C. Solutions to Problems August 20, 2025
C.1 Calculus and Vectors ¶ Solutions to Practice Problems from Chapter 1.9
Problem 1-1: Question
v = C τ ( 1 − e − t / τ ) + v 0 v = C\tau (1 - e^{-t/\tau}) + v_{0} v = C τ ( 1 − e − t / τ ) + v 0 Problem 1-2: Question
23 c , − 7 c k ^ 23\mathrm{c}, -7c\hat{k} 23 c , − 7 c k ^ Problem 1-3: Question
cos θ = 2 k k 2 + 1 \cos \theta = \frac{2k}{k^{2}+ 1} cos θ = k 2 + 1 2 k Problem 1-4: Question
a ) − 48 , b ) 10 , c ) 9 c − 9 − 2 c 2 , d ) c = 3 , 3 2 \mathrm{a}) -48, \mathrm{b}) 10, \mathrm{c}) 9c - 9 - 2c^{2}, \mathrm{d}) c = 3, \frac{3}{2} a ) − 48 , b ) 10 , c ) 9 c − 9 − 2 c 2 , d ) c = 3 , 2 3 Problem 1-5: Question
a ) 4 ı ^ − 26 ȷ ^ − 57 k ^ \mathrm{a}) 4\hat{\imath} - 26\hat{\jmath} - 57\hat{k} a ) 4 ^ − 26 ^ − 57 k ^ b ) ( − 45 s − 6 ) ı ^ + 42 s ȷ ^ + ( 15 s − 12 ) k ^ \mathrm{b}) (-45s - 6)\hat{\imath} + 42s\hat{\jmath} + (15s - 12)\hat{k} b ) ( − 45 s − 6 ) ^ + 42 s ^ + ( 15 s − 12 ) k ^ c ) ( 45 s + 6 ) ı ^ − 42 s ȷ ^ + ( − 15 s + 12 ) k ^ \mathrm{c}) (45s + 6)\hat{\imath} - 42s\hat{\jmath} + (-15s + 12)\hat{k} c ) ( 45 s + 6 ) ^ − 42 s ^ + ( − 15 s + 12 ) k ^ d ) ( 27 s + 6 ) ı ^ + 75 s ȷ ^ + ( − 9 s − 27 ) k ^ \mathrm{d}) (27s + 6)\hat{\imath} + 75s\hat{\jmath} + (-9s - 27)\hat{k} d ) ( 27 s + 6 ) ^ + 75 s ^ + ( − 9 s − 27 ) k ^ Problem 1-6: Question
a ) 4 ı ^ − 7 ȷ ^ − k ^ \mathrm{a}) 4\hat{\imath} - 7\hat{\jmath} - \hat{k} a ) 4 ^ − 7 ^ − k ^ b ) − 7 ı ^ + 4 ȷ ^ − k ^ \mathrm{b}) -7\hat{\imath} + 4\hat{\jmath} - \hat{k} b ) − 7 ^ + 4 ^ − k ^ c ) 7 ı ^ − 4 ȷ ^ + k ^ \mathrm{c}) 7\hat{\imath} - 4\hat{\jmath} + \hat{k} c ) 7 ^ − 4 ^ + k ^ Problem 1-7: Question
a ) ( 5 , tan − 1 ( 4 3 ) ) \mathrm{a}) (5,\tan ^{-1}(\frac{4}{3})) a ) ( 5 , tan − 1 ( 3 4 )) b ) θ = π 4 o r θ = 4 5 ∘ \mathrm{b}) \theta = \frac{\pi}{4} \mathrm{or} \theta = 45^{\circ} b ) θ = 4 π or θ = 4 5 ∘ c ) r = sin θ − cos θ cos 2 θ \mathrm{c}) r = \frac{\sin \theta - \cos \theta}{\cos ^{2}\theta} c ) r = cos 2 θ sin θ − cos θ Problem 1-8: Question
v ⃗ = 2 e 2 t r ^ + 2 t e 2 t θ ^ , a ⃗ = ( 4 e 2 t − 2 t 2 e 2 t ) r ^ + ( 2 e 2 t + 8 t e 2 t ) θ ^ \vec{v} = 2e^{2t}\hat{r} + 2te^{2t}\hat{\theta}, \vec{a} = (4e^{2t}- 2t^{2}e^{2t})\hat{r} + (2e^{2t}+ 8te^{2t})\hat{\theta} v = 2 e 2 t r ^ + 2 t e 2 t θ ^ , a = ( 4 e 2 t − 2 t 2 e 2 t ) r ^ + ( 2 e 2 t + 8 t e 2 t ) θ ^ Problem 1-9: Question
θ = π 4 \theta = \frac{\pi}{4} θ = 4 π Problem 1-10: Question
a ) v ⃗ = k r r ^ + c r θ ^ \mathrm{a}) \vec{v} = kr\hat{r} + cr\hat{\theta} a ) v = k r r ^ + cr θ ^ b ) a ⃗ = ( k 2 − c 2 ) r ( t ) r ^ + ( 2 k c ) r ( t ) θ ^ \mathrm{b}) \vec{a} = (k^{2}- c^{2})r(t)\hat{r} + (2kc)r(t)\hat{\theta} b ) a = ( k 2 − c 2 ) r ( t ) r ^ + ( 2 k c ) r ( t ) θ ^ c ) cos φ = k k 2 + c 2 = c o n s t a n t \mathrm{c}) \cos \varphi = \frac{k}{\sqrt{k^{2}+ c^{2}}} = constant c ) cos φ = k 2 + c 2 k = co n s t an t Problem 1-11: Question
a ) x , b ) 1 + x + x 2 , c ) − x − 1 2 x 2 \mathrm{a}) x, \mathrm{b}) 1 + x + x^{2}, \mathrm{c}) -x - \frac{1}{2} x^{2} a ) x , b ) 1 + x + x 2 , c ) − x − 2 1 x 2 Problem 1-12: Question
