391_PartUniversity Physics Solution

# 391_PartUniversity Physics Solution - 12-12 Chapter 12(b...

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12-12 Chapter 12 (b) The x -component of the gravitational force on the sphere is 2 2 (/) , (1 ( / )) ( ) x UG m M L x G m M F x LL x x L x ∂− =− = ∂+ + with the minus sign indicating an attractive force. As x L >> , the denominator in the above expression approaches 2 x , and x F 2 / GmM x →− , as expected. EVALUATE: When x is much larger than L the rod can be treated as a point mass, and our results for U and x F do reduce to the correct expression when x L >> . 12.41. IDENTIFY: Find the potential due to a small segment of the ring and integrate over the entire ring to find the total U . (a) SET UP: Divide the ring up into small segments dM , as indicated in Figure 12.41. Figure 12.41 EXECUTE: The gravitational potential energy of dM and m is / . dU GmdM r = − The total gravitational potential energy of the ring and particle is /. Ud m d M r == ∫∫ But 22 rx a =+ is the same for all segments of the ring, so Gm GmM GmM M rr x a + (b) EVALUATE: When , x a >> 2 x ax x + →= and / . m M x = − This is the gravitational potential energy of two point masses separated by a distance x . This is the expected result. (c) IDENTIFY and SET UP: Use / x Fd U d x with ( ) Ux from part (a) to calculate . x F EXECUTE: x dU d GmM F dx dx x a ⎛⎞ ⎜⎟ + ⎝⎠ 1 2 3 2 1 () ( ( 2 ) ) 2 x d F GmM x a GmM x x a dx −− + = + 3 2 /( ) ; x FG m M x x a + the minus sign means the force is attractive. EVALUATE: (d) For , x a >> 3 2 2 3 23 ( ) x x +→ = Then 32 // . x F GmMx x GmM x This is the force between two point masses separated by a distance x and is the expected result. (e) For 0, x = M m a Each small segment of the ring is the same distance from the center and the potential is the same as that due to a point charge of mass M located at a distance a . For 0, x = 0. x F = When the particle is at the center of the ring, symmetrically placed segments of the ring exert equal and opposite forces and the total force exerted by the ring is zero. 12.42. IDENTIFY: At the equator the object has inward acceleration 2 E v R and the reading w of the balance is related to the true weight 0 w (the gravitational force exerted by the earth) by 2 0 E mv ww R −= . At the North Pole, rad 0 a = and 0 . = SET UP: As shown in Section 12.7, 465 m/s v = . 6 E 6.38 10 m R . EXECUTE: 0 875 N w = and 0 89.29 kg w m g . 0 6 E (465 m/s) 875 N (89.29 kg) 872 N mv R =− = = × EVALUATE: The rotation of the earth causes the scale reading to be slightly less than the true weight, since there must be a net inward force on the object. 12.43. IDENTIFY and SET UP: Ate the north pole, g0 0 , Fwm g where 0 g is given by Eq.(12.4) applied to Neptune. At the equator, the apparent weight is given by Eq.(12.28). The orbital speed v is obtained from the rotational period using Eq.(12.12).

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Gravitation 12-13 EXECUTE: (a) 21 1 2 2 2 6 7 2 0 / (6.673 10 N m /kg )(1.0 10 kg)/(2.5 10 m) gG m R == ×⋅ × × 2 10.7 m/s . = This agrees with the value of g given in the problem.
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391_PartUniversity Physics Solution - 12-12 Chapter 12(b...

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