Q64P Question:聽(II) A diver (such as... [FREE SOLUTION]

91影视

Chapter 8: Q64P (page 198)

Question:(II) A diver (such as the one shown in Fig. 8鈥�28) can reduce her moment of inertia by a factor of about 3.5 when changing from the straight position to the tuck position. If she makes 2.0 rotations in 1.5 s when in the tuck position, what is her angular speed (rev/s) when in the straight position?

Short Answer

Expert verified

The angular speed of the diver in the straight position is \(0.38\;{\rm{rev/s}}\).

Step by step solution

Given data

The factor by which the moment of inertia reduces is \(\frac{{I'}}{I} = 3.5\).

The number of rotations is \(\omega = 2\;{\rm{rev}}\).

The time is \(t = 1.5\;{\rm{s}}\).

Understanding conservation of angular momentum

In this problem, the external force acting on the diver is the force due to gravity passing through the center of the mass. No other type of force takes place. So, the angular momentum is conserved.

Determine the angular speed of the diver

he relation of angular momentum can be written as:

\(\begin{aligned}{c}L &= L'\\I\omega &= I'\omega '\\\omega ' &= \left( {\frac{I}{{I'}}} \right)\omega \end{aligned}\)

Here, \(L\) and \(L'\) are the initial and final angular momentum, and \(\omega '\) is the final angular velocity of the diver.

On plugging the values in the above relation, you get:

\(\begin{aligned}{l}\omega ' &= \left( {\frac{1}{{3.5}}} \right)\left( {\frac{{2\;{\rm{rev}}}}{{1.5\;{\rm{s}}}}} \right)\\\omega ' &= 0.38\;{\rm{rev/s}}\end{aligned}\)

Thus, \(\omega ' = 0.38\;{\rm{rev/s}}\) is the final angular speed of the diver.