Chapter 11: Q52P (page 465)

A diver dives from a high platform (Figure 11.100). When he leaves the platform, he tucks tightly and performs three complete revolutions in the air, then straightens out with his body fully extended before entering the water. He is in the air for a total time of $1.4 s$ .What is his angular speed $ω$ just as he enters the water? Give a numerical answer. Be explicit about the details of your model, and include (brief) explanations. You will need to estimate some quantities.

Expert verified

The angular speed of the person just as he enters the water is $13.46$ rad/s.

Step by step solution

Angular speed measures how fast the central angle of a rotating body changes with respect to time.

Angular speed is defined as the rate of change of angular displacement.

Number of revolutions in air $n = 3$

Total time in the air, $t = 1.4 s$

Angle covered in one revolution is $2 π rad$ .

Angle covered in 3 revolutions is,

$\begin{array}{rcl} θ & = & 3 (2 π) \\ = & 6 π \end{array}$

Use the relation between angular speed, angle covered and time taken,

$ω = \frac{θ}{t}$ ….. (1)

Here, $ω$ is theangular speed, $θ$ is anangle covered, and $t$ istime taken.

Substitute all known numerical values in equation (1), you get the angular speed of the person just as he enters the water as

$\begin{array}{rcl} ω & = & \frac{θ}{t} \\ = & \frac{6 π}{1.4} \\ = & 13.46 rad / s \end{array}$

Hence, the angular speed of the person is 13.46 rad/s.

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