a ) 1 , b ) 1 + 5 x , c ) ln 2 + 3 2 x \mathrm{a}) 1, \mathrm{b}) 1 + 5x, \mathrm{c}) \ln 2 + \frac{3}{2} x a ) 1 , b ) 1 + 5 x , c ) ln 2 + 2 3 x C.2 Newtonian Review ¶ Solutions to Practice Problems from Chapter 2.8
Problem 2-1: Question
m 1 : T − m 1 g = m 1 y ¨ 1 m_{1}: T - m_{1}g = m_{1}\ddot{y}_{1} m 1 : T − m 1 g = m 1 y ¨ 1 m 2 : T − m 2 g = m 2 y ¨ 2 m_{2}: T - m_{2}g = m_{2}\ddot{y}_{2} m 2 : T − m 2 g = m 2 y ¨ 2 p u l l e y : F − m p g − 2 T = m p a F = 0 pulley : F - m_{p}g - 2T = m_{p}a_{F}= 0 p u ll ey : F − m p g − 2 T = m p a F = 0 Problem 2-2: Question
v min = 2 g sin 2 θ ( h − y 0 ) v_{\min}= \sqrt{\frac{2g}{\sin ^{2}\theta} (h - y_{0})} v m i n = sin 2 θ 2 g ( h − y 0 ) Problem 2-4: Question
v = ( F 0 m ) t − ( α 3 m ) t 3 + v 0 v = \bigg(\frac{F_{0}}{m} \bigg)t - \bigg(\frac{\alpha}{3m} \bigg)t^{3}+ v_{0} v = ( m F 0 ) t − ( 3 m α ) t 3 + v 0 Problem 2-5: Question
a ) v = − k 2 x 2 + v 0 b ) x ( t ) = 2 v 0 k ( e 2 k v 0 t − 1 ) ( 1 + e 2 k v 0 t ) \begin{aligned}
\mathrm{a}) v &= - \frac{k}{2} x^{2}+ v_{0} \\
\mathrm{b}) x(t) &= \sqrt{\frac{2v_{0}}{k}} \frac{\big(e^{\sqrt{2kv_{0}}t}- 1\big)}{\big(1 + e^{\sqrt{2kv_{0}}t}\big)}
\end{aligned} a ) v b ) x ( t ) = − 2 k x 2 + v 0 = k 2 v 0 ( 1 + e 2 k v 0 t ) ( e 2 k v 0 t − 1 ) Problem 2-6: Question
a) ∑ F = − b m v + m g \sum F = -bmv + mg ∑ F = − bm v + m g
b ) g b c ) y = g b 2 e − b t + g b t − g b 2 \begin{aligned}
\mathrm{b}) \frac{g}{b} \\
\mathrm{c}) y &= \frac{g}{b^{2}} e^{-bt}+ \frac{g}{b} t - \frac{g}{b^{2}}
\end{aligned} b ) b g c ) y = b 2 g e − b t + b g t − b 2 g Problem 2-7: Question
b ) θ = tan − 1 μ \mathrm{b}) \theta = \tan ^{-1}\mu b ) θ = tan − 1 μ c ) F e x t = μ M g 1 + μ 2 \mathrm{c}) F_{ext}= \frac{\mu Mg}{\sqrt{1 + \mu ^{2}}} c ) F e x t = 1 + μ 2 μ M g Problem 2-8: Question
b ) ( m − ρ V ) g − α v = m d v d t \mathrm{b}) (m - \rho V)g - \alpha v = m \frac{\mathrm{d}v}{\mathrm{d}t} b ) ( m − ρ V ) g − αv = m d t d v c ) v ( t ) = ( m − ρ V ) g α ( 1 − e − α t / m ) \mathrm{c}) v(t) = \frac{(m - \rho V)g}{\alpha} \Big(1 - e^{-\alpha t/m}\Big) c ) v ( t ) = α ( m − ρ V ) g ( 1 − e − α t / m ) d ) v = ( m − ρ V ) g α \mathrm{d}) v = \frac{(m - \rho V)g}{\alpha} d ) v = α ( m − ρ V ) g Problem 2-9: Question
b ) t r = m 2 v 0 μ g m 1 \mathrm{b}) t_{r}= \frac{m_{2}v_{0}}{\mu gm_{1}} b ) t r = μg m 1 m 2 v 0 c ) v r = m 1 v 0 m 2 + m 1 \mathrm{c}) v_{r}= \frac{m_{1}v_{0}}{m_{2}+ m_{1}} c ) v r = m 2 + m 1 m 1 v 0 Problem 2-10: Question
b) lower pulley goes down, M M M M 2 M_{2} M 2 M 1 M_{1} M 1
c) − 1 2 ( y ¨ 1 + y ¨ 2 ) - \frac{1}{2} (\ddot{y}_{1}+ \ddot{y}_{2}) − 2 1 ( y ¨ 1 + y ¨ 2 ) y ¨ 1 \ddot{y}_{1} y ¨ 1 M 1 M_{1} M 1 y ¨ 2 \ddot{y}_{2} y ¨ 2 M 2 M_{2} M 2
d ) y ¨ 2 = 21 29 g \mathrm{d}) \ddot{y}_{2}= \frac{21}{29} g d ) y ¨ 2 = 29 21 g C.3 Simple Harmonic Motion ¶ Solutions to Practice Problems from Chapter 3.9
Problem 3-1: Question
Quarter the mass ( m / 4 ) (m/4) ( m /4 )
Problem 3-2: Question
Looking up the gravitational acceleration for each planet, some examples:
Mercury: 0.375 m, Earth: 1 m, Mars: 0.376 m, Jupiter: 2.5 m, Neptune: 1.13 m
Problem 3-3: Question
