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Question Number 214047 by ajfour last updated on 25/Nov/24
Commented by ajfour last updated on 25/Nov/24
The L shaped rigid scale can turn about the pivot on a frictionless table. The small mass m hits and after rebounding happens to find the other end. Find θ, if α is measured. (Take coefficient of restitution e)
$$\:{The}\:{L}\:{shaped}\:{rigid}\:{scale}\:{can}\:{turn}\: \\ $$$${about}\:{the}\:{pivot}\:{on}\:{a}\:{frictionless}\:{table}. \\ $$$$\:{The}\:{small}\:{mass}\:{m}\:{hits}\:{and}\:{after}\: \\ $$$${rebounding}\:{happens}\:{to}\:{find}\:{the} \\ $$$$\:{other}\:{end}.\:{Find}\:\theta,\:{if}\:\alpha\:{is}\:{measured}. \\ $$$$\left({Take}\:{coefficient}\:{of}\:{restitution}\:\boldsymbol{{e}}\right) \\ $$
Answered by mr W last updated on 25/Nov/24
Commented by mr W last updated on 27/Nov/24
u=velocity of ball before collision v=velocity of ball after collision ω=angular velocity of L−frame after collision u sin θ=v sin φ ...(i) e=((v cos φ+((ωl)/2))/(u cos θ)) ⇒eu cos θ=v cos φ+((lω)/2) ...(ii) mu cos θ×(l/2)=((Ml^2 ω)/3)−mv cos φ×(l/2) ⇒u cos θ+v cos φ=((2Mlω)/(3m)) ...(iii) t=(α/ω) (l/2)−l sin α=v sin φ×t ((lω)/α)((1/2)−sin α)=v sin φ ⇒((1/(2α))−((sin α)/α))lω=u sin θ ...(iv) l cos α=v cos φ ×t ((lω cos α)/α)=v cos φ ...(v) we have 5 equations for 6 variables: u, v, ω, θ, φ, α (iii): u cos θ+((lω cos α)/α)=((2Mlω)/(3m)) (((2M)/(3m))−((cos α)/α))lω=u cos θ ...(vi) (v) into (ii): eu cos θ=((lω cos α)/α)+((lω)/2) ⇒(((cos α)/α)+(1/2))((lω)/e)=u cos θ from (vi): (((cos α)/α)+(1/2))((lω)/e)=(((2M)/(3m))−((cos α)/α))lω (1+(1/e))((cos α)/α)=((2M)/(3m))−(1/(2e)) ⇒((cos α)/α)=((((2M)/(3m))−(1/(2e)))/(1+(1/e))) ...(vii) since α<(π/6), ((((2M)/(3m))−(1/(2e)))/(1+(1/e)))>(((√3)×6)/(2×π)) ⇒ (M/m)>((9(√3)(1+(1/e)))/(2π))+(3/(4e)) (iv)/(vi): (((1/(2α))−((sin α)/α))/(((2M)/(3m))−((cos α)/α)))=tan θ ⇒θ=tan^(−1) (((1/2)−sin α)/(((2αM)/(3m))−cos α)) ...(viii) from (iv): ((lω)/u)=((α sin θ)/((1/2)−sin α)) ...(ix) (i)^2 +(v)^2 : (u sin θ)^2 +(((lω cos α)/α))^2 =v^2 (v^2 /u^2 )=sin^2 θ+(((lω)/u))^2 (((cos α)/α))^2 ⇒(v/u)=sin θ(√(1+((α/((1/2)−sin α)))^2 (((((2M)/(3m))−(1/(2e)))/(1+(1/e))))^2 )) ...(x) (i)/(v): ((u sin θ)/((lω cos α)/α))=tan φ tan φ=((sin θ)/((((lω)/u))(((cos α)/α))))=((((1/2)−sin α)/α))(((1+(1/e))/(((2M)/(3m))−(1/(2e))))) ⇒φ=tan^(−1) ((((1/2)−sin α)/α))(((1+(1/e))/(((2M)/(3m))−(1/(2e))))) ...(xi)
