Cap 09 halliday

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1. We use Eq. 9-5 to solve for ( x3 , y3 ). (a) The x coordinates of the system’s center of mass is: xcom = m1 x1 + m2 x2 + m3 x3 (2.00 kg)(−1.20 m) + ( 4.00 kg )( 0.600 m ) + ( 3.00 kg ) x3 = m1 + m2 + m3 2.00 kg + 4.00 kg + 3.00 kg

= −0.500 m. Solving the equation yields x3 = –1.50 m. (b) The y coordinates of the system’s center of mass is:
ycom = m1 y1 + m2 y2 + m3 y3 (2.00 kg)(0.500 m) +( 4.00 kg )( −0.750 m ) + ( 3.00 kg ) y3 = m1 + m2 + m3 2.00 kg + 4.00 kg + 3.00 kg

= −0.700 m.

Solving the equation yields y3 = –1.43 m.

2. Our notation is as follows: x1 = 0 and y1 = 0 are the coordinates of the m1 = 3.0 kg particle; x2 = 2.0 m and y2 = 1.0 m are the coordinates of the m2 = 4.0 kg particle; and, x3 = 1.0 m and y3 = 2.0 m are the coordinates of the m3 = 8.0 kg particle.(a) The x coordinate of the center of mass is xcom = m1 x1 + m2 x2 + m3 x3 0 + ( 4.0 kg )( 2.0 m ) + ( 8.0 kg )(1.0 m ) = = 1.1 m. m1 + m2 + m3 3.0 kg + 4.0 kg + 8.0 kg

(b) The y coordinate of the center of mass is ycom = m1 y1 + m2 y2 + m3 y3 0 + ( 4.0 kg )(1.0 m ) + ( 8.0 kg )( 2.0 m ) = = 1.3 m. m1 + m2 + m3 3.0 kg + 4.0 kg + 8.0 kg

(c) As the mass of m3, the topmost particle, isincreased, the center of mass shifts toward that particle. As we approach the limit where m3 is infinitely more massive than the others, the center of mass becomes infinitesimally close to the position of m3.

3. Since the plate is uniform, we can split it up into three rectangular pieces, with the mass of each piece being proportional to its area and its center of mass being at its geometric center.We’ll refer to the large 35 cm × 10 cm piece (shown to the left of the y axis in Fig. 9-38) as section 1; it has 63.6% of the total area and its center of mass is at (x1 ,y1) = (−5.0 cm, −2.5 cm). The top 20 cm × 5 cm piece (section 2, in the first quadrant) has 18.2% of the total area; its center of mass is at (x2,y2) = (10 cm, 12.5 cm). The bottom 10 cm x 10 cm piece (section 3) also has 18.2%of the total area; its center of mass is at (x3,y3) = (5 cm, −15 cm). (a) The x coordinate of the center of mass for the plate is xcom = (0.636)x1 + (0.182)x2 + (0.182)x3 = – 0.45 cm . (b) The y coordinate of the center of mass for the plate is ycom = (0.636)y1 + (0.182)y2 + (0.182)y3 = – 2.0 cm .

4. We will refer to the arrangement as a “table.” We locate the coordinate origin at the left endof the tabletop (as shown in Fig. 9-37). With +x rightward and +y upward, then the center of mass of the right leg is at (x,y) = (+L, –L/2), the center of mass of the left leg is at (x,y) = (0, –L/2), and the center of mass of the tabletop is at (x,y) = (L/2, 0). (a) The x coordinate of the (whole table) center of mass is xcom = M ( + L ) + M ( 0 ) + 3M ( + L / 2 ) = 0.5L . M + M + 3M

With L =22 cm, we have xcom = 11 cm. (b) The y coordinate of the (whole table) center of mass is ycom = or ycom = – 4.4 cm. M ( − L / 2 ) + M ( − L / 2 ) + 3M ( 0 ) L =− , M + M + 3M 5

From the coordinates, we see that the whole table center of mass is a small distance 4.4 cm directly below the middle of the tabletop.

5. (a) By symmetry the center of mass is located on the axis of symmetry of themolecule – the y axis. Therefore xcom = 0. (b) To find ycom, we note that 3mHycom = mN(yN – ycom), where yN is the distance from the nitrogen atom to the plane containing the three hydrogen atoms:

yN = Thus,
ycom =

(10.14 ×10

− 11

m ) − ( 9.4 ×10− 11 m ) = 3.803 ×10− 11 m.
2 2

(14.0067 ) ( 3.803 ×10−11 m ) mN yN = = 3.13 ×10− 11 m 14.0067 + 3 (1.00797 ) mN + 3mH

where Appendix Fhas been used to find the masses.

6. The centers of mass (with centimeters understood) for each of the five sides are as follows: (x1 , y1 , z1 ) = (0, 20, 20) (x2 , y2 , z2 ) = (20, 0, 20) (x3 , y3 , z3 ) = (20, 20, 0) (x4 , y4 , z4 ) = (40, 20, 20) (x5 , y5 , z5 ) = (20, 40, 20) for the side in the yz plane for the side in the xz plane for the side in the xy plane for the remaining side...
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