Question

A diver comes off a board with arms straight up and legs straight down, giving her a moment of inertia about her rotation axis of . She then tucks into a small ball, decreasing this moment of inertia to . While tucked, she makes two complete revolutions in 1.3 .


If she hadn't tucked at all, how many revolutions would she have made in the 1.7 from board to water?
Express your answer using two significant figures.

=_____________rev

Answer 1

Concepts and reason

The concept required to solve the given problem is law of conservation of angular momentum.

Initially, find the angular speed while she tucked i.e. ${\omega _2}$ . Use the law of conservation of angular momentum, re-arrange that for ${\omega _1}$ , substitute the values in the expression and find ${\omega _1}$ . Find the number of revolutions made by her in 1.7 sec.

Fundamentals

The law of conservation of angular momentum states that the angular momentum of a body is conserved unless an external torque acts on the system.

${L_i} = {L_f}$

Here, i and f represent initial and final states of a body.

The moment of inertial value in the straight position is,

${I_1} = 18{\rm{ kg/}}{{\rm{m}}^2}$

The moment of inertial value when she tucks is,

${I_2} = 3.6{\rm{ kg/}}{{\rm{m}}^2}$

Angular speed while she tucked is calculated as:

${\omega _2} = \frac{N}{t}$

Here, N is the number of revolutions and t is the time taken by her to complete that revolutions.

Substitute 2 rev for N and 1.3 sec for t in equation ${\omega _2} = \frac{N}{t}$ .

$\begin{array}{c}\\{\omega _2} = \frac{{2{\rm{ rev}}}}{{1.3{\rm{ sec}}}}\\\\ = 1.54{\rm{ rev/sec}}\\\end{array}$

The law of conservation of angular momentum states that the angular momentum of a body is conserved unless an external torque acts on the system.

${L_i} = {L_f}$

The above equation can be reduced as:

${I_1}{\omega _1} = {I_2}{\omega _2}$

Here, ${I_1}$ and ${I_2}$ are the initial and final moment of inertia and ${\omega _1}$ and ${\omega _2}$ are the initial and final angular speed of the system.

Re-arrange equation ${I_1}{\omega _1} = {I_2}{\omega _2}$ for ${\omega _1}$ .

${\omega _1} = \left( {\frac{{{I_2}}}{{{I_1}}}} \right){\omega _2}$

Substitute 1.54 rev/sec for ${\omega _2}$ , $18{\rm{ kg/}}{{\rm{m}}^2}$ for ${I_1}$ and $3.6{\rm{ kg/}}{{\rm{m}}^2}$ for ${I_2}$ in equation ${\omega _1} = \left( {\frac{{{I_2}}}{{{I_1}}}} \right){\omega _2}$ .

$\begin{array}{c}\\{\omega _1} = \left( {\frac{{3.6{\rm{ kg/}}{{\rm{m}}^2}}}{{18{\rm{ kg/}}{{\rm{m}}^2}}}} \right)\left( {1.54{\rm{ rev/sec}}} \right)\\\\ = 0.308{\rm{ rev/sec}}\\\end{array}$

Number of revolutions made in 1.7 sec are,

$N = {\omega _1}\left( {1.7{\rm{ sec}}} \right)$

Substitute 0.308 rev/sec for ${\omega _1}$ in $N = {\omega _1}\left( {1.7{\rm{ sec}}} \right)$ .

$\begin{array}{c}\\N = \left( {0.308{\rm{ rev/sec}}} \right)\left( {1.7{\rm{ sec}}} \right)\\\\ = 0.52{\rm{ rev}}\\\end{array}$

Ans:

The number of revolutions is 0.52.