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Question-23272




Question Number 23272 by ajfour last updated on 28/Oct/17
Commented by ajfour last updated on 28/Oct/17
If the mounted wheels are kept  spinning at an angular velocity  𝛚 (as shown in diagram), find  the time period of small oscillations  of a rod of mass M kept on them,   if it be displaced slightly from its  equilibrium position. Assume  coefficient of friction between  rod and wheels to be 𝛍.
$${If}\:{the}\:{mounted}\:{wheels}\:{are}\:{kept} \\ $$$${spinning}\:{at}\:{an}\:{angular}\:{velocity} \\ $$$$\boldsymbol{\omega}\:\left({as}\:{shown}\:{in}\:{diagram}\right),\:{find} \\ $$$${the}\:{time}\:{period}\:{of}\:{small}\:{oscillations} \\ $$$${of}\:{a}\:{rod}\:{of}\:{mass}\:\boldsymbol{{M}}\:{kept}\:{on}\:{them},\: \\ $$$${if}\:{it}\:{be}\:{displaced}\:{slightly}\:{from}\:{its} \\ $$$${equilibrium}\:{position}.\:{Assume} \\ $$$${coefficient}\:{of}\:{friction}\:{between} \\ $$$${rod}\:{and}\:{wheels}\:{to}\:{be}\:\boldsymbol{\mu}. \\ $$
Answered by mrW1 last updated on 29/Oct/17
Commented by mrW1 last updated on 29/Oct/17
N_1 =(((d/2)+x)/d)×Mg=((1/2)+(x/d))Mg  F_1 =μN_1 =μMg(1+(x/d))     (direction ←)  N_2 =(((d/2)−x)/d)×Mg=((1/2)−(x/d))Mg  F_2 =μN_2 =μMg(1−(x/d))     (direction →)    F=ΣF=F_2 −F_1 =−((2μMg)/d)×x=−kx  period T=2π(√(M/k))=2π(√(d/(2μg)))
$$\mathrm{N}_{\mathrm{1}} =\frac{\frac{\mathrm{d}}{\mathrm{2}}+\mathrm{x}}{\mathrm{d}}×\mathrm{Mg}=\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{x}}{\mathrm{d}}\right)\mathrm{Mg} \\ $$$$\mathrm{F}_{\mathrm{1}} =\mu\mathrm{N}_{\mathrm{1}} =\mu\mathrm{Mg}\left(\mathrm{1}+\frac{\mathrm{x}}{\mathrm{d}}\right)\:\:\:\:\:\left(\mathrm{direction}\:\leftarrow\right) \\ $$$$\mathrm{N}_{\mathrm{2}} =\frac{\frac{\mathrm{d}}{\mathrm{2}}−\mathrm{x}}{\mathrm{d}}×\mathrm{Mg}=\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{x}}{\mathrm{d}}\right)\mathrm{Mg} \\ $$$$\mathrm{F}_{\mathrm{2}} =\mu\mathrm{N}_{\mathrm{2}} =\mu\mathrm{Mg}\left(\mathrm{1}−\frac{\mathrm{x}}{\mathrm{d}}\right)\:\:\:\:\:\left(\mathrm{direction}\:\rightarrow\right) \\ $$$$ \\ $$$$\mathrm{F}=\Sigma\mathrm{F}=\mathrm{F}_{\mathrm{2}} −\mathrm{F}_{\mathrm{1}} =−\frac{\mathrm{2}\mu\mathrm{Mg}}{\mathrm{d}}×\mathrm{x}=−\mathrm{kx} \\ $$$$\mathrm{period}\:\mathrm{T}=\mathrm{2}\pi\sqrt{\frac{\mathrm{M}}{\mathrm{k}}}=\mathrm{2}\pi\sqrt{\frac{\mathrm{d}}{\mathrm{2}\mu\mathrm{g}}} \\ $$
Commented by ajfour last updated on 29/Oct/17
Thanks a lot, Sir.
$${Thanks}\:{a}\:{lot},\:{Sir}. \\ $$

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