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Question Number 73090 by ajfour last updated on 06/Nov/19

Commented by ajfour last updated on 06/Nov/19

$$Findx\left(t\right),x\left(0\right)=a,{\omega }_0=0,$$$$\alpha isconstant.$$

Answered by mr W last updated on 06/Nov/19

$$\omega =\alpha t$$$$A=\frac{{dx}^2}{{dt}^2}$$$$N=\sqrt{{\left(mg\right)}^2+{\left(mx\alpha \right)}^2}=m\sqrt{g^2+{\alpha }^2{x}^2}$$$$mA=mx{\omega }^2-\mu N$$$$A=x{\omega }^2-\mu \sqrt{g^2+{\alpha }^2{x}^2}$$$$\Rightarrow \frac{{dx}^2}{{dt}^2}={\alpha }^2{t}^{2}x-\mu \sqrt{g^2+{\alpha }^2{x}^2}$$$$i{don}^{\prime }tknowhowtosolveevenifthere$$$$isnofriction,i.e.\mu =0.$$

Commented by ajfour last updated on 06/Nov/19

$$alrightsir,hastodowith$$$$coriolisacceleration,perhaps,$$$$i^{\prime }llreviseandupdateyoutoo..$$

Answered by ajfour last updated on 12/Nov/19

Commented by mr W last updated on 12/Nov/19

$$thankyousir!$$$$iftheangularaccelerationiszero,$$$$wecansolveevenwithfriction.$$$$$$$$butthesolutionfor\frac{d^{2}r}{{dt}^2}={\alpha }^2{rt}^{2}needs$$$$paraboliccylinderfunctionwhichi$$$${don}^{\prime }tunderstand.$$

Commented by ajfour last updated on 12/Nov/19

$$buthere\left(iforgot\right)\frac{d\omega }{dt}=\alpha \left(constant\right)$$$$\frac{d^{2}r}{{dt}^2}={\alpha }^2{rt}^2$$$$dunnowhowtosolve..$$

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