a ) x ¨ = 3 g L x = 0 \mathrm{a}) \ddot{x} = \frac{3g}{L} x = 0 a ) x ¨ = L 3 g x = 0 c ) L = 0.745 m \mathrm{c}) L = 0.745 \mathrm{m} c ) L = 0.745 m Problem 3-4: Question
ω 0 = ( k 1 + k 2 ) m \omega _{0}= \sqrt{\frac{(k_{1}+ k_{2})}{m}} ω 0 = m ( k 1 + k 2 ) Problem 3-5: Question
a ) T = 2 π L g \mathrm{a}) T = 2\pi \sqrt{\frac{L}{g}} a ) T = 2 π g L b) T = 2 π L ( g + y ¨ ) → ∞ T = 2\pi\sqrt{\dfrac{L}{(g + \ddot{y})}} \rightarrow \infty T = 2 π ( g + y ¨ ) L → ∞ y ¨ = − g \ddot{y} = -g y ¨ = − g
Problem 3-6: Question
a ) x = A cos ( t k m + g L ) \mathrm{a}) x = A\cos \Bigg(t\sqrt{\frac{k}{m} + \frac{g}{L}} \Bigg) a ) x = A cos ( t m k + L g ) v = − A k m + g L sin ( t k m + g L ) v = -A\sqrt{\frac{k}{m} + \frac{g}{L}} \sin \Bigg(t\sqrt{\frac{k}{m} + \frac{g}{L}} \Bigg) v = − A m k + L g sin ( t m k + L g ) a = − A ( k m + g L ) cos ( t k m + g L ) a = -A\Bigg(\frac{k}{m} + \frac{g}{L} \Bigg)\cos \Bigg(t\sqrt{\frac{k}{m} + \frac{g}{L}} \Bigg) a = − A ( m k + L g ) cos ( t m k + L g ) Problem 3-7: Question
b ) x 0 = − m g sin θ k \mathrm{b}) x_{0}= - \frac{mg\sin \theta}{k} b ) x 0 = − k m g sin θ c ) 0 = x ¨ + k m ( x − x 0 ) \mathrm{c}) 0 = \ddot{x} + \frac{k}{m} (x - x_{0}) c ) 0 = x ¨ + m k ( x − x 0 ) d ) T = 2 π m k \mathrm{d}) T = 2\pi \sqrt{\frac{m}{k}} d ) T = 2 π k m Problem 3-8: Question
b ) x ¨ + 4 k m x = 0 \mathrm{b}) \ddot{x} + \frac{4k}{m} x = 0 b ) x ¨ + m 4 k x = 0 c ) ω 0 = 4 k m , T = 2 π m 4 k \mathrm{c}) \omega _{0}= \sqrt{\frac{4k}{m}}, T = 2\pi \sqrt{\frac{m}{4k}} c ) ω 0 = m 4 k , T = 2 π 4 k m d ) x = A cos ( t 4 k m ) , v = − A 4 k m sin ( t 4 k m ) , a = − A 4 k m cos ( t 4 k m ) \mathrm{d}) x = A\cos \Bigg(t\sqrt{\frac{4k}{m}} \Bigg), v = -A\sqrt{\frac{4k}{m}} \sin \Bigg(t\sqrt{\frac{4k}{m}} \Bigg), a = -A\frac{4k}{m}\cos \Bigg(t\sqrt{\frac{4k}{m}} \Bigg) d ) x = A cos ( t m 4 k ) , v = − A m 4 k sin ( t m 4 k ) , a = − A m 4 k cos ( t m 4 k ) Problem 3-10: Question
b ) 0 = x ¨ + ( k 1 k 2 + k 3 ( k 2 + k 1 ) m ( k 1 + k 2 ) ) x \mathrm{b}) 0 = \ddot{x} + \Bigg(\frac{k_{1}k_{2}+ k_{3}(k_{2}+ k_{1})}{m(k_{1}+ k_{2})} \Bigg)x b ) 0 = x ¨ + ( m ( k 1 + k 2 ) k 1 k 2 + k 3 ( k 2 + k 1 ) ) x c ) T = 2 π m ( k 1 + k 2 ) k 1 k 2 + k 3 ( k 1 + k 2 ) \mathrm{c}) T = 2\pi \sqrt{\frac{m(k_{1}+ k_{2})}{k_{1}k_{2}+ k_{3}(k_{1}+ k_{2})}} c ) T = 2 π k 1 k 2 + k 3 ( k 1 + k 2 ) m ( k 1 + k 2 ) C.4 Introduction to Non-Inertial and Rotating Frames ¶ Solutions to Practice Problems from Chapter 4.8
Problem 4-1: Question
a) The mass m m m x − x- x −
b) The mass m m m x − x- x −
Problem 4-2: Question
a) 515N [down]
b) 860N [down]
Problem 4-3: Question
θ = 26.6 \theta = 26.6 θ = 26.6
Problem 4-4: Question
a ) a = g tan θ , b ) g e f f = g / cos θ \mathrm{a}) a = g\tan \theta, \mathrm{b}) g_{eff}= g/\cos \theta a ) a = g tan θ , b ) g e ff = g / cos θ Problem 4-5: Question
a) higher (fictitious force points down the incline)
b ) g e f f = 1.05 g \mathrm{b}) g_{eff}= 1.05g b ) g e ff = 1.05 g Problem 4-6: Question
T a c c = 0.97 T 0 T_{acc}= 0.97T_{0} T a cc = 0.97 T 0 Problem 4-7: Question
