Physics 4A Solutions to Chapter 15 Hoework Chapter 15 Questions:, 8, 1 Exercises & Probles 6, 5, 31, 41, 59, 7, 73, 88, 90 Answers to Questions: Q 15- (a) toward -x (b) toward +x (c) between -x and 0 (d) between -x and 0 (e) decreasing () increasing Q 15-8 (a) A, B, C (b) C, B, A Q 15-1 (a) π, -180 (b) π/, -90 (c) +π/, +90 Answers to Probles: P 5-6 (a) The angular requency ω is given by ω = π = π/t, where is the requency and T is the period. The relationship = 1/T was used to obtain the last or. Thus ω = π/(1.00 10 5 s) = 6.8 10 5 rad/s. (b) The axiu speed v and axiu displaceent x are related by v = ωx, so x v 1.00 10 = = 5 ω 6.8 10 3 /s rad / s 3 = 1.59 10. P 5-5 (a) We interpret the proble as asking or the equilibriu position; that is, the block is gently lowered until orces balance (as opposed to being suddenly released and allowed to oscillate). I the aount the spring is stretched is x, then we exaine orce-coponents along the incline surace and ind g sin θ (14.0 N)sin 40.0 kx = g sin θ x = = = 0.0750 k 10 N/ at equilibriu. The calculator is in degrees ode in the above calculation. The distance ro the top o the incline is thereore (0.450 + 0.75) = 0.55.
(b) Just as with a vertical spring, the eect o gravity (or one o its coponents) is siply to shit the equilibriu position; it does not change the characteristics (such as the period) o siple haronic otion. Thus, Eq. 15-13 applies, and we obtain 14.0 N 9.80 /s T = π = 0.686 s. 10 N/ P 5-31 (a) Equation 15-1 (divided by π) yields = 1 π k = 1 1000 N/ 5 π 500kg =. Hz.. (b) With x 0 = 0.500, we have U 0 1 = kx = 15 J. 0 (c) With v 0 = 10.0 /s, the initial kinetic energy is K 0 1 = v = 50 J. 0 (d) Since the total energy E = K 0 + U 0 = 375 J is conserved, then consideration o the energy at the turning point leads to = 1 E E kx x = = 0.866. k P 5-41 (a) A unior disk pivoted at its center has a rotational inertia o 1 Mr, where M is its ass and r is its radius. The disk o this proble rotates about a point that is displaced ro its center by r + L, where L is the length o the rod, so, according to the parallel-axis theore, its rotational 1 1 inertia is Mr + M( L+ r). The rod is pivoted at one end and has a rotational inertia o L /3, where is its ass. The total rotational inertia o the disk and rod is 1 1 I = Mr + M( L+ r) + L 3 1 1 = (0.500 kg)(0.100 ) + (0.500 kg)(0.500 + 0.100 ) + (0.70 kg)(0.500 ) 3 = 0.05 kg. (b) We put the origin at the pivot. The center o ass o the disk is d = L+ r= 0.500 +0.100 = 0.600
away and the center o ass o the rod is r = L / = ( 0. 500 )/ = 0. 50 away, on the sae line. The distance ro the pivot point to the center o ass o the disk rod syste is a a a a M + 0.500 0.600 + 0.70 0.50 d r kg kg d = = M+ 0.500 kg + 0.70 kg (c) The period o oscillation is = 0.477. T I 0.05 kg = π = π = 1.50 s. + gd (0.500 kg + 0.70 kg)(9.80 /s )(0.447 ) ( M ) P 5-59 (a) We want to solve e bt/ = 1/3 or t. We take the natural logarith o both sides to obtain bt/ = ln(1/3). Thereore, t = (/b) ln(1/3) = (/b) ln 3. Thus, (b) The angular requency is a = 1.50 kg t 3 = 14.3. 0.30 kg / s ln s k b 800. N/ 0. 30 kg / s ω = = = 31. rad / s. 4 150. kg 4150. kg The period is T = π/ω = (π)/(.31 rad/s) =.7 s and the nuber o oscillations is a a t/t = (14.3 s)/(.7 s) = 5.7. The displaceent x(t) as a unction o tie is shown below. The aplitude, exponentially with tie. xe bt/, decreases
