A sling that is 3 meters long, sweeps out an area of 50 square meters every 5 seconds. a. What is its angular velocity? (in rad/sec) b. Wh

Question

A sling that is 3 meters long, sweeps out an area of 50 square meters every 5 seconds. a. What is its angular velocity? (in rad/sec) b. What is its speed? (in m/sec) c. How does doubling the length of the sling affect the previous two answers?

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Sigridomena 5 years 2021-08-15T02:52:06+00:00 1 Answers 177 views 1

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  1. huyenthanh
    0
    2021-08-15T02:53:10+00:00
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    a) 2.22 rad/s

    b) 6.66 m/s

    c) 0.56 rad/s, 3.36 m/s

    Explanation:

    a)

    The angular velocity of an object in rotation is equal to the rate of change of the angular displacement:

    \omega=\frac{\Delta \theta}{\Delta \theta}

    where

    \Delta \theta is the angular displacement

    \Delta t is the time elapsed

    First of all, we calculate the full area of the circle. The radius is the length of the sling:

    r = 3 m

    So the total area is

    A=\pi r^2 = \pi(3)^2=28.3 m^2

    This area corresponds to an angle of 2\pi radians.

    Here the area swept out by the sling is

    A'=50 m^2

    So the angle corresponding to this area is

    \Delta \theta = 2\pi \frac{A'}{A}=2\pi \frac{50}{28.3}=11.1 rad

    While the time elapsed is

    \Delta t=5 s

    Therefore, the angular velocity is:

    \omega=\frac{11.1}{5}=2.22 rad/s

    b)

    For an object in rotational motion, the relationship between angular velocity and linear speed is:

    v=\omega r

    where

    v is the linear speed

    \omega is the angular velocity

    r is the radius

    In this problem:

    \omega=2.22 rad/s is the angular velocity

    r = 3 m is the radius (length of the sling)

    So, the speed is:

    v=(2.22)(3)=6.66 m/s

    c)

    In this case, the length of the sling is doubled, so

    r = 6 m

    First of all, we can re-calculate the area of the full circle:

    A=\pi r^2=\pi (6)^2=113.0 m^2

    Therefore, the angular displacement of the sling this time is:

    \Delta \theta=2\pi \frac{A'}{A}=2\pi \frac{50}{113}=2.78 rad

    And so, the angular velocity this time is

    \omega=\frac{\Delta \theta}{\Delta t}=\frac{2.78}{5}=0.56 rad/s

    The linear speed instead will be given by

    v=\omega r

    And by substituting the new values, we get:

    v=(0.56)(6)=3.36 m/s

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