a c e n t = v 0 2 R , a C o r = 0 a_{cent}= \frac{v_{0}^{2}}{R}, a_{Cor}= 0 a ce n t = R v 0 2 , a C or = 0 Problem 4-8: Question
a) No fictitious forces act (assuming the very center of the room is the rotation axis)
b) Centrifugal force
c) Centrifugal and Coriolis forces
Problem 4-9: Question
a ) F f i c = − m a z ^ \mathrm{a}) F_{fic}= -ma\hat{z} a ) F f i c = − ma z ^ b ) g e f f = − 9.8 m s − 2 z ^ \mathrm{b}) g_{eff}= -9.8\mathrm{m} \mathrm{s}^{-2}\hat{z} b ) g e ff = − 9.8 m s − 2 z ^ c) v min = 15.65 m s − 1 v_{\min}= 15.65\mathrm{m} \mathrm{s}^{-1} v m i n = 15.65 m s − 1 t = 0.64 s t = 0.64\mathrm{s} t = 0.64 s
d) No forces
e) v min = 15.65 m s − 1 v_{\min}= 15.65\mathrm{m} \mathrm{s}^{-1} v m i n = 15.65 m s − 1 t = 0.64 s t = 0.64\mathrm{s} t = 0.64 s
f) Inertial: horizontal and Non-Inertial: parabolic
C.5 Applications of Non-Inertial and Rotating Frames ¶ Solutions to Practice Problems from Chapter 5.8
Problem 5-1: Question
a ) F C o r \mathrm{a}) F_{Cor} a ) F C or b ) F c e n t \mathrm{b}) F_{cent} b ) F ce n t c) F c e n t F_{cent} F ce n t F a z F_{az} F a z
d) F c e n t F_{cent} F ce n t F C o r F_{Cor} F C or
e) F c e n t , F C o r , F a z F_{cent}, F_{Cor}, F_{az} F ce n t , F C or , F a z F t r a n s F_{trans} F t r an s
Problem 5-2: Question
a ) 8.84 ω 2 r ^ \mathrm{a}) 8.84\omega ^{2}\hat{r} a ) 8.84 ω 2 r ^ b) 4.7 s − 1 4.7 \mathrm{s}^{-1} 4.7 s − 1
Problem 5-3: Question
The azimuthal force F a z F_{az} F a z y ′ − y^{\prime }- y ′ − F c e n t F_{cent} F ce n t x ′ − x^{\prime }- x ′ −
Problem 5-3: Question
a) azimuthal and centrifugal forces
b ) 0.014 s − 2 \mathrm{b}) 0.014 \mathrm{s}^{-2} b ) 0.014 s − 2 c ) F c e n t = 343 N , F a z = 12 N \mathrm{c}) F_{cent}= 343 \mathrm{N}, F_{az}= 12 \mathrm{N} c ) F ce n t = 343 N , F a z = 12 N Problem 5-5: Question
a) − x ^ -\hat{x} − x ^
b ) 0.0005 \mathrm{b}) 0.0005 b ) 0.0005 Problem 5-6: Question
θ \theta θ
Problem 5-7: Question
East
Problem 5-8: Question
a) F ⃗ c e n t = m ω 2 x ′ x ^ ′ \vec{F}_{cent}= m\omega ^{2}x^{\prime }\hat{x}^{\prime } F ce n t = m ω 2 x ′ x ^ ′ F ⃗ C o r = − 2 m ω x ˙ ′ y ^ ′ \vec{F}_{Cor}= -2m\omega \dot{x}^{\prime }\hat{y}^{\prime } F C or = − 2 mω x ˙ ′ y ^ ′
b ) F ⃗ I = 2 m ω x ˙ ′ y ^ ′ \mathrm{b}) \vec{F}_{I}= 2m\omega \dot{x}^{\prime }\hat{y}^{\prime } b ) F I = 2 mω x ˙ ′ y ^ ′ c) The centrifugal force F c e n t F_{cent} F ce n t x x x F C o r F_{Cor} F C or y y y F I F_{I} F I y y y
d ) x ( t ) = L 2 [ e ω t − e − ω t ] \mathrm{d}) x(t) = \frac{L}{2} [e^{\omega t}- e^{-\omega t}] d ) x ( t ) = 2 L [ e ω t − e − ω t ] C.6 Momentum and Variable Mass ¶ Solutions to Practice Problems from Chapter 6.8
Problem 6-1: Question
v ⃗ c m = ı ^ + ( 2 3 ) ȷ ^ + ( 1 3 ) k ^ \vec{v}_{cm}= \hat{\imath} + \Big(\frac{2}{3} \Big)\hat{\jmath} + \Big(\frac{1}{3} \Big)\hat{k} v c m = ^ + ( 3 2 ) ^ + ( 3 1 ) k ^ Problem 6-2: Question
a ) M 1 M 2 v 1 ı ^ \mathrm{a}) \frac{M_{1}}{M_{2}} v_{1}\hat{\imath} a ) M 2 M 1 v 1 ^ b ) M 1 M 1 + M 2 v 1 ı ^ \mathrm{b}) \frac{M_{1}}{M_{1}+M_{2}} v_{1}\hat{\imath} b ) M 1 + M 2 M 1 v 1 ^ Problem 6-3: Question