P 5-7 (a) We use Eq. 15-9 and the parallel-axis theore I = I c + h where h = R = 0.16. For a solid disk o ass, the rotational inertia about its center o ass is I c = R /. Thereore, R /+ R 3R T = π = π = 0.873s. gr g (b) We seek a value o r R such that π R + r gr = π 3R g and are led to the quadratic orula: Thus, our result is r = 0.16/ = 0.0630. a or. 3R± 3R 8R R r = = R 4 P 5-73 (a) The spring stretches until the agnitude o its upward orce on the block equals the agnitude o the downward orce o gravity: ky = g, where y = 0.096 is the elongation o the spring at equilibriu, k is the spring constant, and = 1.3 kg is the ass o the block. Thus (b) The period is given by k = g/y = (1.3 kg)(9.8 /s )/(0.096 ) = 1.33 10 N/. T 1 π 1.3kg = = = π = π = 0.6 s. ω k 133 N / (c) The requency is = 1/T = 1/0.6 s = 1.6 Hz. (d) The block oscillates in siple haronic otion about the equilibriu point deterined by the orces o the spring and gravity. It is started ro rest 5.0 c below the equilibriu point so the aplitude is 5.0 c. (e) The block has axiu speed as it passes the equilibriu point. At the initial position, the block is not oving but it has potential energy, 1 1 U ( 1.3 kg)( 9.8 /s )( 0.146 ) ( 133 N / )( 0.146 ) i = gyi + kyi = + = 0.44 J.
When the block is at the equilibriu point, the elongation o the spring is y = 9.6 c and the potential energy is 1 1 U ( 1.3 kg)( 9.8 /s )( 0.096 ) ( 133 N / )( 0.096 ) = gy+ ky = + = 0.61 J. We write the equation or conservation o energy as U = U + 1 v and solve or v: ( Ui U ) ( + ) 0.44 J 0.61J v = = = 0.51 /s. 1.3kg i P 5-88 (a) The Hooke s law orce (o agnitude (100)(0.30) = 30 N) is directed upward and the weight (0 N) is downward. Thus, the net orce is 10 N upward. (b) The equilibriu position is where the upward Hooke s law orce balances the weight, which corresponds to the spring being stretched (ro unstretched length) by 0 N/100 N/ = 0.0. Thus, relative to the equilibriu position, the block (at the instant described in part (a)) is at what one ight call the botto turning point (since v = 0) at x = x where the aplitude is x = 0.30 0.0 = 0.10. (c) Using Eq. 15-13 with = W/g.0 kg, we have T =π = 090. s. k (d) The axiu kinetic energy is equal to the axiu potential energy 1 a a K = = 1 U 100 N/ 0.10 = 0.50 J. kx. Thus, P 5-90 Since the particle has zero speed (oentarily) at x 0, then it ust be at its turning point; thus, x o = x = 0.37 c. It is straightorward to iner ro this that the phase constant φ in Eq. 15- is zero. Also, = 0.5 Hz is given, so we have ω = π = π/ rad/s. The variable t is understood to take values in seconds. (a) The period is T = 1/ = 4.0 s. (b) As noted above, ω = π/ rad/s.
(c) The aplitude, as observed above, is 0.37 c. (d) Equation 15-3 becoes x = (0.37 c) cos(πt/). (e) The derivative o x is v = (0.37 c/s)(π/) sin(πt/) ( 0.58 c/s) sin(πt/). () Fro the previous part, we conclude v = 0.58 c/s. (g) The acceleration-aplitude is a = ω x = 0.91 c/s. (h) Making sure our calculator is in radians ode, we ind x = (0.37) cos(π(3.0)/) = 0. It is iportant to avoid rounding o the value o π in order to get precisely zero, here. (i) With our calculator still in radians ode, we obtain v = (0.58 c/s)sin(π(3.0)/) = 0.58 c/s.