Question Number 29322 by ajfour last updated on 07/Feb/18
Commented by ajfour last updated on 07/Feb/18
$${Find}\:{acceleration}\:{of}\:{sphere}\:{as} \\ $$$${a}\:{function}\:{of}\:{angle}\:\theta. \\ $$$${System}\:{is}\:{released}\:{at}\:\theta=\:\theta_{\mathrm{0}} \:\left(<\mathrm{90}°\right)\:; \\ $$$$\left({All}\:{surfaces}\:{are}\:{frictionless}\right). \\ $$
Commented by ajfour last updated on 07/Feb/18
$${Please}\:{solve}\:{or}\:{check}\:{my}\:{solution}\: \\ $$$${mrW}\:\:{Sir}. \\ $$
Commented by ajfour last updated on 07/Feb/18
Commented by ajfour last updated on 07/Feb/18
![x=Rcot (θ/2) v=−((ωR)/2)cosec^2 (θ/2) ⇒ v^2 =((𝛚^2 R^2 )/4)cosec^4 (θ/2) ....(i) a=(dv/dt)= −((αR)/2)cosec^2 (θ/2) +((ω^2 R)/2)cosec^2 (θ/2)cot (θ/2) ...(ii) Nsin θ = ma ....(iii) NRcot (θ/2)−((MgLcos θ)/2) =(((ML^2 )/3))α ⇒ α=(3/(ML^2 ))(((maRcot (θ/2))/(sin θ))−((MgLcos θ)/2)) .....(iv) ((MgL)/2)(sin θ_0 −sin θ)=(1/2)mv^2 +(1/2)(((ML^2 )/3))ω^2 using (i): ((MgL)/2)(sin θ_0 −sin θ)=((mω^2 R^2 )/8)cosec^4 (θ/2) +((ML^2 ω^2 )/6) ω^2 =((((MgL)/2)(sin θ_0 −sin θ))/(((ML^2 )/6)+((mR^2 )/8)cosec^4 (θ/2))) ....(A) Using eq. (A), (iii), (iv) in eq.(ii) a=(R/2)cosec^2 (θ/2)((3/(ML^2 )))(((MgLcos θ)/2)−((maRcot (θ/2))/(sin θ))) +(R/2)cosec^2 (θ/2)cot (θ/2)[((((MgL)/2)(sin θ_0 −sin θ))/(((ML^2 )/6)+((mR^2 )/8)cosec^4 (θ/2)))] a[1+((3mR^2 cosec^2 (θ/2)cot (θ/2))/(2ML^2 sin θ))]= (R/2)cosec^2 (θ/2)[((3gcos θ)/(2L))+((((MgL)/2)(sin θ_0 −sin θ)cot (θ/2))/(((ML^2 )/6)+((mR^2 )/8)cosec^4 (θ/2)))] .](https://www.tinkutara.com/question/Q29333.png)
$${x}={R}\mathrm{cot}\:\frac{\theta}{\mathrm{2}} \\ $$$${v}=−\frac{\omega{R}}{\mathrm{2}}\mathrm{cosec}\:^{\mathrm{2}} \frac{\theta}{\mathrm{2}} \\ $$$$\Rightarrow\:\:\boldsymbol{{v}}^{\mathrm{2}} =\frac{\boldsymbol{\omega}^{\mathrm{2}} \boldsymbol{{R}}^{\mathrm{2}} }{\mathrm{4}}\mathrm{cosec}\:^{\mathrm{4}} \frac{\theta}{\mathrm{2}}\:\:\:….\left({i}\right) \\ $$$${a}=\frac{{dv}}{{dt}}=\:−\frac{\alpha{R}}{\mathrm{2}}\mathrm{cosec}\:^{\mathrm{2}} \frac{\theta}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:+\frac{\omega^{\mathrm{2}} {R}}{\mathrm{2}}\mathrm{cosec}\:^{\mathrm{2}} \frac{\theta}{\mathrm{2}}\mathrm{cot}\:\frac{\theta}{\mathrm{2}}\:\:\:\:\:…\left({ii}\right) \\ $$$${N}\mathrm{sin}\:\theta\:=\:{ma}\:\:\:\:….\left({iii}\right) \\ $$$${NR}\mathrm{cot}\:\frac{\theta}{\mathrm{2}}−\frac{{MgL}\mathrm{cos}\:\theta}{\mathrm{2}}\:=\left(\frac{{ML}^{\mathrm{2}} }{\mathrm{3}}\right)\alpha \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\Rightarrow\:\alpha=\frac{\mathrm{3}}{{ML}^{\mathrm{2}} }\left(\frac{{maR}\mathrm{cot}\:\frac{\theta}{\mathrm{2}}}{\mathrm{sin}\:\theta}−\frac{{MgL}\mathrm{cos}\:\theta}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…..