Question 767: Centre of Mass and Rotational Motion

Question

A ring of mass M and radius R is rotating with angular speed $$\omega$$ about a fixed vertical axis passing through its centre O with two point masses each of mass $${M \over 8}$$ at rest at O. These masses can move radially outwards along two massless rods fixed on the ring as shown in the figure. At some instant, the angular speed of the system is $${8 \over 9}$$$$\omega$$ and one of the masses is at a distance of $${3 \over 5}$$R from O. At this instant, the distance of the other mass from O is

Accepted Answer

Ring has mass M and radius R. Initial angular speed of ring = $$\omega$$. Two point masses, each of mass are at rest at O. Initial angular momentum of ring and point masses system,
$${L_i} = {I_R}\omega  + {I_m}\omega  + {I_m}\omega $$
$$ = M{R^2}\omega  + 0 + 0 = M{R^2}\omega $$
After some time, situation is changed as shown in the figure.
Angular speed of the system, $$\omega ' = {8 \over 9}\omega $$; $$OA = {{3R} \over 5}$$; OB = r = ?
Moment of inertia about O of point mass at A,
$${I_A} = {M \over 8} 	imes {{9{R^2}} \over {25}}$$
Moment of inertia about O of point mass at B, $${I_B} = {M \over 8}{r^2}$$
Fina angular momentum of the system
$${L_f} = M{R^2}\omega ' + {I_A}\omega ' + {I_B}\omega '$$
$$ = M{R^2} 	imes {{8\omega } \over 9} + {M \over 8} 	imes {{9{R^2}} \over {25}} 	imes {{8\omega } \over 9} + {M \over 8}{r^2} 	imes {{8\omega } \over 9}$$
As there is no external torque acting on the system so its angular momentum will be conserved, L
i
= L
f
$$M{R^2}\omega  = M{R^2} 	imes {{8\omega } \over 9} + {M \over 8} 	imes {{9{R^2}} \over {25}} 	imes {{8\omega } \over 9} + {M \over 8}{r^2} 	imes {{8\omega } \over 9}$$
$${R^2} = {{8{R^2}} \over 9} + {{{R^2}} \over {25}} + {{{r^2}} \over 9} \Rightarrow {{{r^2}} \over 9} = {{16} \over {225}}{R^2}$$ $$	herefore$$ $$r = {4 \over 5}R$$

Answer

$${2 \over 3}$$R

Answer

$${1 \over 3}$$R

Answer

$${3 \over 5}$$R

Answer

$${4 \over 5}$$R

Centre of Mass and Rotational Motion

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