$${u}={velocity}\:{of}\:{ball}\:{before}\:{collision} \\ $$$${v}={velocity}\:{of}\:{ball}\:{after}\:{collision} \\ $$$$\omega={angular}\:{velocity}\:{of}\:{L}−{frame} \\ $$$$\:\:\:\:\:\:\:{after}\:{collision} \\ $$$${u}\:\mathrm{sin}\:\theta={v}\:\mathrm{sin}\:\phi\:\:\:…\left({i}\right) \\ $$$${e}=\frac{{v}\:\mathrm{cos}\:\phi+\frac{\omega{l}}{\mathrm{2}}}{{u}\:\mathrm{cos}\:\theta} \\ $$$$\Rightarrow{eu}\:\mathrm{cos}\:\theta={v}\:\mathrm{cos}\:\phi+\frac{{l}\omega}{\mathrm{2}}\:\:\:…\left({ii}\right) \\ $$$${mu}\:\mathrm{cos}\:\theta×\frac{{l}}{\mathrm{2}}=\frac{{Ml}^{\mathrm{2}} \omega}{\mathrm{3}}−{mv}\:\mathrm{cos}\:\phi×\frac{{l}}{\mathrm{2}} \\ $$$$\Rightarrow{u}\:\mathrm{cos}\:\theta+{v}\:\mathrm{cos}\:\phi=\frac{\mathrm{2}{Ml}\omega}{\mathrm{3}{m}}\:\:\:…\left({iii}\right) \\ $$$${t}=\frac{\alpha}{\omega} \\ $$$$\frac{{l}}{\mathrm{2}}−{l}\:\mathrm{sin}\:\alpha={v}\:\mathrm{sin}\:\phi×{t} \\ $$$$\frac{{l}\omega}{\alpha}\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{sin}\:\alpha\right)={v}\:\mathrm{sin}\:\phi \\ $$$$\Rightarrow\left(\frac{\mathrm{1}}{\mathrm{2}\alpha}−\frac{\mathrm{sin}\:\alpha}{\alpha}\right){l}\omega={u}\:\mathrm{sin}\:\theta\:\:\:…\left({iv}\right) \\ $$$${l}\:\mathrm{cos}\:\alpha={v}\:\mathrm{cos}\:\phi\:×{t} \\ $$$$\frac{{l}\omega\:\mathrm{cos}\:\alpha}{\alpha}={v}\:\mathrm{cos}\:\phi\:\:\:…\left({v}\right) \\ $$$${we}\:{have}\:\mathrm{5}\:{equations}\:{for}\:\mathrm{6}\:{variables}: \\ $$$${u},\:{v},\:\omega,\:\theta,\:\phi,\:\alpha \\ $$$$ \\ $$$$\left({iii}\right): \\ $$$${u}\:\mathrm{cos}\:\theta+\frac{{l}\omega\:\mathrm{cos}\:\alpha}{\alpha}=\frac{\mathrm{2}{Ml}\omega}{\mathrm{3}{m}} \\ $$$$\left(\frac{\mathrm{2}{M}}{\mathrm{3}{m}}−\frac{\mathrm{cos}\:\alpha}{\alpha}\right){l}\omega={u}\:\mathrm{cos}\:\theta\:\:\:…\left({vi}\right) \\ $$$$\left({v}\right)\:{into}\:\left({ii}\right): \\ $$$${eu}\:\mathrm{cos}\:\theta=\frac{{l}\omega\:\mathrm{cos}\:\alpha}{\alpha}+\frac{{l}\omega}{\mathrm{2}} \\ $$$$\Rightarrow\left(\frac{\mathrm{cos}\:\alpha}{\alpha}+\frac{\mathrm{1}}{\mathrm{2}}\right)\frac{{l}\omega}{{e}}={u}\:\mathrm{cos}\:\theta \\ $$$${from}\:\left({vi}\right): \\ $$$$\left(\frac{\mathrm{cos}\:\alpha}{\alpha}+\frac{\mathrm{1}}{\mathrm{2}}\right)\frac{{l}\omega}{{e}}=\left(\frac{\mathrm{2}{M}}{\mathrm{3}{m}}−\frac{\mathrm{cos}\:\alpha}{\alpha}\right){l}\omega \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{{e}}\right)\frac{\mathrm{cos}\:\alpha}{\alpha}=\frac{\mathrm{2}{M}}{\mathrm{3}{m}}−\frac{\mathrm{1}}{\mathrm{2}{e}} \\ $$$$\Rightarrow\frac{\mathrm{cos}\:\alpha}{\alpha}=\frac{\frac{\mathrm{2}{M}}{\mathrm{3}{m}}−\frac{\mathrm{1}}{\mathrm{2}{e}}}{\mathrm{1}+\frac{\mathrm{1}}{{e}}}\:\:\:…\left({vii}\right) \\ $$$${since}\:\alpha<\frac{\pi}{\mathrm{6}},\:\:\frac{\frac{\mathrm{2}{M}}{\mathrm{3}{m}}−\frac{\mathrm{1}}{\mathrm{2}{e}}}{\mathrm{1}+\frac{\mathrm{1}}{{e}}}>\frac{\sqrt{\mathrm{3}}×\mathrm{6}}{\mathrm{2}×\pi} \\ $$$$\Rightarrow\:\frac{{M}}{{m}}>\frac{\mathrm{9}\sqrt{\mathrm{3}}\left(\mathrm{1}+\frac{\mathrm{1}}{{e}}\right)}{\mathrm{2}\pi}+\frac{\mathrm{3}}{\mathrm{4}{e}} \\ $$$$\left({iv}\right)/\left({vi}\right): \\ $$$$\frac{\frac{\mathrm{1}}{\mathrm{2}\alpha}−\frac{\mathrm{sin}\:\alpha}{\alpha}}{\frac{\mathrm{2}{M}}{\mathrm{3}{m}}−\frac{\mathrm{cos}\:\alpha}{\alpha}}=\mathrm{tan}\:\theta \\ $$$$\Rightarrow\theta=\mathrm{tan}^{−\mathrm{1}} \frac{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{sin}\:\alpha}{\frac{\mathrm{2}\alpha{M}}{\mathrm{3}{m}}−\mathrm{cos}\:\alpha}\:\:\:…\left({viii}\right) \\ $$$${from}\:\left({iv}\right): \\ $$$$\frac{{l}\omega}{{u}}=\frac{\alpha\:\mathrm{sin}\:\theta}{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{sin}\:\alpha}\:\:\:…\left({ix}\right) \\ $$$$\left({i}\right)^{\mathrm{2}} +\left({v}\right)^{\mathrm{2}} : \\ $$$$\left({u}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} +\left(\frac{{l}\omega\:\mathrm{cos}\:\alpha}{\alpha}\right)^{\mathrm{2}} ={v}^{\mathrm{2}} \\ $$$$\frac{{v}^{\mathrm{2}} }{{u}^{\mathrm{2}} }=\mathrm{sin}^{\mathrm{2}} \:\theta+\left(\frac{{l}\omega}{{u}}\right)^{\mathrm{2}} \left(\frac{\mathrm{cos}\:\alpha}{\alpha}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\frac{{v}}{{u}}=\mathrm{sin}\:\theta\sqrt{\mathrm{1}+\left(\frac{\alpha}{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{sin}\:\alpha}\right)^{\mathrm{2}} \left(\frac{\frac{\mathrm{2}{M}}{\mathrm{3}{m}}−\frac{\mathrm{1}}{\mathrm{2}{e}}}{\mathrm{1}+\frac{\mathrm{1}}{{e}}}\right)^{\mathrm{2}} }\:\:\:…\left({x}\right) \\ $$$$\left({i}\right)/\left({v}\right): \\ $$$$\frac{{u}\:\mathrm{sin}\:\theta}{\frac{{l}\omega\:\mathrm{cos}\:\alpha}{\alpha}}=\mathrm{tan}\:\phi \\ $$$$\mathrm{tan}\:\phi=\frac{\mathrm{sin}\:\theta}{\left(\frac{{l}\omega}{{u}}\right)\left(\frac{\mathrm{cos}\:\alpha}{\alpha}\right)}=\left(\frac{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{sin}\:\alpha}{\alpha}\right)\left(\frac{\mathrm{1}+\frac{\mathrm{1}}{{e}}}{\frac{\mathrm{2}{M}}{\mathrm{3}{m}}−\frac{\mathrm{1}}{\mathrm{2}{e}}}\right)\: \\ $$$$\Rightarrow\phi=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{sin}\:\alpha}{\alpha}\right)\left(\frac{\mathrm{1}+\frac{\mathrm{1}}{{e}}}{\frac{\mathrm{2}{M}}{\mathrm{3}{m}}−\frac{\mathrm{1}}{\mathrm{2}{e}}}\right)\:\:\:…\left({xi}\right) \\ $$
Commented by mr W last updated on 27/Nov/24
example: (M/m)=8, e=0.75 ⇒α≈0.3370≈19.31° ⇒θ≈0.1958≈11.22° ⇒φ≈0.2462≈14.11° ⇒(v/u)≈0.7986 ⇒((lω)/u)≈0.3873
$${example}: \\ $$$$\frac{{M}}{{m}}=\mathrm{8},\:{e}=\mathrm{0}.\mathrm{75} \\ $$$$\Rightarrow\alpha\approx\mathrm{0}.\mathrm{3370}\approx\mathrm{19}.\mathrm{31}° \\ $$$$\Rightarrow\theta\approx\mathrm{0}.\mathrm{1958}\approx\mathrm{11}.\mathrm{22}° \\ $$$$\Rightarrow\phi\approx\mathrm{0}.\mathrm{2462}\approx\mathrm{14}.\mathrm{11}° \\ $$$$\Rightarrow\frac{{v}}{{u}}\approx\mathrm{0}.\mathrm{7986} \\ $$$$\Rightarrow\frac{{l}\omega}{{u}}\approx\mathrm{0}.\mathrm{3873} \\ $$
Commented by ajfour last updated on 26/Nov/24
s^2 =l^2 +(l^2 /4)−(l^2 /2)sin α s^2 =(l^2 /4)(3−2sin α)=v^2 ((α^2 /ω^2 )) ⇒ (s/(αl))=(v/(ωl))=((√(3−2sin α))/(2α)) mu((l/2)cos θ)=((Ml^2 ω)/3)−mv((l/2))cos φ usin θ=vsin φ & m((u/v))((l/2)cos θ)=((Ml^2 )/3)((ω/v))−mcos φ((l/2)) ((sin φcos θ)/(sin θ))+cos φ=((2M)/(3m))(((αl)/s)) cot θ((l/2)−lsin α)+lcos α=(((2M)/(3m)))αl cot θ=(((((2M)/(3m)))α−cos α)/((1/2)−sin α)) tan θ = (((1/2)−sin α)/((((2M)/(3m)))α−cos α))
$${s}^{\mathrm{2}} ={l}^{\mathrm{2}} +\frac{{l}^{\mathrm{2}} }{\mathrm{4}}−\frac{{l}^{\mathrm{2}} }{\mathrm{2}}\mathrm{sin}\:\alpha \\ $$$${s}^{\mathrm{2}} =\frac{{l}^{\mathrm{2}} }{\mathrm{4}}\left(\mathrm{3}−\mathrm{2sin}\:\alpha\right)={v}^{\mathrm{2}} \left(\frac{\alpha^{\mathrm{2}} }{\omega^{\mathrm{2}} }\right) \\ $$$$\Rightarrow\:\:\frac{{s}}{\alpha{l}}=\frac{{v}}{\omega{l}}=\frac{\sqrt{\mathrm{3}−\mathrm{2sin}\:\alpha}}{\mathrm{2}\alpha} \\ $$$${mu}\left(\frac{{l}}{\mathrm{2}}\mathrm{cos}\:\theta\right)=\frac{{Ml}^{\mathrm{2}} \omega}{\mathrm{3}}−{mv}\left(\frac{{l}}{\mathrm{2}}\right)\mathrm{cos}\:\phi \\ $$$${u}\mathrm{sin}\:\theta={v}\mathrm{sin}\:\phi \\ $$$$\& \\ $$$${m}\left(\frac{{u}}{{v}}\right)\left(\frac{{l}}{\mathrm{2}}\mathrm{cos}\:\theta\right)=\frac{{Ml}^{\mathrm{2}} }{\mathrm{3}}\left(\frac{\omega}{{v}}\right)−{m}\mathrm{cos}\:\phi\left(\frac{{l}}{\mathrm{2}}\right) \\ $$$$\frac{\mathrm{sin}\:\phi\mathrm{cos}\:\theta}{\mathrm{sin}\:\theta}+\mathrm{cos}\:\phi=\frac{\mathrm{2}{M}}{\mathrm{3}{m}}\left(\frac{\alpha{l}}{{s}}\right) \\ $$$$\mathrm{cot}\:\theta\left(\frac{{l}}{\mathrm{2}}−{l}\mathrm{sin}\:\alpha\right)+{l}\mathrm{cos}\:\alpha=\left(\frac{\mathrm{2}{M}}{\mathrm{3}{m}}\right)\alpha{l} \\ $$$$\mathrm{cot}\:\theta=\frac{\left(\frac{\mathrm{2}{M}}{\mathrm{3}{m}}\right)\alpha−\mathrm{cos}\:\alpha}{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{sin}\:\alpha} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{tan}\:\theta\:=\:\frac{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{sin}\:\alpha}{\left(\frac{\mathrm{2}{M}}{\mathrm{3}{m}}\right)\alpha−\mathrm{cos}\:\alpha}\:\:\: \\ $$
Commented by ajfour last updated on 26/Nov/24
yeah! And I got the same, another way, Sir.
Commented by mr W last updated on 26/Nov/24
can you please check following: α is not independent. it is given through ((cos α)/α)=((((2M)/(3m))−(1/(2e)))/(1+(1/e)))
$${can}\:{you}\:{please}\:{check}\:{following}: \\ $$$$\alpha\:{is}\:{not}\:{independent}.\:{it}\:{is}\:{given} \\ $$$${through} \\ $$$$\frac{\mathrm{cos}\:\alpha}{\alpha}=\frac{\frac{\mathrm{2}{M}}{\mathrm{3}{m}}−\frac{\mathrm{1}}{\mathrm{2}{e}}}{\mathrm{1}+\frac{\mathrm{1}}{{e}}} \\ $$
Commented by ajfour last updated on 26/Nov/24
hinge reaction is also there, i think frame is not free, so this should not apply.
$${hinge}\:{reaction}\:{is}\:{also}\:{there},\:{i}\:{think} \\ $$$${frame}\:{is}\:{not}\:{free},\:{so}\:{this}\:{should}\:{not} \\ $$$${apply}. \\ $$
Commented by mr W last updated on 26/Nov/24
see above
$${see}\:{above} \\ $$
Commented by ajfour last updated on 27/Nov/24
Nice sir, e decides all angles for this case!
$${Nice}\:{sir},\:{e}\:{decides}\:{all}\:{angles}\: \\ $$$${for}\:{this}\:{case}! \\ $$
Commented by mr W last updated on 27/Nov/24
yes! beside it′s possible only if (M/m)>((9(√3)(1+1/e))/(3π))+(3/(4e))
$${yes}! \\ $$$${beside}\:{it}'{s}\:{possible}\:{only}\:{if} \\ $$$$\frac{{M}}{{m}}>\frac{\mathrm{9}\sqrt{\mathrm{3}}\left(\mathrm{1}+\mathrm{1}/{e}\right)}{\mathrm{3}\pi}+\frac{\mathrm{3}}{\mathrm{4}{e}} \\ $$

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