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Question-219769

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Question Number 219769 by mr W last updated on 01/May/25
Commented by mr W last updated on 01/May/25
a ball with mass m and an uniform rod with mass M and length L are released from the rest at the same heigth h as shown. (L>h) case 1: the floor is slippery case 2: the floor is rough enough for each of the cases find the time each object takes to reach the floor. do the objects collide with each other?
$${a}\:{ball}\:{with}\:{mass}\:\boldsymbol{{m}}\:{and}\:{an}\:{uniform} \\ $$$${rod}\:{with}\:{mass}\:\boldsymbol{{M}}\:{and}\:{length}\:\boldsymbol{{L}}\:{are} \\ $$$${released}\:{from}\:{the}\:{rest}\:{at}\:{the}\:{same} \\ $$$${heigth}\:\boldsymbol{{h}}\:{as}\:{shown}.\:\left({L}>{h}\right) \\ $$$${case}\:\mathrm{1}:\:{the}\:{floor}\:{is}\:{slippery} \\ $$$${case}\:\mathrm{2}:\:{the}\:{floor}\:{is}\:{rough}\:{enough} \\ $$$${for}\:{each}\:{of}\:{the}\:{cases}\:{find}\:{the}\:{time} \\ $$$${each}\:{object}\:{takes}\:{to}\:{reach}\:{the}\:{floor}. \\ $$$${do}\:{the}\:{objects}\:{collide}\:{with}\:{each}\: \\ $$$${other}? \\ $$
Answered by vnm last updated on 02/May/25
case 1. according to my calculations, the fall time of the rod is (1/( (√(12g))))∫_0 ^h (√((4l^2 −3x^2 )/((l^2 −x^2 )(h−x))))dx the fall time of the ball is (√((2h)/g)) if (h/l)≈0.944615 the rod and the ball will fall simultaneously.
$$ \\ $$$$ \\ $$$$\mathrm{case}\:\mathrm{1}. \\ $$$$\mathrm{according}\:\mathrm{to}\:\mathrm{my}\:\mathrm{calculations}, \\ $$$$\mathrm{the}\:\mathrm{fall}\:\mathrm{time}\:\mathrm{of}\:\mathrm{the}\:\mathrm{rod}\:\mathrm{is} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{12}{g}}}\int_{\mathrm{0}} ^{{h}} \sqrt{\frac{\mathrm{4}{l}^{\mathrm{2}} −\mathrm{3}{x}^{\mathrm{2}} }{\left({l}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)\left({h}−{x}\right)}}\mathrm{d}{x} \\ $$$$\mathrm{the}\:\mathrm{fall}\:\mathrm{time}\:\mathrm{of}\:\mathrm{the}\:\mathrm{ball}\:\mathrm{is} \\ $$$$\sqrt{\frac{\mathrm{2}{h}}{{g}}} \\ $$$$\mathrm{if}\:\frac{{h}}{{l}}\approx\mathrm{0}.\mathrm{944615}\:\mathrm{the}\:\mathrm{rod}\:\mathrm{and}\:\mathrm{the}\:\mathrm{ball}\:\mathrm{will} \\ $$$$\mathrm{fall}\:\mathrm{simultaneously}. \\ $$
Commented by mr W last updated on 02/May/25
correct! thanks!
$${correct}!\:{thanks}! \\ $$
Answered by mr W last updated on 02/May/25
Commented by mr W last updated on 03/May/25
case 2: sin θ_0 =(h/L) ω=−(dθ/dt) ((Mg)/2)(h−L sin θ)=(1/2)(((ML^2 )/3))ω^2 sin θ_0 −sin θ=((Lω^2 )/(3g)) ⇒ω^2 =((3g(sin θ_0 −sin θ))/L) ω=−(dθ/dt)=(√((3g(sin θ_0 −sin θ))/L)) dt=−(√(L/(3g (sin θ_0 −sin θ)))) dθ ⇒T_(Rod) =(√(L/(3g)))∫_0 ^θ_0 (dθ/( (√(sin θ_0 −sin θ)))) T_(Ball) =(√((2h)/g))=(√((2L sin θ_0 )/g)) (T_(Rod) /T_(Ball) )=(1/( (√(6 sin θ_0 ))))∫_0 ^θ_0 (dθ/( (√(sin θ_0 −sin θ)))) examples: θ=(π/3): (T_(Rod) /T_(Bal) )≈1.14488 θ=(π/4): (T_(Rod) /T_(Bal) )≈0.974489 θ=(π/6): (T_(Rod) /T_(Bal) )≈0.880394