a ) 8 ı ^ \mathrm{a}) 8\hat{\imath} a ) 8 ^ b ) 4 ı ^ + ȷ ^ \mathrm{b}) 4\hat{\imath} + \hat{\jmath} b ) 4 ^ + ^ Problem 6-4: Question
a ) I ⃗ = − 325 N s \mathrm{a}) \vec{I} = -325 \mathrm{N} \mathrm{s} a ) I = − 325 Ns b ) ∣ F ⃗ ∣ = 1625 N \mathrm{b}) |\vec{F}| = 1625 \mathrm{N} b ) ∣ F ∣ = 1625 N Problem 6-5: Question
u ⃗ = m M v 2 ı ^ \vec{u} = \frac{m}{M} \frac{v}{2} \hat{\imath} u = M m 2 v ^ Problem 6-6: Question
∣ v ⃗ ∣ = v 0 4 cos 2 θ + 1 4 |\vec{v}| = v_{0}\sqrt{4\cos ^{2}\theta + \frac{1}{4}} ∣ v ∣ = v 0 4 cos 2 θ + 4 1 Problem 6-7: Question
a ) v ⃗ f = m p ( m p + M ) v ⃗ 0 \mathrm{a}) \vec{v}_{f}= \frac{m_{p}}{(m_{p}+ M)} \vec{v}_{0} a ) v f = ( m p + M ) m p v 0 b ) A = m p ( m p + M ) v 0 L g \mathrm{b}) A = \frac{m_{p}}{(m_{p}+ M)} v_{0}\sqrt{\frac{L}{g}} b ) A = ( m p + M ) m p v 0 g L c ) θ max = m p ( m p + M ) v 0 1 L g \mathrm{c}) \theta _{\max}= \frac{m_{p}}{(m_{p}+ M)} v_{0}\sqrt{\frac{1}{Lg}} c ) θ m a x = ( m p + M ) m p v 0 Lg 1 d ) 31.91 m s − 1 \mathrm{d}) 31.91 \mathrm{m} \mathrm{s}^{-1} d ) 31.91 m s − 1 Problem 6-8: Question
0.625 m s − 1 0.625 \mathrm{m} \mathrm{s}^{-1} 0.625 m s − 1 Problem 6-9: Question
a ) v ⃗ c m = m 1 u ⃗ ( m 1 + m 2 ) \mathrm{a}) \vec{v}_{cm}= \frac{m_{1}\vec{u}}{(m_{1}+ m_{2})} a ) v c m = ( m 1 + m 2 ) m 1 u b ) d v d t = − k ( m 1 + m 2 ) x \mathrm{b}) \frac{\mathrm{d}v}{\mathrm{d}t} = - \frac{k}{(m_{1}+ m_{2})} x b ) d t d v = − ( m 1 + m 2 ) k x c ) A = m 1 2 u 2 k ( m 1 + m 2 ) \mathrm{c}) A = \sqrt{\frac{m_{1}^{2}u^{2}}{k(m_{1}+ m_{2})}} c ) A = k ( m 1 + m 2 ) m 1 2 u 2 Problem 6-10: Question
a ) M + λ x \mathrm{a}) M + \lambda x a ) M + λ x b ) v = M v 0 M + λ x \mathrm{b}) v = \frac{Mv_{0}}{M + \lambda x} b ) v = M + λ x M v 0 c ) T = λ v 2 \mathrm{c}) T = \lambda v^{2} c ) T = λ v 2 Problem 6-11: Question
b ) M = v 0 v M 0 \mathrm{b}) M = \frac{v_{0}}{v} M_{0} b ) M = v v 0 M 0 c ) v = v 0 2 M 0 2 ρ A v 0 t + M 0 \mathrm{c}) v = \sqrt{\frac{v_{0}^{2}M_{0}}{2\rho Av_{0}t + M_{0}}} c ) v = 2 ρ A v 0 t + M 0 v 0 2 M 0 C.7 Torques and Angular Momentum ¶ Solutions to Practice Problems from Chapter 7.9
Problem 7-1: Question
a ) 3 2 M R 2 \mathrm{a}) \frac{3}{2} MR^{2} a ) 2 3 M R 2 b ) 1 3 M R L 2 + 1 2 M S R 2 + M S ( L + R ) 2 \mathrm{b}) \frac{1}{3} M_{R}L^{2}+ \frac{1}{2} M_{S}R^{2}+ M_{S}(L + R)^{2} b ) 3 1 M R L 2 + 2 1 M S R 2 + M S ( L + R ) 2 c ) 1 3 M R L 2 + 1 2 M P ( ℓ 2 + w 2 ) + M P ( L + 1 2 ℓ ) 2 \mathrm{c}) \frac{1}{3} M_{R}L^{2}+ \frac{1}{2} M_{P}(\ell ^{2}+ w^{2}) + M_{P}(L + \frac{1}{2} \ell)^{2} c ) 3 1 M R L 2 + 2 1 M P ( ℓ 2 + w 2 ) + M P ( L + 2 1 ℓ ) 2 d ) 1 12 M R L 2 + 1 2 M C ( R 1 2 + R 2 2 ) + M C ( 1 2 L + R 2 ) 2 \mathrm{d}) \frac{1}{12} M_{R}L^{2}+ \frac{1}{2} M_{C}(R_{1}^{2}+ R_{2}^{2}) + M_{C}(\frac{1}{2} L + R_{2})^{2} d ) 12 1 M R L 2 + 2 1 M C ( R 1 2 + R 2 2 ) + M C ( 2 1 L + R 2 ) 2 Problem 7-2: Question
5.6 × 1 0 14 5.6 \times 10^{14} 5.6 × 1 0 14
Problem 7-3: Question
a) τ ⃗ = r F \vec{\tau} = rF τ = r F
b) τ ⃗ = r F sin θ \vec{\tau} = rF \sin \theta τ = r F sin θ
c) τ ⃗ = R m g \vec{\tau} = Rmg τ = R m g
d) τ ⃗ = R ( m 1 − m 2 ) g \vec{\tau} = R(m_{1}- m_{2})g τ = R ( m 1 − m 2 ) g
