Question

A 50-gm mouse falls onto the outer edge of a phonograph turnable of radius of 20cm rotating at 33 rev/min. How much work it does to walk into the center post? Assume that the angular velocity of the turnable does not change.

Answer 1

To walk into the center post mouse must overcome the centrifugal force:

$(1) F=m\cdot r\cdot w^2$

This force acts along the radius along which the mouse will travel to the center. The force of the mouse legs will be directed against the radius and, taking into account the vector equation for working $\Delta W=\vec F\cdot \Delta \vec { l}$ on each small segment of the radius, will be written as:

(2) $dW=\vec F\cdot d\vec {r}=F\cdot dr \cdot cos180\degree=-F\cdot dr$

In order to find a job when moving completely, it is necessary to integrate (2) from the outer edge radius $R=20cm=0.2m$ to the center post $r=0$ .

(3) $W=\int_R^0 (-F)\cdot dr=-\int_R^0 mw^2\cdot r \cdot dr=-mw^2[\frac {R^2}{2}]_R^0=\frac{m(wR)^2}{2}$

This expression (3) shows that the mouse must stop, i.e. compensate its kinetic energy of rotation with the original speed $V=wR=\frac{33 }{60 s}\cdot 2\pi \cdot 0.2 m=0.69 ms^{-1}$ . When calculating, we converted all the values to the SI system and took into account that in one revolution there is $2\pi$ radians.

Thus from (3) we have

(4) $W=\frac{mV^2}{2}=\frac{0.05 kg\cdot (0.69 ms^{-1})^2 }{2}=1.2\cdot 10^{-2} J$

Answer:  The work mouse does to walk into the center post is $1.2\cdot 10^{-2} J$