\left({iv}\right) \\ $$$$\frac{{MgL}}{\mathrm{2}}\left(\mathrm{sin}\:\theta_{\mathrm{0}} −\mathrm{sin}\:\theta\right)=\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{ML}^{\mathrm{2}} }{\mathrm{3}}\right)\omega^{\mathrm{2}} \\ $$$${using}\:\left({i}\right): \\ $$$$\frac{{MgL}}{\mathrm{2}}\left(\mathrm{sin}\:\theta_{\mathrm{0}} −\mathrm{sin}\:\theta\right)=\frac{{m}\omega^{\mathrm{2}} {R}^{\mathrm{2}} }{\mathrm{8}}\mathrm{cosec}\:^{\mathrm{4}} \frac{\theta}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{{ML}^{\mathrm{2}} \omega^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\omega^{\mathrm{2}} =\frac{\frac{{MgL}}{\mathrm{2}}\left(\mathrm{sin}\:\theta_{\mathrm{0}} −\mathrm{sin}\:\theta\right)}{\frac{{ML}^{\mathrm{2}} }{\mathrm{6}}+\frac{{mR}^{\mathrm{2}} }{\mathrm{8}}\mathrm{cosec}\:^{\mathrm{4}} \frac{\theta}{\mathrm{2}}}\:\:\:….\left({A}\right) \\ $$$${Using}\:\:{eq}.\:\left({A}\right),\:\left({iii}\right),\:\left({iv}\right)\:{in}\:{eq}.\left(\boldsymbol{{ii}}\right) \\ $$$${a}=\frac{{R}}{\mathrm{2}}\mathrm{cosec}\:^{\mathrm{2}} \frac{\theta}{\mathrm{2}}\left(\frac{\mathrm{3}}{{ML}^{\mathrm{2}} }\right)\left(\frac{{MgL}\mathrm{cos}\:\theta}{\mathrm{2}}−\frac{{maR}\mathrm{cot}\:\frac{\theta}{\mathrm{2}}}{\mathrm{sin}\:\theta}\right) \\ $$$$\:\:\:\:\:+\frac{{R}}{\mathrm{2}}\mathrm{cosec}\:^{\mathrm{2}} \frac{\theta}{\mathrm{2}}\mathrm{cot}\:\frac{\theta}{\mathrm{2}}\left[\frac{\frac{{MgL}}{\mathrm{2}}\left(\mathrm{sin}\:\theta_{\mathrm{0}} −\mathrm{sin}\:\theta\right)}{\frac{{ML}^{\mathrm{2}} }{\mathrm{6}}+\frac{{mR}^{\mathrm{2}} }{\mathrm{8}}\mathrm{cosec}\:^{\mathrm{4}} \frac{\theta}{\mathrm{2}}}\right] \\ $$$$ \\ $$$${a}\left[\mathrm{1}+\frac{\mathrm{3}{mR}^{\mathrm{2}} \mathrm{cosec}\:^{\mathrm{2}} \frac{\theta}{\mathrm{2}}\mathrm{cot}\:\frac{\theta}{\mathrm{2}}}{\mathrm{2}{ML}^{\mathrm{2}} \mathrm{sin}\:\theta}\right]= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\frac{{R}}{\mathrm{2}}\mathrm{cosec}\:^{\mathrm{2}} \frac{\theta}{\mathrm{2}}\left[\frac{\mathrm{3}{g}\mathrm{cos}\:\theta}{\mathrm{2}{L}}+\frac{\frac{{MgL}}{\mathrm{2}}\left(\mathrm{sin}\:\theta_{\mathrm{0}} −\mathrm{sin}\:\theta\right)\mathrm{cot}\:\frac{\theta}{\mathrm{2}}}{\frac{{ML}^{\mathrm{2}} }{\mathrm{6}}+\frac{{mR}^{\mathrm{2}} }{\mathrm{8}}\mathrm{cosec}\:^{\mathrm{4}} \frac{\theta}{\mathrm{2}}}\right]\:\:. \\ $$
Commented by mrW2 last updated on 07/Feb/18
$${solution}\:{is}\:{correct}\:{sir}! \\ $$