Question 195: Centre of Mass and Rotational Motion

Question

Two masses m and $${m \over 2}$$ are connected at the two ends of a massless rigid rod of length l. The rod is suspended by a thin wire of torsional constant k, at the centre of mass of the rod-mass system(see figure). Because of torsional constant k, the restoring torque is $$	au $$ = k$$	heta $$ for angular displacement $$	heta $$. If the rod is rotated by $$	heta $$0 and released, the tension in it when it passes through its mean position will be :

Accepted Answer

$$ 	herefore $$   $${r_1} = \left( {{{{n \over 2}} \over {m + {m \over 2}}}} ight)l$$
$${r_1} = {l \over 3}$$
When rod is rotated by $$	heta $$
0
and released, then it look this.
After roating $$	heta $$
0
and released, when the rod go through mean position it will have maximum speed.
V
max
$$=$$ A$$\omega $$
$$=$$ r
1
$$	heta $$
0
$$\omega $$
$$=$$ $$\left( {{l \over 3}{	heta _0}} ight)\omega $$
We know,
$$\omega $$ $$ = \sqrt {{K \over {m I}}} $$
and    $${m I} = m{\left( {{l \over 3}} ight)^2} + {m \over 2}{\left( {{{2l} \over 3}} ight)^2}$$
$$ = {{m{l^2}} \over 9} + {{4m{l^2}} \over {18}}$$
$$ = {{2m{l^2} + 4m{l^2}} \over {18}}$$
$$ = {{m{l^2}} \over 3}$$
$$ 	herefore $$   $$\omega  = \sqrt {{{3K} \over {m{l^2}}}} $$
$$ 	herefore $$   Tension in the rod when it passes through the mean position is,
T = $${{mv_{\max }^2} \over {{r_1}}}$$
$$ = {{m{{\left( {{l \over 3}{	heta _0}\omega } ight)}^2}} \over {{l \over 3}}}$$
$$ = {{m{{\left( {{l \over 3}} ight)}^2}	heta _0^2{\omega ^2}} \over {{\ell  \over 3}}}$$
$$ = m\left( {{l \over 3}} ight)	heta _0^2\left( {{{3K} \over {m{l^2}}}} ight)$$
$$ = {{K	heta _0^2} \over l}$$

Answer

$${{3k{	heta _0}^2} \over l}$$

Answer

$${{2k{	heta _0}^2} \over l}$$

Answer

$${{k{	heta _0}^2} \over l}$$

Answer

$${{k{	heta _0}^2} \over {2l}}$$

Centre of Mass and Rotational Motion

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