$$\underline{\boldsymbol{{case}}\:\mathrm{2}:} \\ $$$$\mathrm{sin}\:\theta_{\mathrm{0}} =\frac{{h}}{{L}} \\ $$$$\omega=−\frac{{d}\theta}{{dt}} \\ $$$$\frac{{Mg}}{\mathrm{2}}\left({h}−{L}\:\mathrm{sin}\:\theta\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{ML}^{\mathrm{2}} }{\mathrm{3}}\right)\omega^{\mathrm{2}} \\ $$$$\mathrm{sin}\:\theta_{\mathrm{0}} −\mathrm{sin}\:\theta=\frac{{L}\omega^{\mathrm{2}} }{\mathrm{3}{g}} \\ $$$$\Rightarrow\omega^{\mathrm{2}} =\frac{\mathrm{3}{g}\left(\mathrm{sin}\:\theta_{\mathrm{0}} −\mathrm{sin}\:\theta\right)}{{L}} \\ $$$$\omega=−\frac{{d}\theta}{{dt}}=\sqrt{\frac{\mathrm{3}{g}\left(\mathrm{sin}\:\theta_{\mathrm{0}} −\mathrm{sin}\:\theta\right)}{{L}}} \\ $$$${dt}=−\sqrt{\frac{{L}}{\mathrm{3}{g}\:\left(\mathrm{sin}\:\theta_{\mathrm{0}} −\mathrm{sin}\:\theta\right)}}\:{d}\theta \\ $$$$\Rightarrow{T}_{{Rod}} =\sqrt{\frac{{L}}{\mathrm{3}{g}}}\int_{\mathrm{0}} ^{\theta_{\mathrm{0}} } \frac{{d}\theta}{\:\sqrt{\mathrm{sin}\:\theta_{\mathrm{0}} −\mathrm{sin}\:\theta}} \\ $$$${T}_{{Ball}} =\sqrt{\frac{\mathrm{2}{h}}{{g}}}=\sqrt{\frac{\mathrm{2}{L}\:\mathrm{sin}\:\theta_{\mathrm{0}} }{{g}}} \\ $$$$\frac{{T}_{{Rod}} }{{T}_{{Ball}} }=\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}\:\mathrm{sin}\:\theta_{\mathrm{0}} }}\int_{\mathrm{0}} ^{\theta_{\mathrm{0}} } \frac{{d}\theta}{\:\sqrt{\mathrm{sin}\:\theta_{\mathrm{0}} −\mathrm{sin}\:\theta}} \\ $$$${examples}: \\ $$$$\theta=\frac{\pi}{\mathrm{3}}:\:\frac{{T}_{{Rod}} }{{T}_{{Bal}} }\approx\mathrm{1}.\mathrm{14488} \\ $$$$\theta=\frac{\pi}{\mathrm{4}}:\:\frac{{T}_{{Rod}} }{{T}_{{Bal}} }\approx\mathrm{0}.\mathrm{974489} \\ $$$$\theta=\frac{\pi}{\mathrm{6}}:\:\frac{{T}_{{Rod}} }{{T}_{{Bal}} }\approx\mathrm{0}.\mathrm{880394} \\ $$
Commented by mr W last updated on 03/May/25
case 1: sin θ_0 =(h/L) ω=−(dθ/dt) y_C =((L sin θ)/2) v_C =((Lω cos θ)/2) ((Mg)/2)(h−L sin θ)=(M/2)(((Lω cos θ)/2))^2 +(1/2)(((ML^2 )/(12)))ω^2 (sin θ_0 −sin θ)=((Lω^2 (3 cos^2 θ+1))/(12g)) ⇒ω^2 =((12g(sin θ_0 −sin θ))/(L(3 cos^2 θ+1))) ω=−(dθ/dt)=(√((12g(sin θ_0 −sin θ))/(L(3 cos^2 θ+1)))) dt=−(√((L (3 cos^2 θ+1))/(12g (sin θ_0 −sin θ)))) dθ ⇒T_(Rod) =(√(L/(12g)))∫_0 ^θ_0 (√((3 cos^2 θ+1)/(sin θ_0 −sin θ))) dθ T_(Ball) =(√((2h)/g))=(√((2L sin θ_0 )/g)) (T_(Rod) /T_(Ball) )=(1/(2(√(6 sin θ_0 ))))∫_0 ^θ_0 (√((3 cos^2 θ+1)/(sin θ_0 −sin θ))) dθ examples: θ=((5π)/(11)): (T_(Rod) /T_(Bal) )≈1.20497 >1 θ=((2π)/5): (T_(Rod) /T_(Bal) )≈1.0 θ=(π/3): (T_(Rod) /T_(Bal) )≈0.916019 <1 θ=(π/6): (T_(Rod) /T_(Bal) )≈0.833142 <1 θ=(π/(10)): (T_(Rod) /T_(Bal) )≈0.822078 <1
