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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 25/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 5 variables: u, v, θ, φ, ω (iii): u cos θ+((lω cos α)/α)=((2Mlω)/(3m)) (((2M)/(3m))−((cos α)/α))lω=u cos θ ...(vi) (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)) +++++++++++++++++++
$${u}={velocity}\:{of}\:{ball}\:{before}\:{collision} \\ $$$${v}={velocity}\:{of}\:{ball}\:{after}\:{collision} \\ $$$$\omega={angular}\:{velocity}\:{of}\:{L}−{frame} \\ $$$$\:\:\:\:\:\:\:{after}\:{collision} \\ $$$${u}\:\mathrm{sin}\:\theta={v}\:\mathrm{sin}\:\phi\:\:\:…\left({i}\right) \\ $$$${e}=\frac{{v}\:\mathrm{cos}\:\phi+\frac{\omega{l}}{\mathrm{2}}}{{u}\:\mathrm{cos}\:\theta} \\ $$$$\Rightarrow{eu}\:\mathrm{cos}\:\theta={v}\:\mathrm{cos}\:\phi+\frac{{l}\omega}{\mathrm{2}}\:\:\:…\left({ii}\right) \\ $$$${mu}\:\mathrm{cos}\:\theta×\frac{{l}}{\mathrm{2}}=\frac{{Ml}^{\mathrm{2}} \omega}{\mathrm{3}}−{mv}\:\mathrm{cos}\:\phi×\frac{{l}}{\mathrm{2}} \\ $$$$\Rightarrow{u}\:\mathrm{cos}\:\theta+{v}\:\mathrm{cos}\:\phi=\frac{\mathrm{2}{Ml}\omega}{\mathrm{3}{m}}\:\:\:…\left({iii}\right) \\ $$$${t}=\frac{\alpha}{\omega} \\ $$$$\frac{{l}}{\mathrm{2}}−{l}\:\mathrm{sin}\:\alpha={v}\:\mathrm{sin}\:\phi×{t} \\ $$$$\frac{{l}\omega}{\alpha}\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{sin}\:\alpha\right)={v}\:\mathrm{sin}\:\phi \\ $$$$\Rightarrow\left(\frac{\mathrm{1}}{\mathrm{2}\alpha}−\frac{\mathrm{sin}\:\alpha}{\alpha}\right){l}\omega={u}\:\mathrm{sin}\:\theta\:\:\:…\left({iv}\right) \\ $$$${l}\:\mathrm{cos}\:\alpha={v}\:\mathrm{cos}\:\phi\:×{t} \\ $$$$\frac{{l}\omega\:\mathrm{cos}\:\alpha}{\alpha}={v}\:\mathrm{cos}\:\phi\:\:\:…\left({v}\right) \\ $$$${we}\:{have}\:\mathrm{5}\:{equations}\:{for}\:\mathrm{5}\:{variables}: \\ $$$${u},\:{v},\:\theta,\:\phi,\:\omega \\ $$$$ \\ $$$$\left({iii}\right): \\ $$$${u}\:\mathrm{cos}\:\theta+\frac{{l}\omega\:\mathrm{cos}\:\alpha}{\alpha}=\frac{\mathrm{2}{Ml}\omega}{\mathrm{3}{m}} \\ $$$$\left(\frac{\mathrm{2}{M}}{\mathrm{3}{m}}−\frac{\mathrm{cos}\:\alpha}{\alpha}\right){l}\omega={u}\:\mathrm{cos}\:\theta\:\:\:…\left({vi}\right) \\ $$$$\left({iv}\right)/\left({vi}\right): \\ $$$$\frac{\frac{\mathrm{1}}{\mathrm{2}\alpha}−\frac{\mathrm{sin}\:\alpha}{\alpha}}{\frac{\mathrm{2}{M}}{\mathrm{3}{m}}−\frac{\mathrm{cos}\:\alpha}{\alpha}}=\mathrm{tan}\:\theta \\ $$$$\Rightarrow\theta=\mathrm{tan}^{−\mathrm{1}} \frac{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{sin}\:\alpha}{\frac{\mathrm{2}\alpha{M}}{\mathrm{3}{m}}−\mathrm{cos}\:\alpha} \\ $$$$+++++++++++++++++++ \\ $$
Commented by mr W last updated on 25/Nov/24
i′m wondering that θ is independent from e.
$${i}'{m}\:{wondering}\:{that}\:\theta\:{is}\:{independent} \\ $$$${from}\:{e}. \\ $$

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