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Question Number 200137 by ajfour last updated on 14/Nov/23
Commented by ajfour last updated on 14/Nov/23
Find range of M (x coordinate of point where it hits x axis), if 2M is dropped from height h and hits left gear elastically.
$${Find}\:{range}\:{of}\:{M}\:\left({x}\:{coordinate}\:{of}\right. \\ $$$$\left.{point}\:{where}\:{it}\:{hits}\:{x}\:{axis}\right),\:{if}\:\mathrm{2}{M}\:{is} \\ $$$${dropped}\:{from}\:{height}\:{h}\:{and}\:{hits} \\ $$$${left}\:{gear}\:{elastically}. \\ $$
Commented by mr W last updated on 14/Nov/23
distance from 2M to left gear is ...?
$${distance}\:{from}\:\mathrm{2}{M}\:{to}\:{left}\:{gear}\:{is}\:…? \\ $$
Commented by ajfour last updated on 14/Nov/23
right at periphery, sir..
$${right}\:{at}\:{periphery},\:{sir}.. \\ $$
Commented by ajfour last updated on 15/Nov/23
Commented by ajfour last updated on 16/Nov/23
If impulse J acts for time △t, then (J−F△t)R=I_1 ω_1 (F△t)r=I_2 ω_2 ⇒ ((I_1 ω_1 )/R)+((I_2 ω_2 )/r)=J but ω_1 R=ω_2 r ⇒ ω_2 (((I_1 r)/R^2 )+(I_2 /r))=J=(2M)((√(2gh))+v) (√(2gh))=v+ω_1 (2a) ⇒ v+(√(2gh))=2(√(2gh))−2ω_1 a ⇒ ω_2 (((I_1 r)/R^2 )+(I_2 /r))=J=4M((√(2gh))−((ω_2 a)/2)) if I_1 =(1/2)(8M)(2a)^2 and I_2 =(1/2)(4M)a^2 +Ma^2 then ω_2 (4Ma+3Ma+2Ma)=4M(√(2gh)) ω_2 =((4(√(2gh)))/(9a)) say ω_2 a=((4(√(2gh)))/9)=u u_y =(u/2) v_x =((√3)/2)u y=xtan α−((gx^2 )/(2u^2 cos^2 α)) y=−acos α=−((a(√3))/2) −((a(√3))/2)=(x/( (√3)))−((2gx^2 )/(3u^2 )) (((27)/(16h)))x^2 −(x/( (√3)))−((a(√3))/2)=0 Range=x−a(1+sin α) for h=4 , a=1, α=30° ((27(√3)x^2 )/(64))−x−(3/2)=0 x=((32)/(27(√3)))+(√(((1024)/(3×729))+((3×64)/(2×27(√3))))) =((32)/(27(√3))){1+(√(1+((81(√3))/(32))))} Range=x−(a+(a/2))=x−(3/2) ≈ 0.772
$${If}\:{impulse}\:{J}\:{acts}\:{for}\:{time}\:\bigtriangleup{t},\:{then} \\ $$$$\left({J}−{F}\bigtriangleup{t}\right){R}={I}_{\mathrm{1}} \omega_{\mathrm{1}} \\ $$$$\left({F}\bigtriangleup{t}\right){r}={I}_{\mathrm{2}} \omega_{\mathrm{2}} \\ $$$$\Rightarrow\:\:\frac{{I}_{\mathrm{1}} \omega_{\mathrm{1}} }{{R}}+\frac{{I}_{\mathrm{2}} \omega_{\mathrm{2}} }{{r}}={J} \\ $$$${but}\:\:\omega_{\mathrm{1}} {R}=\omega_{\mathrm{2}} {r} \\ $$$$\Rightarrow\:\:\omega_{\mathrm{2}} \left(\frac{{I}_{\mathrm{1}} {r}}{{R}^{\mathrm{2}} }+\frac{{I}_{\mathrm{2}} }{{r}}\right)={J}=\left(\mathrm{2}{M}\right)\left(\sqrt{\mathrm{2}{gh}}+{v}\right) \\ $$$$\sqrt{\mathrm{2}{gh}}={v}+\omega_{\mathrm{1}} \left(\mathrm{2}{a}\right) \\ $$$$\Rightarrow\:\:{v}+\sqrt{\mathrm{2}{gh}}=\mathrm{2}\sqrt{\mathrm{2}{gh}}−\mathrm{2}\omega_{\mathrm{1}} {a} \\ $$$$\Rightarrow\:\:\omega_{\mathrm{2}} \left(\frac{{I}_{\mathrm{1}} {r}}{{R}^{\mathrm{2}} }+\frac{{I}_{\mathrm{2}} }{{r}}\right)={J}=\mathrm{4}{M}\left(\sqrt{\mathrm{2}{gh}}−\frac{\omega_{\mathrm{2}} {a}}{\mathrm{2}}\right) \\ $$$$ \\ $$$${if}\:\:\:{I}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{8}{M}\right)\left(\mathrm{2}{a}\right)^{\mathrm{2}} \\ $$$${and}\:\:{I}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{4}{M}\right){a}^{\mathrm{2}} \:+{Ma}^{\mathrm{2}} \:\:\:{then} \\ $$$$\omega_{\mathrm{2}} \left(\mathrm{4}{Ma}+\mathrm{3}{Ma}+\mathrm{2}{Ma}\right)=\mathrm{4}{M}\sqrt{\mathrm{2}{gh}} \\ $$$$\omega_{\mathrm{2}} =\frac{\mathrm{4}\sqrt{\mathrm{2}{gh}}}{\mathrm{9}{a}}\:\:\:{say}\:\omega_{\mathrm{2}} {a}=\frac{\mathrm{4}\sqrt{\mathrm{2}{gh}}}{\mathrm{9}}={u} \\ $$$${u}_{{y}} =\frac{{u}}{\mathrm{2}}\:\:\:\:\:{v}_{{x}} =\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{u} \\ $$$${y}={x}\mathrm{tan}\:\alpha−\frac{{gx}^{\mathrm{2}} }{\mathrm{2}{u}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \alpha} \\ $$$${y}=−{a}\mathrm{cos}\:\alpha=−\frac{{a}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$−\frac{{a}\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{{x}}{\:\sqrt{\mathrm{3}}}−\frac{\mathrm{2}{gx}^{\mathrm{2}} }{\mathrm{3}{u}^{\mathrm{2}} } \\ $$$$\left(\frac{\mathrm{27}}{\mathrm{16}{h}}\right){x}^{\mathrm{2}} −\frac{{x}}{\:\sqrt{\mathrm{3}}}−\frac{{a}\sqrt{\mathrm{3}}}{\mathrm{2}}=\mathrm{0} \\ $$$${Range}={x}−{a}\left(\mathrm{1}+\mathrm{sin}\:\alpha\right)\:\: \\ $$$${for}\:{h}=\mathrm{4}\:\:,\:{a}=\mathrm{1},\:\alpha=\mathrm{30}° \\ $$$$\frac{\mathrm{27}\sqrt{\mathrm{3}}{x}^{\mathrm{2}} }{\mathrm{64}}−{x}−\frac{\mathrm{3}}{\mathrm{2}}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{32}}{\mathrm{27}\sqrt{\mathrm{3}}}+\sqrt{\frac{\mathrm{1024}}{\mathrm{3}×\mathrm{729}}+\frac{\mathrm{3}×\mathrm{64}}{\mathrm{2}×\mathrm{27}\sqrt{\mathrm{3}}}} \\ $$$$\:\:=\frac{\mathrm{32}}{\mathrm{27}\sqrt{\mathrm{3}}}\left\{\mathrm{1}+\sqrt{\mathrm{1}+\frac{\mathrm{81}\sqrt{\mathrm{3}}}{\mathrm{32}}}\right\} \\ $$$${Range}={x}−\left({a}+\frac{{a}}{\mathrm{2}}\right)={x}−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\:\:\approx\:\mathrm{0}.\mathrm{772}\: \\ $$
Answered by mr W last updated on 17/Nov/23
Commented by ajfour last updated on 16/Nov/23
Thanks sir, great persistence! i was commiting lot errors...
$${Thanks}\:{sir},\:{great}\:{persistence}!\:{i}\:{was} \\ $$$${commiting}\:{lot}\:{errors}… \\ $$
Commented by mr W last updated on 17/Nov/23
thanks for confirming sir!
