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The-Figure-shows-a-system-consisting-of-i-a-ring-of-outer-radius-3R-rolling-clockwise-without-slipping-on-a-horizontal-surface-with-angular-speed-and-ii-an-inner-disc-of-radius-2R-rotating-anti-

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Question Number 20891 by Tinkutara last updated on 06/Sep/17
The Figure shows a system consisting of (i) a ring of outer radius 3R rolling clockwise without slipping on a horizontal surface with angular speed ω and (ii) an inner disc of radius 2R rotating anti-clockwise with angular speed ω/2. The ring and disc are separated by frictionless ball bearing. The system is in the x-z plane. The point P on the inner disc is at a distance R from the origin, where OP makes an angle 30° with the horizontal. Then with respect to the horizontal surface (a) The point O has a linear velocity 3Rωi^∧ (b) The point P has a linear velocity ((11)/4)Rωi^∧ + ((√3)/4)Rωk^∧
$$\mathrm{The}\:\mathrm{Figure}\:\mathrm{shows}\:\mathrm{a}\:\mathrm{system}\:\mathrm{consisting} \\ $$$$\mathrm{of}\:\left({i}\right)\:\mathrm{a}\:\mathrm{ring}\:\mathrm{of}\:\mathrm{outer}\:\mathrm{radius}\:\mathrm{3}{R}\:\mathrm{rolling} \\ $$$$\mathrm{clockwise}\:\mathrm{without}\:\mathrm{slipping}\:\mathrm{on}\:\mathrm{a} \\ $$$$\mathrm{horizontal}\:\mathrm{surface}\:\mathrm{with}\:\mathrm{angular}\:\mathrm{speed} \\ $$$$\omega\:\mathrm{and}\:\left({ii}\right)\:\mathrm{an}\:\mathrm{inner}\:\mathrm{disc}\:\mathrm{of}\:\mathrm{radius}\:\mathrm{2}{R} \\ $$$$\mathrm{rotating}\:\mathrm{anti}-\mathrm{clockwise}\:\mathrm{with}\:\mathrm{angular} \\ $$$$\mathrm{speed}\:\omega/\mathrm{2}.\:\mathrm{The}\:\mathrm{ring}\:\mathrm{and}\:\mathrm{disc}\:\mathrm{are} \\ $$$$\mathrm{separated}\:\mathrm{by}\:\mathrm{frictionless}\:\mathrm{ball}\:\mathrm{bearing}. \\ $$$$\mathrm{The}\:\mathrm{system}\:\mathrm{is}\:\mathrm{in}\:\mathrm{the}\:{x}-{z}\:\mathrm{plane}.\:\mathrm{The} \\ $$$$\mathrm{point}\:{P}\:\mathrm{on}\:\mathrm{the}\:\mathrm{inner}\:\mathrm{disc}\:\mathrm{is}\:\mathrm{at}\:\mathrm{a}\:\mathrm{distance} \\ $$$${R}\:\mathrm{from}\:\mathrm{the}\:\mathrm{origin},\:\mathrm{where}\:{OP}\:\mathrm{makes}\:\mathrm{an} \\ $$$$\mathrm{angle}\:\mathrm{30}°\:\mathrm{with}\:\mathrm{the}\:\mathrm{horizontal}.\:\mathrm{Then} \\ $$$$\mathrm{with}\:\mathrm{respect}\:\mathrm{to}\:\mathrm{the}\:\mathrm{horizontal}\:\mathrm{surface} \\ $$$$\left({a}\right)\:\mathrm{The}\:\mathrm{point}\:{O}\:\mathrm{has}\:\mathrm{a}\:\mathrm{linear}\:\mathrm{velocity} \\ $$$$\mathrm{3}{R}\omega\overset{\wedge} {{i}} \\ $$$$\left({b}\right)\:\mathrm{The}\:\mathrm{point}\:{P}\:\mathrm{has}\:\mathrm{a}\:\mathrm{linear}\:\mathrm{velocity} \\ $$$$\frac{\mathrm{11}}{\mathrm{4}}{R}\omega\overset{\wedge} {{i}}\:+\:\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{R}\omega\overset{\wedge} {{k}} \\ $$
Commented by Tinkutara last updated on 06/Sep/17
Commented by ajfour last updated on 06/Sep/17
Commented by Tinkutara last updated on 06/Sep/17
But since radius at O is 0, so linear velocity = rω = 0(ω) = 0? Why not?
$$\mathrm{But}\:\mathrm{since}\:\mathrm{radius}\:\mathrm{at}\:{O}\:\mathrm{is}\:\mathrm{0},\:\mathrm{so}\:\mathrm{linear} \\ $$$$\mathrm{velocity}\:=\:{r}\omega\:=\:\mathrm{0}\left(\omega\right)\:=\:\mathrm{0}?\:\mathrm{Why}\:\mathrm{not}? \\ $$
Commented by ajfour last updated on 06/Sep/17
let velocity of point O be v_0 . as the bottommost point does not slip its velocity= v_0 −ω(3R)=0 ⇒ v_0 ^→ =3ωRi^� velocity of point P be v_P ; then v_P ^→ =v_0 ^→ +((ωR)/2)(−sin 30°i^� +cos 30°k^� ) =(3ωR−((ωR)/4))i^� +((ωR)/2)×((√3)/2)k^� =((11ωR)/4)i^� +((ωR(√3))/4)k^� . both are true.
