The center of oscillation of a physical pendulum has this interesting property: It an impulse (assumed horizontal and in the plane of oscillation) acts at the center of oscillation, no oscillations are felt at the point of support. Baseball players (and players of many other sports) know that unless the ball hits the bat at this point (called the “sweet spot” by athletes), the oscillations due to the impact will sting their hands. To prove this property, let the stick in Fig. 15-I3 a simulate a baseball bat. Suppose that a horizontal force F → (due to impact with the ball) acts toward the right at P , the center of oscillation. The batter is assumed to hold the bat at O , the pivot point of the stick, (a) What acceleration does the point O undergo as a result of F → ? (b) What angular acceleration is produced by F → about the center of mass of the stick? (c) As a result of the angular acceleration in (b), what linear acceleration does point O undergo? (d) Considering the magnitudes and directions of the accelerations in (a) and (c), convince yourself that P is indeed the “sweet spot.”
The center of oscillation of a physical pendulum has this interesting property: It an impulse (assumed horizontal and in the plane of oscillation) acts at the center of oscillation, no oscillations are felt at the point of support. Baseball players (and players of many other sports) know that unless the ball hits the bat at this point (called the “sweet spot” by athletes), the oscillations due to the impact will sting their hands. To prove this property, let the stick in Fig. 15-I3 a simulate a baseball bat. Suppose that a horizontal force F → (due to impact with the ball) acts toward the right at P , the center of oscillation. The batter is assumed to hold the bat at O , the pivot point of the stick, (a) What acceleration does the point O undergo as a result of F → ? (b) What angular acceleration is produced by F → about the center of mass of the stick? (c) As a result of the angular acceleration in (b), what linear acceleration does point O undergo? (d) Considering the magnitudes and directions of the accelerations in (a) and (c), convince yourself that P is indeed the “sweet spot.”
The center of oscillation of a physical pendulum has this interesting property: It an impulse (assumed horizontal and in the plane of oscillation) acts at the center of oscillation, no oscillations are felt at the point of support. Baseball players (and players of many other sports) know that unless the ball hits the bat at this point (called the “sweet spot” by athletes), the oscillations due to the impact will sting their hands. To prove this property, let the stick in Fig. 15-I3a simulate a baseball bat. Suppose that a horizontal force
F
→
(due to impact with the ball) acts toward the right at P, the center of oscillation. The batter is assumed to hold the bat at O, the pivot point of the stick, (a) What acceleration does the point O undergo as a result of
F
→
? (b) What angular acceleration is produced by
F
→
about the center of mass of the stick? (c) As a result of the angular acceleration in (b), what linear acceleration does point O undergo? (d) Considering the magnitudes and directions of the accelerations in (a) and (c), convince yourself that P is indeed the “sweet spot.”
Study of body parts and their functions. In this combined field of study, anatomy refers to studying the body structure of organisms, whereas physiology refers to their function.
Consider the situation in the figure below; a neutral conducting ball hangs from the ceiling by an insulating string, and a charged insulating rod is going to be placed nearby.
A. First, if the rod was not there, what statement best describes the charge distribution of the ball?
1) Since it is a conductor, all the charges are on the outside of the ball. 2) The ball is neutral, so it has no positive or negative charges anywhere. 3) The positive and negative charges are separated from each other, but we don't know what direction the ball is polarized. 4) The positive and negative charges are evenly distributed everywhere in the ball.
B. Now, when the rod is moved close to the ball, what happens to the charges on the ball?
1) There is a separation of charges in the ball; the side closer to the rod becomes positively charged, and the opposite side becomes negatively charged. 2) Negative charge is drawn from the ground (via the string), so the ball acquires a net negative charge. 3)…
answer question 5-9
AMPS
VOLTS
OHMS
5) 50 A
110 V
6) .08 A
39 V
7) 0.5 A
60
8) 2.5 A
110 V
Chapter 15 Solutions
Fundamentals of Physics Extended 10e Binder Ready Version + WileyPLUS Registration Card
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