Question

A Juggler is juggling a uniform rod one end of which is coated in tar and burning. He is holding the rod by the opposite end and throws it up so that, at the moment of release, it is horizontal, its CM is traveling vertically up at speed 120 and it is rotating with angular velocity coo. To catch it, he wants to arrange that when it returns to his hand it will have made an integer number of complete rotations. What would 120 be, if the rod is to have made exactly n rotations when it returns to his hand

Answer 1

The vertical speed of the rod varies inversely as the rotational speed for a given number of complete rotations

The equation that gives the vertical velocity, v₀ is  $\, \underline{v_0= \dfrac{ g \cdot \pi \cdot n}{ \omega_0 }}$

Reason:

From a similar question, the given parameters appear to be correctly given as follows;

The vertical speed of the rod = v₀

Angular velocity of the = ω₀

Required;

The value of, v₀, so that the rod has made exactly, n, number of turns when it returns to his hand

Where;

n = An integer

Solution;

The time the rod spends in the air is given as follows;

$Height, \ h = u \cdot t - \dfrac{1}{2} \cdot g \cdot t^2$

Where;

g = Acceleration due to gravity

t = Time of motion

When the rod returns to his hand, we have, h = 0, therefore;

$0 = u \cdot t - \dfrac{1}{2} \cdot g \cdot t^2$

$u \cdot t = \dfrac{1}{2} \cdot g \cdot t^2$

$t^2 = \dfrac{u \cdot t }{\dfrac{1}{2} \cdot g } = \dfrac{2 \cdot u \cdot t }{g }$

$Time, \ t = \dfrac{2 \cdot u }{g }$

We have;

$Angular \ velocity, \ \omega_0 = \dfrac{ 2 \cdot \pi \cdot n}{t} \ (required)$

$\omega_0 = \dfrac{ 2 \cdot \pi \cdot n}{ \dfrac{2 \cdot v_0 }{g }} = \dfrac{ g \cdot \pi \cdot n}{ v_0 }$

Therefore;

$The \ vertical \ velocity, \, \underline{v_0= \dfrac{ g \cdot \pi \cdot n}{ \omega_0 }}$

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