$$\underline{\boldsymbol{{case}}\:\mathrm{1}:} \\ $$$$\mathrm{sin}\:\theta_{\mathrm{0}} =\frac{{h}}{{L}} \\ $$$$\omega=−\frac{{d}\theta}{{dt}} \\ $$$${y}_{{C}} =\frac{{L}\:\mathrm{sin}\:\theta}{\mathrm{2}} \\ $$$${v}_{{C}} =\frac{{L}\omega\:\mathrm{cos}\:\theta}{\mathrm{2}} \\ $$$$\frac{{Mg}}{\mathrm{2}}\left({h}−{L}\:\mathrm{sin}\:\theta\right)=\frac{{M}}{\mathrm{2}}\left(\frac{{L}\omega\:\mathrm{cos}\:\theta}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{ML}^{\mathrm{2}} }{\mathrm{12}}\right)\omega^{\mathrm{2}} \\ $$$$\left(\mathrm{sin}\:\theta_{\mathrm{0}} −\mathrm{sin}\:\theta\right)=\frac{{L}\omega^{\mathrm{2}} \left(\mathrm{3}\:\mathrm{cos}^{\mathrm{2}} \:\theta+\mathrm{1}\right)}{\mathrm{12}{g}} \\ $$$$\Rightarrow\omega^{\mathrm{2}} =\frac{\mathrm{12}{g}\left(\mathrm{sin}\:\theta_{\mathrm{0}} −\mathrm{sin}\:\theta\right)}{{L}\left(\mathrm{3}\:\mathrm{cos}^{\mathrm{2}} \:\theta+\mathrm{1}\right)} \\ $$$$\omega=−\frac{{d}\theta}{{dt}}=\sqrt{\frac{\mathrm{12}{g}\left(\mathrm{sin}\:\theta_{\mathrm{0}} −\mathrm{sin}\:\theta\right)}{{L}\left(\mathrm{3}\:\mathrm{cos}^{\mathrm{2}} \:\theta+\mathrm{1}\right)}} \\ $$$${dt}=−\sqrt{\frac{{L}\:\left(\mathrm{3}\:\mathrm{cos}^{\mathrm{2}} \:\theta+\mathrm{1}\right)}{\mathrm{12}{g}\:\left(\mathrm{sin}\:\theta_{\mathrm{0}} −\mathrm{sin}\:\theta\right)}}\:{d}\theta \\ $$$$\Rightarrow{T}_{{Rod}} =\sqrt{\frac{{L}}{\mathrm{12}{g}}}\int_{\mathrm{0}} ^{\theta_{\mathrm{0}} } \sqrt{\frac{\mathrm{3}\:\mathrm{cos}^{\mathrm{2}} \:\theta+\mathrm{1}}{\mathrm{sin}\:\theta_{\mathrm{0}} −\mathrm{sin}\:\theta}}\:{d}\theta \\ $$$${T}_{{Ball}} =\sqrt{\frac{\mathrm{2}{h}}{{g}}}=\sqrt{\frac{\mathrm{2}{L}\:\mathrm{sin}\:\theta_{\mathrm{0}} }{{g}}} \\ $$$$\frac{{T}_{{Rod}} }{{T}_{{Ball}} }=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{6}\:\mathrm{sin}\:\theta_{\mathrm{0}} }}\int_{\mathrm{0}} ^{\theta_{\mathrm{0}} } \sqrt{\frac{\mathrm{3}\:\mathrm{cos}^{\mathrm{2}} \:\theta+\mathrm{1}}{\mathrm{sin}\:\theta_{\mathrm{0}} −\mathrm{sin}\:\theta}}\:{d}\theta \\ $$$${examples}: \\ $$$$\theta=\frac{\mathrm{5}\pi}{\mathrm{11}}:\:\frac{{T}_{{Rod}} }{{T}_{{Bal}} }\approx\mathrm{1}.\mathrm{20497}\:>\mathrm{1} \\ $$$$\theta=\frac{\mathrm{2}\pi}{\mathrm{5}}:\:\frac{{T}_{{Rod}} }{{T}_{{Bal}} }\approx\mathrm{1}.\mathrm{0} \\ $$$$\theta=\frac{\pi}{\mathrm{3}}:\:\frac{{T}_{{Rod}} }{{T}_{{Bal}} }\approx\mathrm{0}.\mathrm{916019}\:<\mathrm{1} \\ $$$$\theta=\frac{\pi}{\mathrm{6}}:\:\frac{{T}_{{Rod}} }{{T}_{{Bal}} }\approx\mathrm{0}.\mathrm{833142}\:<\mathrm{1} \\ $$$$\theta=\frac{\pi}{\mathrm{10}}:\:\frac{{T}_{{Rod}} }{{T}_{{Bal}} }\approx\mathrm{0}.\mathrm{822078}\:<\mathrm{1} \\ $$
Commented by mr W last updated on 02/May/25

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