Problem 7-4: Question
a) L ⃗ i = ℓ m v 0 2 \vec{L}_{i}= \dfrac{\ell mv_{0}}{2} L i = 2 ℓ m v 0 L ⃗ f = [ 1 12 M ℓ 2 + 1 4 m ℓ 2 ] ω \vec{L}_{f}= \Big[\frac{1}{12} M\ell ^{2}+ \frac{1}{4} m\ell ^{2}\Big]\omega L f = [ 12 1 M ℓ 2 + 4 1 m ℓ 2 ] ω
b) ω = 6 m v 0 M ℓ + 3 m ℓ \omega = \frac{6mv_{0}}{M\ell + 3m\ell} ω = M ℓ + 3 m ℓ 6 m v 0
Problem 7-5: Question
h = L 2 2 h = \frac{L}{2 \sqrt{2}} h = 2 2 L Problem 7-6: Question
a ) I 0 = 1 2 M R 2 + m s 2 \mathrm{a}) I_{0}= \frac{1}{2} MR^{2}+ ms^{2} a ) I 0 = 2 1 M R 2 + m s 2 b ) T = 2 π R 2 + 2 s 2 2 g s \mathrm{b}) T = 2\pi \sqrt{\frac{R^{2}+ 2s^{2}}{2gs}} b ) T = 2 π 2 g s R 2 + 2 s 2 c ) s = R 2 \mathrm{c}) s = \frac{R}{\sqrt{2}} c ) s = 2 R Problem 7-7: Question
ω 0 = 4.1 s − 1 \omega _{0}= 4.1 \mathrm{s}^{-1} ω 0 = 4.1 s − 1 Problem 7-8: Question
a ) h = 3 L + R 4 \mathrm{a}) h = \frac{3L + R}{4} a ) h = 4 3 L + R b ) I = 1 3 M L 2 + 2 5 M R 2 + M ( L + 1 2 R ) 2 \mathrm{b}) I = \frac{1}{3} ML^{2}+ \frac{2}{5} MR^{2}+ M(L + \frac{1}{2} R)^{2} b ) I = 3 1 M L 2 + 5 2 M R 2 + M ( L + 2 1 R ) 2 c ) T = 2 π 2 3 M L 2 + 4 5 M R 2 + 2 M ( L + 1 2 R ) 2 M g ( 3 L + R ) \mathrm{c}) T = 2\pi \sqrt{\frac{\frac{2}{3} ML^{2}+ \frac{4}{5} MR^{2}+ 2M(L + \frac{1}{2} R)^{2}}{Mg(3L + R)}} c ) T = 2 π M g ( 3 L + R ) 3 2 M L 2 + 5 4 M R 2 + 2 M ( L + 2 1 R ) 2 Problem 7-9: Question
b ) τ = R F f = I α \mathrm{b}) \tau = RF_{f}= I\alpha b ) τ = R F f = I α c ) x ¨ = 2 3 g sin θ \mathrm{c}) \ddot{x} = \frac{2}{3} g\sin \theta c ) x ¨ = 3 2 g sin θ d ) θ = tan − 1 ( 3 μ ) \mathrm{d}) \theta = \tan ^{-1}(3\mu) d ) θ = tan − 1 ( 3 μ ) C.8 Work and Energy ¶ Solutions to Practice Problems from Chapter 8.10
Problem 8-5: Question
Only if c = − 1 c = -1 c = − 1
Problem 8-6: Question
a ) F ⃗ = − ( 2 ı ^ + 6 y ȷ ^ + 8 z k ^ ) \mathrm{a}) \vec{F} = -(2\hat{\imath} + 6y\hat{\jmath} + 8z\hat{k}) a ) F = − ( 2 ^ + 6 y ^ + 8 z k ^ ) b ) F ⃗ = − ( 2 x y 2 ı ^ + 2 x 2 y ȷ ^ + 3 z 2 k ^ ) \mathrm{b}) \vec{F} = -(2xy^{2}\hat{\imath} + 2x^{2}y\hat{\jmath} + 3z^{2}\hat{k}) b ) F = − ( 2 x y 2 ^ + 2 x 2 y ^ + 3 z 2 k ^ ) Problem 8-7: Question
x = ± A B x = \pm \sqrt{\frac{A}{B}} x = ± B A Problem 8-8: Question
v = 2 m ( a + b + c ) v = \sqrt{\frac{2}{m} (a + b + c)} v = m 2 ( a + b + c ) Problem 8-9: Question
a ) 0.85 J , b ) 11.54 J , c ) 12.39 J , d ) 31.20 J , e ) 32.89 m / s \mathrm{a}) 0.85 \mathrm{J}, \mathrm{b}) 11.54 \mathrm{J}, \mathrm{c}) 12.39 \mathrm{J}, \mathrm{d}) 31.20 \mathrm{J}, \mathrm{e}) 32.89 \mathrm{m/s} a ) 0.85 J , b ) 11.54 J , c ) 12.39 J , d ) 31.20 J , e ) 32.89 m/s Problem 8-10: Question
a) W 1 W_{1} W 1 W 2 = − 1 W_{2}= -1 W 2 = − 1
b) W 1 W_{1} W 1 W 2 W_{2} W 2
c) F ⃗ 1 \vec{F}_{1} F 1 F ⃗ 2 \vec{F}_{2} F 2 F ⃗ 1 \vec{F}_{1} F 1 F ⃗ 2 \vec{F}_{2} F 2
d) F 1 F_{1} F 1 F 2 F_{2} F 2
C.9 Applications of Energy Conservation ¶ Solutions to Practice Problems from Chapter 9.6
Problem 9-1: Question
Position C C C
Problem 9-2: Question
Problem 9-3: Question
U = 1 2 k A 2 cos 2 ( k m t ) U = \frac{1}{2} kA^{2}\cos ^{2}\Bigg(\sqrt{\frac{k}{m}} t \Bigg) U = 2 1 k A 2 cos 2 ( m k t ) C.10 Central Forces and Motion in Space ¶ Solutions to Practice Problems from Chapter 10.9
