Question-224665

Question Number 224665 by ajfour last updated on 24/Sep/25

Commented by ajfour last updated on 24/Sep/25

At the instant shown rotating frame plane lies in yz plane. Angular acceleration (constant) is 𝛂, angular velocity 𝛚. Find 𝛉 and 𝛗. String length is L. Circle below is projection on ground of the mass m. CP is the xy projection of string length.

$${At}\:{the}\:{instant}\:{shown}\:{rotating} \\ $$$${frame}\:{plane}\:{lies}\:{in}\:{yz}\:{plane}. \\ $$$${Angular}\:{acceleration}\:\left({constant}\right) \\ $$$${is}\:\boldsymbol{\alpha},\:{angular}\:{velocity}\:\boldsymbol{\omega}.\:{Find}\:\boldsymbol{\theta} \\ $$$${and}\:\boldsymbol{\phi}.\:{String}\:{length}\:{is}\:{L}. \\ $$$${Circle}\:{below}\:{is}\:{projection}\:{on} \\ $$$${ground}\:{of}\:{the}\:{mass}\:{m}.\:{CP}\:{is}\:{the} \\ $$$${xy}\:{projection}\:{of}\:{string}\:{length}. \\ $$

Commented by mr W last updated on 25/Sep/25

Commented by mr W last updated on 26/Sep/25

too hard to use classical mechanics! at t=0: ϕ=0, φ=0, θ=0 ω(t)=αt ϕ(t)=((αt^2 )/2) φ(t)=....≠constant θ(t)=...≠constant

$${too}\:{hard}\:{to}\:{use}\:{classical}\:{mechanics}! \\ $$$${at}\:{t}=\mathrm{0}: \\ $$$$\varphi=\mathrm{0},\:\phi=\mathrm{0},\:\theta=\mathrm{0} \\ $$$$\omega\left({t}\right)=\alpha{t} \\ $$$$\varphi\left({t}\right)=\frac{\alpha{t}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\phi\left({t}\right)=….\neq{constant} \\ $$$$\theta\left({t}\right)=…\neq{constant} \\ $$

Commented by ajfour last updated on 25/Sep/25

$${no}\:{sir},\:{its}\:{easy}\:{when}\:{we}\:{resolve} \\ $$$${vetically}\:{and}\:{horizontally}\:{and} \\ $$$${consider}\:{tangential}\:{and} \\ $$$${centripetal}\:{acceleration}\:{with}\:{due} \\ $$$${credence}\:{to}\:{the}\:{geometry}\:{of}\:{the} \\ $$$${question}. \\ $$

Commented by mahdipoor last updated on 25/Sep/25

S(x,y,z) B(−c.sin(ψ),c.cos(ψ),b) , ψ(t)=((αt^2 )/2) ∣SB∣^2 =L^2 ma_S ^→ =W^→ +T^→ m(x^(..) ,y^(..) ,z^(..) )=−mg k^→ + Te_( SB) ^→ m(x^(..) ,y^(..) ,z^(..) )=−mg k^→ + ((T(t))/L) V_(SB) ^→ (t) ((T/(mL))=f(t)) { ((x^(..) =f(x+c.sinψ))),((y^(..) =f(y−c.cosψ))),((z^(..) =−g+f(z−b))),(((x+c.sinψ)^2 +(y−c.cosψ)^2 +(z−b)^2 =L^2 )) :} 4 unkown (x,y,z,f) and 4 eq ⇒ ...

$$\mathrm{S}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right) \\ $$$$\mathrm{B}\left(−\mathrm{c}.\mathrm{sin}\left(\psi\right),\mathrm{c}.\mathrm{cos}\left(\psi\right),\mathrm{b}\right)\:,\:\psi\left(\mathrm{t}\right)=\frac{\alpha\mathrm{t}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\mid\mathrm{SB}\mid^{\mathrm{2}} =\mathrm{L}^{\mathrm{2}} \\ $$$$\mathrm{m}\overset{\rightarrow} {\mathrm{a}}_{\mathrm{S}} =\overset{\rightarrow} {\mathrm{W}}+\overset{\rightarrow} {\mathrm{T}} \\ $$$$\mathrm{m}\left(\overset{..} {\mathrm{x}},\overset{..} {\mathrm{y}},\overset{..} {\mathrm{z}}\right)=−\mathrm{mg}\:\overset{\rightarrow} {\boldsymbol{\mathrm{k}}}\:+\:\mathrm{T}\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\:\mathrm{SB}} \\ $$$$\mathrm{m}\left(\overset{..} {\mathrm{x}},\overset{..} {\mathrm{y}},\overset{..} {\mathrm{z}}\right)=−\mathrm{mg}\:\overset{\rightarrow} {\boldsymbol{\mathrm{k}}}\:+\:\frac{\mathrm{T}\left(\mathrm{t}\right)}{\mathrm{L}}\:\overset{\rightarrow} {\mathrm{V}}_{\mathrm{SB}} \left(\mathrm{t}\right)\:\:\:\:\:\left(\frac{\mathrm{T}}{\mathrm{mL}}=\mathrm{f}\left(\mathrm{t}\right)\right) \\ $$$$\begin{cases}{\overset{..} {\mathrm{x}}=\mathrm{f}\left(\mathrm{x}+\mathrm{c}.\mathrm{sin}\psi\right)}\\{\overset{..} {\mathrm{y}}=\mathrm{f}\left(\mathrm{y}−\mathrm{c}.\mathrm{cos}\psi\right)}\\{\overset{..} {\mathrm{z}}=−\mathrm{g}+\mathrm{f}\left(\mathrm{z}−\mathrm{b}\right)}\\{\left(\mathrm{x}+\mathrm{c}.\mathrm{sin}\psi\right)^{\mathrm{2}} +\left(\mathrm{y}−\mathrm{c}.\mathrm{cos}\psi\right)^{\mathrm{2}} +\left(\mathrm{z}−\mathrm{b}\right)^{\mathrm{2}} =\mathrm{L}^{\mathrm{2}} }\end{cases} \\ $$$$\mathrm{4}\:\mathrm{unkown}\:\left(\mathrm{x},\mathrm{y},\mathrm{z},\mathrm{f}\right)\:\mathrm{and}\:\mathrm{4}\:\mathrm{eq}\:\Rightarrow\:… \\ $$

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