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PhMichael
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These stairs are rotating as a whole while a dog is climbing on them with a velocity v relative to them, as shown.
The angle that these stairs make is 60 degrees, and the radius is R.
1. find the radial component of the friction force between the dog's feet and the stairs.
2. after a while, the dog stop's moving completely relative to the stairs (but the stairs keep on rotating). find the amount of heat produced as a result of the friction which stopped the dog from moving. 1. The attempt at a solution
1. What I've is:
[tex]m \vec{a}_{rel} = \vec{F} -2m \vec{\omega} \times {\vec{v}}_{rel} -m\vec{\omega}\times(\vec{\omega}\times\vec{r}})[/tex]
where
[tex] \vec{v}_{rel} = vcos(60) \hat{\theta} + vsin(60) \hat{z} [/tex]
[tex] \vec{r} = R\hat{r} [/tex]
[tex] \vec{a}_{rel} = -\frac{(vcos(60))^{2}}{R} \hat{r} [/tex]
SO,
[tex] \vec{F}=-\left ( \frac{v^{2}}{4R}+\omega{v}+\omega^{2}R \right )m\hat{r} [/tex]
Does this seem reasonable?
2.
Here I'm a little puzzled as I know that the work done by friction equals the difference of the mechanical energies, where the final K.E. vanishes. Now, how can I treat potential energies?
If I take the K.E. difference relative to the stairs, I get:
[tex] K_{2}-K_{1} = -\frac{1}{2}mv^{2} [/tex]
However, doing this with respect to a fixed frame, we get:
[tex]K_{2}-K_{1}=\frac{1}{2}m \left ( (\omega R)^{2}-\left ( \frac{1}{2}v+\omega R \right )^{2}-\frac{3}{4}v^{2} \right )[/tex]
[tex]K_{2}-K_{1}=-\frac{1}{2}m\left ( v^{2}+\omega v R \right )[/tex]
which is higher than the previous result, WHY? and what is the amount of heat from all of these?! does the rotating of the stairs contribute anything to the amount of heat produced?!
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Homework Statement
These stairs are rotating as a whole while a dog is climbing on them with a velocity v relative to them, as shown.
The angle that these stairs make is 60 degrees, and the radius is R.
1. find the radial component of the friction force between the dog's feet and the stairs.
2. after a while, the dog stop's moving completely relative to the stairs (but the stairs keep on rotating). find the amount of heat produced as a result of the friction which stopped the dog from moving. 1. The attempt at a solution
1. What I've is:
[tex]m \vec{a}_{rel} = \vec{F} -2m \vec{\omega} \times {\vec{v}}_{rel} -m\vec{\omega}\times(\vec{\omega}\times\vec{r}})[/tex]
where
[tex] \vec{v}_{rel} = vcos(60) \hat{\theta} + vsin(60) \hat{z} [/tex]
[tex] \vec{r} = R\hat{r} [/tex]
[tex] \vec{a}_{rel} = -\frac{(vcos(60))^{2}}{R} \hat{r} [/tex]
SO,
[tex] \vec{F}=-\left ( \frac{v^{2}}{4R}+\omega{v}+\omega^{2}R \right )m\hat{r} [/tex]
Does this seem reasonable?
2.
Here I'm a little puzzled as I know that the work done by friction equals the difference of the mechanical energies, where the final K.E. vanishes. Now, how can I treat potential energies?
If I take the K.E. difference relative to the stairs, I get:
[tex] K_{2}-K_{1} = -\frac{1}{2}mv^{2} [/tex]
However, doing this with respect to a fixed frame, we get:
[tex]K_{2}-K_{1}=\frac{1}{2}m \left ( (\omega R)^{2}-\left ( \frac{1}{2}v+\omega R \right )^{2}-\frac{3}{4}v^{2} \right )[/tex]
[tex]K_{2}-K_{1}=-\frac{1}{2}m\left ( v^{2}+\omega v R \right )[/tex]
which is higher than the previous result, WHY? and what is the amount of heat from all of these?! does the rotating of the stairs contribute anything to the amount of heat produced?!
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