Problem 10-1: Question
f ( r ) = ( 3 k 1 r 4 + 2 k 2 r 3 ) r ^ f(r) = \Bigg(\frac{3k_{1}}{r^{4}} + \frac{2k_{2}}{r^{3}} \Bigg)\hat{r} f ( r ) = ( r 4 3 k 1 + r 3 2 k 2 ) r ^ Problem 10-2: Question
F e f f = L 2 + 2 m m r 3 r ^ F_{eff}= \frac{L^{2}+ 2m}{mr^{3}} \hat{r} F e ff = m r 3 L 2 + 2 m r ^ Problem 10-3: Question
U e f f , max = − m 2 L U_{eff,\max}= -\dfrac{m}{2L} U e ff , m a x = − 2 L m
Problem 10-5: Question
L = 2 m k R 3 , v = 2 k R 2 L = 2m\sqrt{k} R^{3}, v = 2\sqrt{k} R^{2} L = 2 m k R 3 , v = 2 k R 2 Problem 10-6: Question
F e f f = L 2 m r 3 − γ r 2 − 3 γ ε r 4 F_{eff}= \frac{L^{2}}{mr^{3}} - \frac{\gamma}{r^{2}} - \frac{3\gamma \varepsilon}{r^{4}} F e ff = m r 3 L 2 − r 2 γ − r 4 3 γ ε Problem 10-2: Question
U e f f = 1 2 L 2 m r 2 + − γ r − 1 2 ε r 2 U_{eff}= \frac{1}{2} \frac{L^{2}}{mr^{2}} + - \frac{\gamma}{r} - \frac{1}{2} \varepsilon r^{2} U e ff = 2 1 m r 2 L 2 + − r γ − 2 1 ε r 2 Problem 10-3: Question
L = 3 A β m R 5 e β R 3 L = \sqrt{3A\beta mR^{5}e^{\beta R^{3}}} L = 3 A β m R 5 e β R 3 Problem 10-4: Question
a) For small values of r , L 2 m r 2 > > A r 2 r, \frac{L^{2}}{mr^{2}} >> Ar^{2} r , m r 2 L 2 >> A r 2 U e f f U_{eff} U e ff r , L 2 m r 2 < < A r 2 r, \frac{L^{2}}{mr^{2}} << Ar^{2} r , m r 2 L 2 << A r 2 U e f f U_{eff} U e ff
b ) r = ( L 2 2 A m ) 1 / 4 \mathrm{b}) r = \Bigg(\frac{L^{2}}{2Am} \Bigg)^{1/4} b ) r = ( 2 A m L 2 ) 1/4 c ) E = 2 A L 2 m \mathrm{c}) E = \sqrt{\frac{2AL^{2}}{m}} c ) E = m 2 A L 2 Problem 10-5: Question
a ) r ˙ = 2 A L m r 3 / 2 \mathrm{a}) \dot{r} = 2\sqrt{A}\frac{L}{mr^{3/2}} a ) r ˙ = 2 A m r 3/2 L b ) E = L 2 2 m [ 4 A r 3 + 1 r 2 ] + U ( r ) \mathrm{b}) E = \frac{L^{2}}{2m} \bigg[\frac{4A}{r^{3}} + \frac{1}{r^{2}} \bigg] + U(r) b ) E = 2 m L 2 [ r 3 4 A + r 2 1 ] + U ( r ) c ) U ( r ) = E − L 2 2 m [ 4 A r 3 + 1 r 2 ] \mathrm{c}) U(r) = E - \frac{L^{2}}{2m} \bigg[\frac{4A}{r^{3}} + \frac{1}{r^{2}} \bigg] c ) U ( r ) = E − 2 m L 2 [ r 3 4 A + r 2 1 ] d ) F r = − L 2 m ( 6 A r 4 + 1 r 3 ) \mathrm{d}) F_{r}= - \frac{L^{2}}{m} \bigg(\frac{6A}{r^{4}} + \frac{1}{r^{3}} \bigg) d ) F r = − m L 2 ( r 4 6 A + r 3 1 ) Problem 10-6: Question
a) For small values of r , 1 r 3 ≫ 1 r 2 r, \frac{1}{r^{3}} \gg \frac{1}{r^{2}} r , r 3 1 ≫ r 2 1 r , 1 r 3 ≪ 1 r 2 r, \frac{1}{r^{3}} \ll \frac{1}{r^{2}} r , r 3 1 ≪ r 2 1
b ) r = 3 A m L 2 \mathrm{b}) r = \frac{3Am}{L^{2}} b ) r = L 2 3 A m c ) U e f f ( r = r 0 ) = 1 54 ( L 6 A 2 m 2 ) \mathrm{c}) U_{eff}(r = r_{0}) = \frac{1}{54} \Bigg(\frac{L^{6}}{A^{2}m^{2}} \Bigg) c ) U e ff ( r = r 0 ) = 54 1 ( A 2 m 2 L 6 ) d) Unstable
C.11 Orbits and Kepler’s Laws ¶ Solutions to Practice Problems from Chapter 11.7
Problem 11-1: Question
a ) 0.206 , b ) 0.078 , c ) 0.967 , d ) 0.041 \mathrm{a}) 0.206, \mathrm{b}) 0.078, \mathrm{c}) 0.967, \mathrm{d}) 0.041 a ) 0.206 , b ) 0.078 , c ) 0.967 , d ) 0.041 Problem 11-2: Question
a ) 0.24 y r , b ) 4.6 y r , c ) 75.5 y r , d ) 289 y r \mathrm{a}) 0.24 \mathrm{yr}, \mathrm{b}) 4.6 \mathrm{yr}, \mathrm{c}) 75.5 \mathrm{yr}, \mathrm{d}) 289 \mathrm{yr} a ) 0.24 yr , b ) 4.6 yr , c ) 75.5 yr , d ) 289 yr Problem 11-3: Question
