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Question Number 216788 by Tawa11 last updated on 20/Feb/25

Commented by Tawa11 last updated on 20/Feb/25

In Figure DIN11, the mass

In Figure DIN11, the mass "m" is located on an inclined plane of length "L" which rotates around the vertical axis shown. The angle between the vertical axis and the plane is "θ", and the coefficient of static friction between the plane and the mass is µₛ. Calculate the range of velocities of the mass for which it does not slide on the inclined plane (that is, the minimum and maximum velocity).

Answered by mr W last updated on 22/Feb/25

Commented by mr W last updated on 22/Feb/25

r=L sin θ for 𝛚_(max) : N=mg sin θ+mω_(max) ^2 r cos θ f=μ_s N=mω_(max) ^2 r sin θ−mg cos θ μ_s (mg sin θ+mω_(max) ^2 r cos θ)=mω_(max) ^2 r sin θ−mg cos θ ω_(max) ^2 L sin θ( sin θ−μ_s cos θ)=g(cos θ+μ_s sin θ) ⇒ω_(max) =(√((g(1+μ_s tan θ))/(L sin θ(tan θ−μ_s )))) for 𝛚_(min) : N=mg sin θ+mω_(min) ^2 r cos θ f=μ_s N=−mω_(max) ^2 r sin θ+mg cos θ μ_s (mg sin θ+mω_(min) ^2 r cos θ)=−mω_(max) ^2 r sin θ+mg cos θ ω_(min) ^2 L sin θ(sin θ+μ_s cos θ)=g(cos θ−μ_s sin θ) ⇒ω_(min) =(√((g(1−μ_s tan θ))/(L sin θ(tan θ+μ_s ))))

$${r}={L}\:\mathrm{sin}\:\theta \\ $$$$\underline{\boldsymbol{{for}}\:\boldsymbol{\omega}_{\boldsymbol{{max}}} :} \\ $$$${N}={mg}\:\mathrm{sin}\:\theta+{m}\omega_{{max}} ^{\mathrm{2}} {r}\:\mathrm{cos}\:\theta \\ $$$${f}=\mu_{{s}} {N}={m}\omega_{{max}} ^{\mathrm{2}} {r}\:\mathrm{sin}\:\theta−{mg}\:\mathrm{cos}\:\theta \\ $$$$\mu_{{s}} \left({mg}\:\mathrm{sin}\:\theta+{m}\omega_{{max}} ^{\mathrm{2}} {r}\:\mathrm{cos}\:\theta\right)={m}\omega_{{max}} ^{\mathrm{2}} {r}\:\mathrm{sin}\:\theta−{mg}\:\mathrm{cos}\:\theta \\ $$$$\omega_{{max}} ^{\mathrm{2}} {L}\:\mathrm{sin}\:\theta\left(\:\mathrm{sin}\:\theta−\mu_{{s}} \mathrm{cos}\:\theta\right)={g}\left(\mathrm{cos}\:\theta+\mu_{{s}} \:\mathrm{sin}\:\theta\right) \\ $$$$\Rightarrow\omega_{{max}} =\sqrt{\frac{{g}\left(\mathrm{1}+\mu_{{s}} \:\mathrm{tan}\:\theta\right)}{{L}\:\mathrm{sin}\:\theta\left(\mathrm{tan}\:\theta−\mu_{{s}} \right)}} \\ $$$$\underline{\boldsymbol{{for}}\:\boldsymbol{\omega}_{\boldsymbol{{min}}} :} \\ $$$${N}={mg}\:\mathrm{sin}\:\theta+{m}\omega_{{min}} ^{\mathrm{2}} {r}\:\mathrm{cos}\:\theta \\ $$$${f}=\mu_{{s}} {N}=−{m}\omega_{{max}} ^{\mathrm{2}} {r}\:\mathrm{sin}\:\theta+{mg}\:\mathrm{cos}\:\theta \\ $$$$\mu_{{s}} \left({mg}\:\mathrm{sin}\:\theta+{m}\omega_{{min}} ^{\mathrm{2}} {r}\:\mathrm{cos}\:\theta\right)=−{m}\omega_{{max}} ^{\mathrm{2}} {r}\:\mathrm{sin}\:\theta+{mg}\:\mathrm{cos}\:\theta \\ $$$$\omega_{{min}} ^{\mathrm{2}} {L}\:\mathrm{sin}\:\theta\left(\mathrm{sin}\:\theta+\mu_{{s}} \:\mathrm{cos}\:\theta\right)={g}\left(\mathrm{cos}\:\theta−\mu_{{s}} \mathrm{sin}\:\theta\right) \\ $$$$\Rightarrow\omega_{{min}} =\sqrt{\frac{{g}\left(\mathrm{1}−\mu_{{s}} \mathrm{tan}\:\theta\right)}{{L}\:\mathrm{sin}\:\theta\left(\mathrm{tan}\:\theta+\mu_{{s}} \right)}} \\ $$

Commented by Tawa11 last updated on 22/Feb/25

God bless you sir. I really appreciate.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$

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