Physics Solution Manual for 1100 and 2101

Reasoning and solution the wheel must rotate f 2200

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: eed v of the waves on the strand of silk is given by v = F = Mg (1) Neither the mass m nor the length L of the silk strand are given, but we know the density ρ of the silk, which is the ratio of the mass m of the strand to its volume V, according to m ρ = (Equation 11.1). The strand is a cylinder, so its volume V is the product of its length V L and its cross-sectional area A = π r2: V = AL = π r 2 L (2) Equation 11.1 and Equation (2) will permit us to determine the mass per unit length m/L of the silk strand. SOLUTION Substituting Equation (1) into v = v= F = mL F (Equation 16.2), we obtain mL Mg mL (3) Squaring both sides of Equation (3) and solving for the mass M of the spider yields v2 = Mg mL or M= v2 ( m L ) g (4) We must now determine the value of the ratio m/L in terms of the density ρ and radius r of m the silk strand. Substituting Equation (2) for the volume V of the silk strand into ρ = V (Equation 11.1), we obtain 844 WAVES AND SOUND ρ= m m 1 m = 2 = 2 V πr L πr L (5) Therefore, the ratio m/L is given by m L = ρπ r 2 (6) Substituting Equation (6) into Equation (4), then, yields the mass M of the spider: ( )( −6 3 v 2 ( m L ) v 2 ρπ r 2 ( 280 m/s ) 1300 kg/m π 4.0 × 10 m = = M= g g 9.80 m/s 2 2 ) 2 = 5.2 × 10 −4 kg 21. REASONING According to Equation 16.2, the speed of the wave at a distance y above the bottom of the rope depends on the tension in the rope at that spot, and the tension is greater near the top than near the bottom. This is because the rope has weight. Consider the section of the rope between the bottom end and a point at a distance y above the bottom end. The part of the rope above this point must support the weight of this section, which is the mass of the section times the magnitude g of the acceleration due to gravity. Since the rope is uniform, the mass of the section is simply the total mass m of the rope times the fraction y/L, which is the length of the section divided by the total length of the rope. Thus, the weight of y the section is m g . It is this weight that determines the tension and, hence, the speed of L the wave. SOLUTION a. According to Equation 16.2, the speed v of the wave is v= F m/ L where F is the tension, m is the total mass of the rope, and L is the length of the rope. At a point y meters above the bottom end, the rope is supporting the weight of the section y beneath that point, which is m g , as discussed in the REASONING. The rope supports L the weight by virtue of the tension in the rope. Since the rope does not accelerate upward or y downward, the tension must be equal to m g , according to Newton’s second law of L motion. Substituting this tension for F in Equation 16.2 reveals that the speed at a point y meters above the bottom end is Chapter 16 Problems y m g L = v= m/ L 845 yg b. Using the expression just derived, we find the following speeds [y = 0.50 m] v = yg = ( 0.50 m ) ( 9.80 m/s2 ) = [y = 2.0 m] v = yg = ( 2.0 m ) ( 9.80 m/s2 ) = 2.2 m/s 4.4 m/s 22. REASONING AND SOLUTION According to Equation 16.2, the tension in the wire initially is F0 = (m/L)v2 = (9.8 × 10–3 kg/m)(46 m/s)2 = 21 N As the temperature is lowered, the wire will attempt to shrink by an amount ∆L = αL∆T (Equation 12.2), where α is the coefficient of thermal expansion. Since the wire cannot shrink, a stress will develop (see Equation 10.17 and Section 10.8), according to Stress = Y∆L/L = Yα∆T where Y is Young’s modulus. This stress corresponds (see Section 10.8) to an additional tension F ′ : F ′ = (Stress)A = YAα∆T = (1.1 × 1011 Pa)(1.1 × 10–6 m2)(17 × 10–6/C°)(14 C°) = 29 N The total tension in the wire at the lower temperature is now F = F0 + F ′ , so that the new speed of the waves on the wire is v= F0 + F ′ 21 N + 29 N = 71 m/s 9.8 ×10−3 kg/m ______________________________________________________________________________ m/ L = 23. REASONING AND SOLUTION If the string has length L, the time required for a wave on the string to travel from the center of the circle to the ball is t= L vwave The speed of the wave is given by text Equation 16.2 (1) 846 WAVES AND SOUND vwave = F (2) mstring / L The tension F in the string provides the centripetal force on the ball, so that F = mballω 2 r = mballω 2 L (3) Eliminating the tension F from Equations (2) and (3) above yields vwave = mballω 2 L mstring / L = mballω 2 L2 mstring =L mballω 2 mstring Substituting this expression for vwave into Equation (1) gives L t= mballω mstring 2 L = mstring mballω 2 = 0.0230 kg (15.0 kg ) (12.0 rad/s ) 2 = 3.26 × 10−3 s ______________________________________________________________________________ 24. REASONING The speed v of the wave is related to its wavelength λ and frequency f according to v = f λ (Equation 16.1), so we will need to determine the wavelength and 2π x frequency. The mathematical description of this wave has the form y = A sin 2π ft + λ (Equation 16.4), which applies to waves moving in the −x direction. Identifying lik...
View Full Document

Ask a homework question - tutors are online