v p = 1.52 v a v_{p}= 1.52v_{a} v p = 1.52 v a Problem 11-4: Question
a A = 16 a u a_{A}= 16 \mathrm{au} a A = 16 au Problem 11-5: Question
a f = 2.945 × 1 0 5 k m a_{f}= 2.945 \times 10^{5}\mathrm{km} a f = 2.945 × 1 0 5 km Problem 11-6: Question
v = G M R s v = \sqrt{\frac{GM}{R_{s}}} v = R s GM Problem 11-7: Question
M B H = 1.2 × 1 0 6 M s u n M_{BH}= 1.2 \times 10^{6}M_{sun} M B H = 1.2 × 1 0 6 M s u n Problem 11-8: Question
v p = 2.3 v 0 v_{p}= 2.3v_{0} v p = 2.3 v 0 Problem 11-9: Question
Height = 1140 km
Problem 11-10: Question
a ) r = p ( 1 + ε ) 1 + ε cos θ \mathrm{a}) r = \frac{p(1 + \varepsilon)}{1 + \varepsilon \cos \theta} a ) r = 1 + ε cos θ p ( 1 + ε ) b ) cos θ = p ( 1 + ε ) − R R ε \mathrm{b}) \cos \theta = \frac{p(1 + \varepsilon) - R}{R\varepsilon} b ) cos θ = Rε p ( 1 + ε ) − R c) If ε \varepsilon ε p = R p = R p = R
d) If ε \varepsilon ε cos θ = 2 p − R R \cos \theta = \dfrac{2p - R}{R} cos θ = R 2 p − R ( ± θ ) (\pm \theta) ( ± θ )
C.12 The Lagrange Method ¶ Solutions to Practice Problems from Chapter 12.8
Problem 12-7: Question
x ¨ = ( M sin θ − m ) g M + m \ddot{x} = \frac{(M \sin \theta - m)g}{M + m} x ¨ = M + m ( M sin θ − m ) g Problem 12-8: Question
a ) L = 3 8 M R θ ˙ 2 − 1 4 M g R θ 2 \mathrm{a}) L = \frac{3}{8} MR\dot{\theta}^{2}- \frac{1}{4} MgR\theta ^{2} a ) L = 8 3 MR θ ˙ 2 − 4 1 M g R θ 2 b ) 0 = θ ¨ + 2 g 3 R θ \mathrm{b}) 0 = \ddot{\theta} + \frac{2g}{3R} \theta b ) 0 = θ ¨ + 3 R 2 g θ Problem 12-9: Question
a ) U g r a v = 1 4 m g a ( 1 − cos 2 θ ) \mathrm{a}) U_{grav}= \frac{1}{4} mga(1 - \cos 2\theta) a ) U g r a v = 4 1 m g a ( 1 − cos 2 θ ) b ) K = 1 4 m a 2 θ ˙ 2 [ 1 + cos 2 θ ] \mathrm{b}) K = \frac{1}{4} ma^{2}\dot{\theta}^{2}[1 + \cos 2\theta] b ) K = 4 1 m a 2 θ ˙ 2 [ 1 + cos 2 θ ] c ) L = 1 4 m a 2 θ ˙ 2 [ 1 + cos 2 θ ] + 1 4 m g a cos 2 θ \mathrm{c}) L = \frac{1}{4} ma^{2}\dot{\theta}^{2}[1 + \cos 2\theta] + \frac{1}{4} mga\cos 2\theta c ) L = 4 1 m a 2 θ ˙ 2 [ 1 + cos 2 θ ] + 4 1 m g a cos 2 θ d ) 0 = a θ ¨ ( 1 + cos 2 θ ) + sin 2 θ ( g − a θ ˙ 2 ) \mathrm{d}) 0 = a\ddot{\theta}(1 + \cos 2\theta) + \sin 2\theta (g - a\dot{\theta}^{2}) d ) 0 = a θ ¨ ( 1 + cos 2 θ ) + sin 2 θ ( g − a θ ˙ 2 ) Problem 12-10: Question
a) The mass will swing from the motion of the pendulum as well as oscillate up and down along the axis of the spring due to the motion of the spring.
b ) U = 1 2 k x 2 − M g ( L + x ) cos θ \mathrm{b}) U = \frac{1}{2} kx^{2}- Mg(L + x)\cos \theta b ) U = 2 1 k x 2 − M g ( L + x ) cos θ c ) K = 1 2 M x ˙ 2 + 1 2 M ( L + x ) 2 θ ˙ 2 \mathrm{c}) K = \frac{1}{2} M\dot{x}^{2}+ \frac{1}{2} M(L + x)^{2}\dot{\theta}^{2} c ) K = 2 1 M x ˙ 2 + 2 1 M ( L + x ) 2 θ ˙ 2 d) M x ¨ = M ( L + x ) θ ˙ 2 − k x + M g cos θ M\ddot{x} = M(L+x)\dot{\theta}^{2}-kx+Mg\cos \theta M x ¨ = M ( L + x ) θ ˙ 2 − k x + M g cos θ M ( L + x ) 2 θ ¨ + 2 M ( L + x ) x ˙ θ ˙ = − M g ( L + x ) sin θ M(L+x)^{2}\ddot{\theta}+2M(L+x)\dot{x}\dot{\theta} = -Mg(L+x)\sin \theta M ( L + x ) 2 θ ¨ + 2 M ( L + x ) x ˙ θ ˙ = − M g ( L + x ) sin θ