Question

QUES NO. 12 on
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Answer 1

Answer on Question #56591, Physics Mechanics Relativity

A smooth uniform rod AB of mass $M$ length $l$ rotates freely with an angular velocity $\omega_0$ in a horizontal plane about a stationary vertical axis passing through its end A. A small sleeve of mass $m$ starts sliding along the rod from the point A. Find the velocity $V$ of the sleeve relative to the rod at the moment it reaches the other end B.

Solution

We have used in non-inertial reference frame is rigidly connected to the rotating shaft. We draw the x-axis along AB. The origin coincides with the point A. The force of inertia

$$f_i = m \frac{dV_x}{dt} = m\omega^2 x$$

where $m$ is the mass of sleeve.

Law of energy conservation

$$\frac{J\omega_0^2}{2} = \frac{J\omega^2}{2} + \frac{m(\omega x)^2}{2} + \frac{mv_x^2}{2}$$

where $J = \frac{1}{3} M l^2$ is the moment of inertia.

Then

$$\frac {M l ^ {2} \omega_ {0} ^ {2}}{6} = \frac {M l ^ {2} \omega^ {2}}{6} + \frac {m (x \omega) ^ {2}}{2} + \frac {m v _ {x} ^ {2}}{2}$$

From (3)

$$\omega^ {2} = \frac {1}{M l ^ {2} + 3 m x ^ {2}} \left[ M l ^ {2} \omega_ {0} ^ {2} - 3 m v _ {x} ^ {2} \right]$$

From (1) and (4)

$$m \frac {d v _ {x}}{d t} = m \omega^ {2} x \Rightarrow \frac {d v _ {x}}{d t} = \omega^ {2} x \Rightarrow \frac {d v _ {x}}{d x} v _ {x}$$ $$\frac {d v _ {x}}{d x} v _ {x} = \frac {1}{M l ^ {2} + 3 m x ^ {2}} \left[ M l ^ {2} \omega_ {0} ^ {2} - 3 m v _ {x} ^ {2} \right] x \Rightarrow \frac {d v _ {x}}{d x} \frac {v _ {x}}{M l ^ {2} \omega_ {0} ^ {2} - 3 m v _ {x} ^ {2}} = \frac {x}{M l ^ {2} + 3 m x ^ {2}} d x$$

Then

$$\frac {1}{6} \frac {d v _ {x}}{d x} \frac {v _ {x}}{M l ^ {2} \omega_ {0} ^ {2} - 3 m v _ {x} ^ {2}} = \frac {1}{6} \frac {x}{M l ^ {2} + 3 m x ^ {2}} d x \Rightarrow$$ $$\frac {1}{6} \frac {\ln \left(\frac {M l ^ {2} \omega_ {0} ^ {2}}{M l ^ {2} \omega_ {0} ^ {2} - 3 m V ^ {2}}\right)}{m} = \frac {1}{6} \frac {\ln \left(\frac {M l ^ {2} + 3 m x ^ {2}}{M l ^ {2}}\right)}{m} \Rightarrow V (x) = \frac {\omega_ {0} x}{\sqrt {1 + \frac {3 m}{M} \cdot \frac {x ^ {2}}{l ^ {2}}}}$$ $$V (x) = \frac {\omega_ {0} x}{\sqrt {1 + \frac {3 m}{M} \cdot \frac {x ^ {2}}{l ^ {2}}}}$$

The velocity $V$ of the sleeve relative to the rod at the moment it reaches the other end B

$$V (l) = \frac {\omega_ {0} l}{\sqrt {1 + \frac {3 m}{M} \cdot \frac {l ^ {2}}{l ^ {2}}}} = \frac {\omega_ {0} l}{\sqrt {1 + \frac {3 m}{M}}}$$

Answer: $V(l) = \frac{\omega_0 l}{\sqrt{1 + \frac{3m}{M}}}$.

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