A spaceship negotiates a circular turn of radius 2925 km at a speed of 29960 km/h. (a) What is the magnitude of the angular speed? 0.0028

Question

A spaceship negotiates a circular turn of radius 2925 km at a speed of 29960 km/h. (a) What is the magnitude of the angular speed? 0.0028 Correct: Your answer is correct. rad/s (b) What is the magnitude of the radial acceleration? 23.68 Correct: Your answer is correct. m/s2 (c) What is the magnitude of the tangential acceleration? m/s2

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Vodka 2 months 2021-07-31T19:10:31+00:00 1 Answers 2 views 0

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    2021-07-31T19:11:44+00:00

    a) 0.0028 rad/s

    b) 23.68 m/s^2

    c) 0 m/s^2

    Explanation:

    a)

    When an object is in circular motion, the angular speed of the object is the rate of change of its angular position. In formula, it is given by

    \omega = \frac{\theta}{t}

    where

    \theta is the angular displacement

    t is the time interval

    The angular speed of an object in circular motion can also be written as

    \omega = \frac{v}{r} (1)

    where

    v is the linear speed of the object

    r is the radius of the orbit

    For the spaceship in this problem we have:

    v=29,960 km/h is the linear speed, converted into m/s,

    v=8322 m/s

    r=2925 km = 2.925\cdot 10^6 m is the radius of the orbit

    Subsituting into eq(1), we find the angular speed of the spaceship:

    \omega=\frac{8322}{2.925\cdot 10^6}=0.0028 rad/s

    b)

    When an object is in circular motion, its direction is constantly changing, therefore the object is accelerating; in particular, there is a component of the acceleration acting towards the  centre of the orbit: this is called centripetal acceleration, or radial acceleration.

    The magnitude of the radial acceleration is given by

    a_r=\omega^2 r

    where

    \omega is the angular speed

    r is the radius of the orbit

    For the spaceship in the problem, we have

    \omega=0.0028 rad/s is the angular speed

    r=2925 km = 2.925\cdot 10^6 m is the radius of the orbit

    Substittuing into the equation above, we find the radial acceleration:

    a_r=(0.0028)^2(2.925\cdot 10^6)=23.68 m/s^2

    c)

    When an object is in circular motion, it can also have a component of the acceleration in the direction tangential to its motion: this component is called tangential acceleration.

    The tangential acceleration is given by

    a_t=\frac{\Delta v}{\Delta t}

    where

    \Delta v is the change in the linear speed

    \Delta  t is the time interval

    In this problem, the spaceship is moving with constant linear speed equal to

    v=8322 m/s

    Therefore, its linear speed is not changing, so the change in linear speed is zero:

    \Delta v=0

    And therefore, the tangential acceleration is zero as well:

    a_t=\frac{0}{\Delta t}=0 m/s^2

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