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Psi-String

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Though I solved this problem, the answer of mine is strange. Could someone check my solution, please? Thanks in advance.

My solution to (a):

Since the normal force N and frictional force f can't do work on the rod, the energy of the rod-earth system is conserved. Hence

[tex] Mg \frac{L}{2}(1-\frac{1}{\sqrt{2}})=\frac{1}{2} I \omega ^2 =\frac{1}{2} \times \frac{1}{3}ML^2 \omega^2 [/tex]

so we can know [tex] a_{C.M.r} =\omega^2 \frac{L}{2}=\frac{3g}{2} (1-\frac{1}{\sqrt{2}})[/tex]

By [tex] \tau = I \alpha [/tex] ,

[tex]Mg(sin\frac{\pi}{4})\frac{L}{2} = (\frac{1}{3}ML^2)\alpha [/tex]

so [tex] \alpha=\frac{3}{2\sqrt{2}} \frac{g}{L} [/tex]

then [tex] a_{C.M.t}= \alpha \frac{L}{2} = \frac{3g}{4\sqrt{2}} [/tex]

So now we have [tex] \vec{a_{C.M}}=(-\frac{3\sqrt{2}}{4}+\frac{9}{8})g \hat{i}+(\frac{3\sqrt{2}}{4}+\frac{3}{8})g \hat{j}[/tex]

Since [tex] \vec{F}_{net}= M \vec{a}_{C.M}=M\vec{g}+\vec{F}_{floor}[/tex]

we find [tex] \vec{F}_{floor} =Mg(-\frac{3\sqrt{2}}{4}+\frac{9}{8}) \hat{i}+Mg(\frac{3\sqrt{2}}{4}+\frac{11}{8}) \hat{j}[/tex]

My solution to (b):

Since the contact between rod and ground is frictionless, we know that x-directional linear momenum is conserved. So the center of mass of the rod will always on y-axis.

From [tex] \frac{L}{2} sin\theta =y [/tex]

we find [tex] \frac{L}{2} \frac{d\theta}{dt} cos\theta = v [/tex] ------------ Eq.1

and [tex]\frac{L}{2} ( \frac{d^2\theta}{dt^2} cos\theta -\frac{d\theta}{dt} \frac{d\theta}{dt} sin\theta) = \frac{L}{2}(\alpha cos \theta -\omega^2 sin\theta) = a [/tex] ---------Eq.2

By the conservation of energy, we have

[tex]Mg\frac{L}{2} (1-sin\theta) = \frac{1}{2}Mv^2+\frac{1}{2} I \omega^2=\frac{1}{2} Mv^2+\frac{1}{24} ML^2 \omega^2 [/tex]

substitute v by Eq.1 and [tex] \theta = \frac{\pi}{4} [/tex] we have [tex] \omega^2 =\frac{24}{5} (1-\frac{1}{\sqrt{2}}) \frac{g}{L} [/tex]

because[tex] \alpha = \frac{3\sqrt{2} g}{L} [/tex] and by Eq.2

[tex] a=\frac{L}{2} ( \frac{3\sqrt{2} g}{L} \times \frac{1}{\sqrt{2}} -\frac{24}{5} (1-\frac{1}{\sqrt{2}}) \times \frac{1}{\sqrt{2}}) = (\frac{3}{2}+\frac{12}{5}-\frac{12\sqrt{2}}{5})g [/tex]

then

[tex] N= mg-ma= mg(1-(\frac{3}{2}+\frac{12}{5}-\frac{12\sqrt{2}}{5})=mg(\frac{24\sqrt{2}-29}{10}) [/tex]

Thanks for help! I'll be very very appriciated!

**Problem:**

A rod with length L and mass M initially stands vertically on the ground. A small perturbation makes it start to falling down. What's the force exerted by the ground on the rod when the rod is inclined 45 degree with respect to the horizon if the contact between rod and ground is (a)rigid or (b) frictionless ?A rod with length L and mass M initially stands vertically on the ground. A small perturbation makes it start to falling down. What's the force exerted by the ground on the rod when the rod is inclined 45 degree with respect to the horizon if the contact between rod and ground is (a)rigid or (b) frictionless ?

My solution to (a):

Since the normal force N and frictional force f can't do work on the rod, the energy of the rod-earth system is conserved. Hence

[tex] Mg \frac{L}{2}(1-\frac{1}{\sqrt{2}})=\frac{1}{2} I \omega ^2 =\frac{1}{2} \times \frac{1}{3}ML^2 \omega^2 [/tex]

so we can know [tex] a_{C.M.r} =\omega^2 \frac{L}{2}=\frac{3g}{2} (1-\frac{1}{\sqrt{2}})[/tex]

By [tex] \tau = I \alpha [/tex] ,

[tex]Mg(sin\frac{\pi}{4})\frac{L}{2} = (\frac{1}{3}ML^2)\alpha [/tex]

so [tex] \alpha=\frac{3}{2\sqrt{2}} \frac{g}{L} [/tex]

then [tex] a_{C.M.t}= \alpha \frac{L}{2} = \frac{3g}{4\sqrt{2}} [/tex]

So now we have [tex] \vec{a_{C.M}}=(-\frac{3\sqrt{2}}{4}+\frac{9}{8})g \hat{i}+(\frac{3\sqrt{2}}{4}+\frac{3}{8})g \hat{j}[/tex]

Since [tex] \vec{F}_{net}= M \vec{a}_{C.M}=M\vec{g}+\vec{F}_{floor}[/tex]

we find [tex] \vec{F}_{floor} =Mg(-\frac{3\sqrt{2}}{4}+\frac{9}{8}) \hat{i}+Mg(\frac{3\sqrt{2}}{4}+\frac{11}{8}) \hat{j}[/tex]

My solution to (b):

Since the contact between rod and ground is frictionless, we know that x-directional linear momenum is conserved. So the center of mass of the rod will always on y-axis.

From [tex] \frac{L}{2} sin\theta =y [/tex]

we find [tex] \frac{L}{2} \frac{d\theta}{dt} cos\theta = v [/tex] ------------ Eq.1

and [tex]\frac{L}{2} ( \frac{d^2\theta}{dt^2} cos\theta -\frac{d\theta}{dt} \frac{d\theta}{dt} sin\theta) = \frac{L}{2}(\alpha cos \theta -\omega^2 sin\theta) = a [/tex] ---------Eq.2

By the conservation of energy, we have

[tex]Mg\frac{L}{2} (1-sin\theta) = \frac{1}{2}Mv^2+\frac{1}{2} I \omega^2=\frac{1}{2} Mv^2+\frac{1}{24} ML^2 \omega^2 [/tex]

substitute v by Eq.1 and [tex] \theta = \frac{\pi}{4} [/tex] we have [tex] \omega^2 =\frac{24}{5} (1-\frac{1}{\sqrt{2}}) \frac{g}{L} [/tex]

because[tex] \alpha = \frac{3\sqrt{2} g}{L} [/tex] and by Eq.2

[tex] a=\frac{L}{2} ( \frac{3\sqrt{2} g}{L} \times \frac{1}{\sqrt{2}} -\frac{24}{5} (1-\frac{1}{\sqrt{2}}) \times \frac{1}{\sqrt{2}}) = (\frac{3}{2}+\frac{12}{5}-\frac{12\sqrt{2}}{5})g [/tex]

then

[tex] N= mg-ma= mg(1-(\frac{3}{2}+\frac{12}{5}-\frac{12\sqrt{2}}{5})=mg(\frac{24\sqrt{2}-29}{10}) [/tex]

Thanks for help! I'll be very very appriciated!

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