S08P2AChap13Sol

S08P2AChap13Sol - CHAPTER 13 Solutions Problem 3. A wheel...

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CHAPTER 13 Solutions Problem 3. A wheel is spinning at 45 rpm with its spin axis vertical. After 15 s, it’s spinning at 60 rpm with its axis horizontal. Find (a) the magnitude of its average angular acceleration and (b) the angle the average angular acceleration vector makes with the horizontal. Solution Suppose that the x -axis is horizontal in the direction of the final angular velocity ( ! f = (60 rpm) î ) and the y -axis is vertical in the direction of the initial angular velocity ( î = ( 45 rpm ) ˆ j ). Equation 13-1 implies that av = ( " f #" i ) = $ t = ( 60 î ! 45 ˆ j ) rpm = 15 s = ( 4 î ! 3 ˆ j ) rpm / s . Its magnitude is av = (4) 2 + ( " 3) 2 rpm/s = 5 rpm/s = 5( # = 30) s " 2 = 0.524 s " 2 , at an angle = tan " 1 ( " 3 4 ) = " 36.9 ° to the x -axis (i.e., below the horizontal). Problem 5. A wheel is spinning with angular speed = 5.0 rad/s, when a constant angular acceleration = 0.85 rad/s 2 is applied at right angles to the initial angular velocity. How long does it take for the angular speed to increase by 10 rad/s? Solution For constant angular acceleration, f = i + t . If is perpendicular to i , f 2 = i 2 + ( t ) 2 , or t = f 2 " i 2 = . Therefore, for i = 5 s " 1 to increase to f = 15 s " 1 under the given conditions requires time t = (15) 2 ! (5) 2 s ! 1 ÷ (0.85 s ! 2 ) = 16.6 s. (Note: f is the total angular velocity of the wheel, not just its spin.) Problem 8. The disk in Fig. 13-28 has radius 35 cm. Give the magnitude and direction of the smallest force that you could apply at point P to produce a torque of magnitude 1.2 N ! m (a) about an axis through the center of the disk and perpendicular to the page and (b) about a vertical axis tangent to the left edge of the disk. Solution Equation 13-2 defines the torque about a point (the origin of the displacement r ); the torque about an axis through that point is the component of the torque in the direction of the axis. This component of torque depends only on the components of r and F perpendicular to the axis, so we can restrict our choice of origin to be on the axis, in a plane perpendicular to the axis and containing the point of application of the force, P . If only the magnitude of the torque is specified, its component along the axis could be positive or negative. (a) Choose an origin at the center of the disk with x -axis to the right, y -axis up, and z - axis out of the page. Then r = (0.35 m) î and the z component of the torque is [ r ! F ] z = (0.35 m) F y = ± 1.2 N " m (see the result of Problem 14). Thus, F y = ± 1 . 2 N ! m = 0 . 35 m = ± 3 . 43 N . (b) Now choose the origin at the tangent point of the disk. Then r = ( 0 . 70 m ) î , [ r ! F ] y = " ( 0 . 70 m ) F z = ± 1 . 2 N # m , and F z = m 1 . 71 N . Of course, the other components of the smallest force in (a) and (b) are zero.
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