The angular acceleration is constant, so this problem just requires substituting given values into Equations 7.7 and 7.8. Solution · (a) Find the angular displacement after 2.00 s, in both radians and revolutions.Use Equation 7.8, setting ωi = 2.00 rad/s, α = 3.5 rad/s2, and t = 2.00 s:image12.jpgConvert radians to revolutions.image13.jpg

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Project Desciption

Why is the launch area for the European Space Agency in South America and not in Europe?

Explanation Satellites are boosted into orbit on top of rockets, which provide the large tangential speed necessary to achieve orbit. Due to its rotation, the surface of Earth is already traveling toward the east at a tangential speed of nearly 1 700 m/s at the equator. This tangential speed is steadily reduced farther north because the distance to the axis of rotation is decreasing. It finally goes to zero at the North Pole. Launching eastward from the equator gives the satellite a starting initial tangential speed of 1 700 m/s, whereas a European launch provides roughly half that speed (depending on the exact latitude).

Goal Apply the rotational kinematics equations in tandem with tangential acceleration and speed.

Problem A compact disc rotates from rest up to an angular speed of 31.4 rad/s in a time of 0.892 s. (a) What is the angular acceleration of the disc, assuming the angular acceleration is uniform? (b) Through what angle does the disc turn while coming up to speed? (c) If the radius of the disc is 4.45 cm, find the tangential speed of a microbe riding on the rim of the disc when t = 0.892 s. (d) What is the magnitude of the tangential acceleration of the microbe at the given time?

Strategy We can solve parts (a) and (b) by applying the kinematic equations for angular speed and angular displacement (Eqs. 7.7 and 7.8). Multiplying the radius by the angular acceleration yields the tangential acceleration at the rim, whereas multiplying the radius by the angular speed gives the tangential speed at that point.

Solution

· (a) Find the angular acceleration of the disc.Apply the angular velocity equation ω = ω i+ αt, taking ωi = 0 at t = 0:image18.jpg

· (b) Through what angle does the disc turn?Use Equation 7.8 for angular displacement, with t = 0.892 s and ωi = 0:image19.jpg

· (c) Find the final tangential speed of a microbe at r = 4.45 cm.Substitute into Equation 7.10:image20.jpg

· (d) Find the tangential acceleration of the microbe at r = 4.45 cm.Substitute into Equation 7.11:image21.jpg

Remarks Because 2π rad = 1 rev, the angular displacement in part (b) corresponds to 2.23 rev. In general, dividing the number of radians by 6 gives a good approximation to the number of revolutions, because 2π ~ 6.

APPLICATION
Phonograph Records and Compact Discs

Before compact discs became the medium of choice for recorded music, phonographs were popular. There are similarities and differences between the rotational motion of phonograph records and that of compact discs. A phonograph record rotates at a constant angular speed. Popular angular speeds were 33⅓ rev/min for long-playing albums (hence the nickname “LP”), 45 rev/min for “singles,” and 78 rev/min used in very early recordings. At the outer edge of the record, the pickup needle (stylus) moves over the vinyl material at a faster tangential speed than when the needle is close to the center of the record. As a result, the sound information is compressed into a smaller length of track near the center of the record than near the outer edge.

CDs, on the other hand, are designed so that the disc moves under the laser pickup at a constant tangential speed. Because the pickup moves radially as it follows the tracks of information, the angular speed of the compact disc must vary according to the radial position of the laser. Because the tangential speed is fixed, the information density (per length of track) anywhere on the disc is the same. Example 7.5 demonstrates numerical calculations for both compact discs and phonograph records.

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