

$u ˆ i + v ˆ j + w ˆ k$ $u hat i + v hat j + w hat k$

Here

$α = arctan (\frac{1}{\sqrt{24}})$ $alpha = arctan(1/sqrt(24))$
$\to ω = (0, cos α, sin α) ω$ $vec omega = (0,cosalpha,sinalpha)omega$
${\to v}_{g} = {\to r}_{g} \times \to ω = \frac{17}{2} a^{2} m ω (0, cos α, sin α)$ $vec v_g = vec r_g xx vec omega = 17/2a^2m omega(0,cosalpha,sinalpha)$

then

$∣ ∣ {\to v}_{g} ∣ ∣ = \frac{17}{2} a^{2} m ω$ $abs(vec v_g)=17/2a^2m omega$

and

(D) $Omega = abs(vec v_g)/abs(vec r_g) = 1/5 a omega$

or

$vec Omega = (0,0,1)Omega$

Now

$vec L_O = J_(omega) vec omega+J_(Omega) vec Omega$

with

$J_(omega)=(ma^2)/2+(4m(2a)^2)/2$

We have also

$vec L_g = J_(omega) vec omega$ and

(B) $norm(vec L_g)=17/2a^2m omega$

and

$J_(Omega)=(ma^2)/4+ml_0^2+((4m)(2a)^2)/4+4m(2l_0)^2$

with $l_0 = l cos alpha$

so

(A) $<< vec L_O, hat k >> = J_(Omega)Omega+J_(omega)<< vec omega, hat k >> = J_(Omega)Omega+J_(omega)omega sin alpha$

and

$norm(vec L_O) = sqrt(<< vec L_O, hat k >>^2+(J_(omega)omega cosalpha)^2 )$

or

(C) $norm(vec L_O) = sqrt((J_(Omega)Omega)^2+2J_(Omega)J_(omega) Omega omega sinalpha+(J_(omega)omega)^2)$

The final numeric results are left to the reader as an exercise.

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The question is based on Rotational Mechanics. Find the correct option (s)?

1 Answer

Explanation:

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