$${thanks}\:{for}\:{confirming}\:{sir}! \\ $$
Commented by mr W last updated on 17/Nov/23
Commented by mr W last updated on 17/Nov/23
Commented by mr W last updated on 17/Nov/23
collision duration Δt v_1 =(√(2gh)) I_1 =((8M×(2a)^2 )/2)=16Ma^2 I_2 =((4Ma^2 )/2)+Ma^2 =3Ma^2 2M×((v_2 +v_1 )/(Δt))=F_1 I_1 (ω_1 /(Δt))=(F_1 −F_2 )R I_2 (ω_2 /(Δt))=F_2 r ω_1 R=ω_2 r ⇒ω_1 =((rω_2 )/R)=(ω_2 /2) I_1 ω_1 =(F_1 Δt−F_2 Δt)R I_1 ω_1 =2M(v_2 +v_1 )R−((I_2 ω_2 R)/r) 2M(v_2 +v_1 )R=I_1 ω_1 +((I_2 ω_2 R)/r) ⇒2M(v_2 +v_1 )R=(((I_1 r)/R)+((I_2 R)/r))ω_2 ⇒2M(v_2 +v_1 )(2a)=(16Ma^2 ×(1/2)+3Ma^2 ×2)ω_2 ⇒v_2 =((7aω_2 )/2)−v_1 ((2Mv_1 ^2 )/2)=((2Mv_2 ^2 )/2)+((I_1 ω_1 ^2 )/2)+((I_2 ω_2 ^2 )/2) ⇒v_1 ^2 =v_2 ^2 +(7/2)a^2 ω_2 ^2 ⇒v_1 ^2 =(((7aω_2 )/2)−v_1 )^2 +(7/2)a^2 ω_2 ^2 ⇒((9aω_2 )/4)−v_1 =0 ⇒ω_2 =((4v_1 )/(9a)) u=ω_2 a=((4v_1 )/9)=((4(√(2gh)))/9) v_2 =(7/2)×((4v_1 )/9)−v_1 =((5v_1 )/9) ⇒rebound height=(((√5)h)/3)≈((3h)/4) −a cos θ=u sin θ t−((gt^2 )/2) ((gt^2 )/2)−u sin θ t−a cos θ=0 ⇒t=((u sin θ+(√(u^2 sin^2 θ+2ga cos θ)))/g) Range+a+a sin θ=u cos θ t ⇒Range=u cos θ t−a(1+sin θ) example: θ=30°, a=1m, h=4m u=((4(√(2×10×4)))/9)=((16(√5))/9) t=((((16(√5))/9)×(1/2)+(√(((16^2 ×5)/9^2 )×(1/4)+2×10×1×((√3)/2))))/(10))≈0.659968 R=((16(√5))/9)×((√3)/2)×0.659968−1×(1+(1/2))≈0.772 m
$${collision}\:{duration}\:\Delta{t} \\ $$$${v}_{\mathrm{1}} =\sqrt{\mathrm{2}{gh}} \\ $$$${I}_{\mathrm{1}} =\frac{\mathrm{8}{M}×\left(\mathrm{2}{a}\right)^{\mathrm{2}} }{\mathrm{2}}=\mathrm{16}{Ma}^{\mathrm{2}} \\ $$$${I}_{\mathrm{2}} =\frac{\mathrm{4}{Ma}^{\mathrm{2}} }{\mathrm{2}}+{Ma}^{\mathrm{2}} =\mathrm{3}{Ma}^{\mathrm{2}} \\ $$$$\mathrm{2}{M}×\frac{{v}_{\mathrm{2}} +{v}_{\mathrm{1}} }{\Delta{t}}={F}_{\mathrm{1}} \\ $$$${I}_{\mathrm{1}} \frac{\omega_{\mathrm{1}} }{\Delta{t}}=\left({F}_{\mathrm{1}} −{F}_{\mathrm{2}} \right){R} \\ $$$${I}_{\mathrm{2}} \frac{\omega_{\mathrm{2}} }{\Delta{t}}={F}_{\mathrm{2}} {r} \\ $$$$\omega_{\mathrm{1}} {R}=\omega_{\mathrm{2}} {r} \\ $$$$\Rightarrow\omega_{\mathrm{1}} =\frac{{r}\omega_{\mathrm{2}} }{{R}}=\frac{\omega_{\mathrm{2}} }{\mathrm{2}} \\ $$$${I}_{\mathrm{1}} \omega_{\mathrm{1}} =\left({F}_{\mathrm{1}} \Delta{t}−{F}_{\mathrm{2}} \Delta{t}\right){R} \\ $$$${I}_{\mathrm{1}} \omega_{\mathrm{1}} =\mathrm{2}{M}\left({v}_{\mathrm{2}} +{v}_{\mathrm{1}} \right){R}−\frac{{I}_{\mathrm{2}} \omega_{\mathrm{2}} {R}}{{r}} \\ $$$$\mathrm{2}{M}\left({v}_{\mathrm{2}} +{v}_{\mathrm{1}} \right){R}={I}_{\mathrm{1}} \omega_{\mathrm{1}} +\frac{{I}_{\mathrm{2}} \omega_{\mathrm{2}} {R}}{{r}} \\ $$$$\Rightarrow\mathrm{2}{M}\left({v}_{\mathrm{2}} +{v}_{\mathrm{1}} \right){R}=\left(\frac{{I}_{\mathrm{1}} {r}}{{R}}+\frac{{I}_{\mathrm{2}} {R}}{{r}}\right)\omega_{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}{M}\left({v}_{\mathrm{2}} +{v}_{\mathrm{1}} \right)\left(\mathrm{2}{a}\right)=\left(\mathrm{16}{Ma}^{\mathrm{2}} ×\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{3}{Ma}^{\mathrm{2}} ×\mathrm{2}\right)\omega_{\mathrm{2}} \\ $$$$\Rightarrow{v}_{\mathrm{2}} =\frac{\mathrm{7}{a}\omega_{\mathrm{2}} }{\mathrm{2}}−{v}_{\mathrm{1}} \\ $$$$\frac{\mathrm{2}{Mv}_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{2}{Mv}_{\mathrm{2}} ^{\mathrm{2}} }{\mathrm{2}}+\frac{{I}_{\mathrm{1}} \omega_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{2}}+\frac{{I}_{\mathrm{2}} \omega_{\mathrm{2}} ^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow{v}_{\mathrm{1}} ^{\mathrm{2}} ={v}_{\mathrm{2}} ^{\mathrm{2}} +\frac{\mathrm{7}}{\mathrm{2}}{a}^{\mathrm{2}} \omega_{\mathrm{2}} ^{\mathrm{2}} \\ $$$$\Rightarrow{v}_{\mathrm{1}} ^{\mathrm{2}} =\left(\frac{\mathrm{7}{a}\omega_{\mathrm{2}} }{\mathrm{2}}−{v}_{\mathrm{1}} \right)^{\mathrm{2}} +\frac{\mathrm{7}}{\mathrm{2}}{a}^{\mathrm{2}} \omega_{\mathrm{2}} ^{\mathrm{2}} \\ $$$$\Rightarrow\frac{\mathrm{9}{a}\omega_{\mathrm{2}} }{\mathrm{4}}−{v}_{\mathrm{1}} =\mathrm{0} \\ $$$$\Rightarrow\omega_{\mathrm{2}} =\frac{\mathrm{4}{v}_{\mathrm{1}} }{\mathrm{9}{a}} \\ $$$${u}=\omega_{\mathrm{2}} {a}=\frac{\mathrm{4}{v}_{\mathrm{1}} }{\mathrm{9}}=\frac{\mathrm{4}\sqrt{\mathrm{2}{gh}}}{\mathrm{9}} \\ $$$${v}_{\mathrm{2}} =\frac{\mathrm{7}}{\mathrm{2}}×\frac{\mathrm{4}{v}_{\mathrm{1}} }{\mathrm{9}}−{v}_{\mathrm{1}} =\frac{\mathrm{5}{v}_{\mathrm{1}} }{\mathrm{9}}\:\Rightarrow{rebound}\:{height}=\frac{\sqrt{\mathrm{5}}{h}}{\mathrm{3}}\approx\frac{\mathrm{3}{h}}{\mathrm{4}} \\ $$$$−{a}\:\mathrm{cos}\:\theta={u}\:\mathrm{sin}\:\theta\:{t}−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\frac{{gt}^{\mathrm{2}} }{\mathrm{2}}−{u}\:\mathrm{sin}\:\theta\:{t}−{a}\:\mathrm{cos}\:\theta=\mathrm{0} \\ $$$$\Rightarrow{t}=\frac{{u}\:\mathrm{sin}\:\theta+\sqrt{{u}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{2}{ga}\:\mathrm{cos}\:\theta}}{{g}} \\ $$$${Range}+{a}+{a}\:\mathrm{sin}\:\theta={u}\:\mathrm{cos}\:\theta\:{t} \\ $$$$\Rightarrow{Range}={u}\:\mathrm{cos}\:\theta\:{t}−{a}\left(\mathrm{1}+\mathrm{sin}\:\theta\right) \\ $$$$ \\ $$$${example}: \\ $$$$\theta=\mathrm{30}°,\:{a}=\mathrm{1}{m},\:{h}=\mathrm{4}{m} \\ $$$${u}=\frac{\mathrm{4}\sqrt{\mathrm{2}×\mathrm{10}×\mathrm{4}}}{\mathrm{9}}=\frac{\mathrm{16}\sqrt{\mathrm{5}}}{\mathrm{9}} \\ $$$${t}=\frac{\frac{\mathrm{16}\sqrt{\mathrm{5}}}{\mathrm{9}}×\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\frac{\mathrm{16}^{\mathrm{2}} ×\mathrm{5}}{\mathrm{9}^{\mathrm{2}} }×\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{2}×\mathrm{10}×\mathrm{1}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}}{\mathrm{10}}\approx\mathrm{0}.\mathrm{659968} \\ $$$${R}=\frac{\mathrm{16}\sqrt{\mathrm{5}}}{\mathrm{9}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\mathrm{0}.\mathrm{659968}−\mathrm{1}×\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)\approx\mathrm{0}.\mathrm{772}\:{m} \\ $$

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