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`        A metre stick is held vertically with one end on the floor and is then allowed to fall.Find the velocity of the other end when it hits the floor, assuming that the end of the stick does not slip.`
one year ago

```							This is an example of the dynamics of a rigid body (rotational motion). It's moment of inertia is (ML^2) / 3. It's rotational kinetic energy is therefore (1/2)Lω^2 = (ML^2 ω^2) / 6. The potential energy is Mgh where h is the height of the center of mass above the floor. When it is held up vertically, ω1 = 0 and h = L/2 so the total energy is: E = [(ML^2 ω^2) / 6] + Mgh1 = 0 + Mgh / 2 Just before the meterstick hits the floor, the angualr velocity is ω2 and h2 = 0. The energy is: E = [(ML^2 ω2^2) / 6] + Mgh2 = [(ML^2 ω2^2) / 6] + 0 Conservation of energy therefore implies: (ML^2 ω2^2) / 6 = Mgh / 2 Solve for ω2: ω2^2 = 3g / L = sq rt [(3*9.81) / 1] = 5.4 radians/s
```
one year ago
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