$${let}\:{velocity}\:{of}\:{point}\:{O}\:{be}\:{v}_{\mathrm{0}} . \\ $$$${as}\:{the}\:{bottommost}\:{point}\:{does}\:{not} \\ $$$${slip}\:\:{its}\:{velocity}=\:{v}_{\mathrm{0}} −\omega\left(\mathrm{3}{R}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\Rightarrow\:\:\overset{\rightarrow} {{v}}_{\mathrm{0}} =\mathrm{3}\omega{R}\hat {{i}} \\ $$$$\:{velocity}\:{of}\:{point}\:{P}\:{be}\:{v}_{{P}} ;\:{then} \\ $$$$\:\:\overset{\rightarrow} {{v}}_{{P}} =\overset{\rightarrow} {{v}}_{\mathrm{0}} +\frac{\omega{R}}{\mathrm{2}}\left(−\mathrm{sin}\:\mathrm{30}°\hat {{i}}+\mathrm{cos}\:\mathrm{30}°\hat {{k}}\right) \\ $$$$\:\:\:\:\:\:\:=\left(\mathrm{3}\omega{R}−\frac{\omega{R}}{\mathrm{4}}\right)\hat {{i}}+\frac{\omega{R}}{\mathrm{2}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\hat {{k}} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{11}\omega{R}}{\mathrm{4}}\hat {{i}}+\frac{\omega{R}\sqrt{\mathrm{3}}}{\mathrm{4}}\hat {{k}}\:. \\ $$$${both}\:{are}\:{true}. \\ $$
Commented by ajfour last updated on 06/Sep/17
that?would be the case if you hang the wheel in air, like a pulley.
$${that}?{would}\:{be}\:{the}\:{case}\:{if}\:{you} \\ $$$${hang}\:{the}\:{wheel}\:{in}\:{air},\:{like}\:{a}\:{pulley}. \\ $$
Commented by ajfour last updated on 06/Sep/17
rolling is rotating plus sliding..
$${rolling}\:{is}\:{rotating}\:{plus}\:{sliding}.. \\ $$
Commented by Tinkutara last updated on 06/Sep/17
Then why v_O ≠ (2R)((ω/2)) = Rω? Why outer radius is taken only?
$$\mathrm{Then}\:\mathrm{why}\:{v}_{{O}} \:\neq\:\left(\mathrm{2}{R}\right)\left(\frac{\omega}{\mathrm{2}}\right)\:=\:{R}\omega?\:\mathrm{Why} \\ $$$$\mathrm{outer}\:\mathrm{radius}\:\mathrm{is}\:\mathrm{taken}\:\mathrm{only}? \\ $$
Commented by ajfour last updated on 06/Sep/17
owing to rotation the speed is zero, but due to translation it is 3ωR.
$${owing}\:{to}\:{rotation}\:{the}\:{speed}\:{is} \\ $$$${zero},\:{but}\:{due}\:{to}\:{translation}\:{it}\:{is} \\ $$$$\mathrm{3}\omega{R}. \\ $$
Commented by Tinkutara last updated on 06/Sep/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$
Commented by ajfour last updated on 06/Sep/17
only the outer radius is used for v_O because it is the outer ring which is in contact with ground. so we have a relation only of the angular velocity of ring and linear velocity of its center. v_O =3ωR. The disc cam have any angular velocity (here ω/2) ..
$${only}\:{the}\:{outer}\:{radius}\:{is}\:{used}\:{for} \\ $$$${v}_{{O}} \:{because}\:{it}\:{is}\:{the}\:{outer}\:{ring}\: \\ $$$${which}\:{is}\:{in}\:{contact}\:{with}\:{ground}. \\ $$$${so}\:{we}\:{have}\:{a}\:{relation}\:{only}\:{of}\:{the} \\ $$$${angular}\:{velocity}\:{of}\:{ring}\:{and} \\ $$$${linear}\:{velocity}\:{of}\:{its}\:{center}. \\ $$$${v}_{{O}} =\mathrm{3}\omega{R}.\:\:{The}\:{disc}\:{cam}\:{have}\:{any} \\ $$$${angular}\:{velocity}\:\left({here}\:\omega/\mathrm{2}\right)\:.. \\ $$
Commented by Tinkutara last updated on 06/Sep/17
Thanks. Now I understand.
$$\mathrm{Thanks}.\:\mathrm{Now}\:\mathrm{I}\:\mathrm{understand}. \\ $$

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