I. Filling the blanks a gypsy moth caterpillar(Porthetria dispar) inches along a crooked branch to a tasty oak leaf, wriggling 15 Tree trunk ty cm horizontally and then 30 cm along a section of the branch inclined at 30 to the horizontal as shown in Figure 1.(a) The initial position vectors -0 15i and the final Branch position vectors 0.153i +0.15 of the caterpillar in Fig 1 SI units using the crook in the branch as the origin for a set of horizontal and vertical coordinate axes (b)If the caterpillar traverses the distance during 1.0 min, its average speed is-0.0075m/s,its average velocity is 0.0068i+0.0025 j(m/s), and the magnitude of its average velocity is 724×10-3m/s Solution (a)The coordinate system are shown in figure. The initial position vectors is P=-0 15i, and the final position vectors is F=0.3cos 30 i+0.3sin 30 j=0. 26i+0.15j s0.15+0.3 (b) Its average speed is v 60=0.0075m/s Its average velocity isvave-A A 10.261+015/+0.15i=000 0.0025jm/s The magnitude of its average velocity is m11008+0010090572103(ms) 2. Use Equation 4.28 for r(o: r(o=[rose(Oli+[rsin e(ol and Equation 4.26 for (t) o(0=@(k to evaluate the expression r(o)[o(txr(o=0 Explain why this result is another way to see that v() is always perpendicular to r(t for circular motion: solution(2). Solution: (O(0xF(=0, (kxfrcos e(o)+(rsin(li=rcos 0(), (0j-rsin 0(), (02 →F()-[o(x-cosi+rsno小 rose(o2(0)- rsin g(n)a2( r A(cos0(o(0) sin e(ocos 6(Do(0=0II. Filling the Blanks 1. A gypsy moth caterpillar (Porthetria dispar) inches along a crooked branch to a tasty oak leaf, wriggling 15 cm horizontally and then 30 cm along a section of the branch inclined at 30o to the horizontal as shown in Figure 1. (a) The initial position vectors i 15ˆ − 0. and the final position vectors i j 15 ˆ 0. 15 3ˆ 0. + of the caterpillar in SI units using the crook in the branch as the origin for a set of horizontal and vertical coordinate axes. (b) If the caterpillar traverses the distance during 1.0 min, its average speed is 0.0075m/s , its average velocity is (m/s) 0025 ˆ 0. 0068ˆ 0. i + j , and the magnitude of its average velocity is 7.24 10 m/s −3 × . Solution: (a) The coordinate system are shown in figure. The initial position vectors is r i 15ˆ = −0. v , and the final position vectors is r i j i j f 15 ˆ 0. 26ˆ 0. 30 ˆ = 0.3cos30 ˆ + 0.3sin = + v o o . (b) Its average speed is 0.0075 m/s 60 0.15 0.3 = + = = t s vave . Its average velocity is i j i j i t r r t r v f i ave 0025 ˆ 0. 0068ˆ 0. 60 15ˆ 0. 15 ˆ 0. 26ˆ 0. = + + + = ∆ − = ∆ ∆ = v v v v m/s. The magnitude of its average velocity is 0.0068 0.0025 7.24 10 (m/s) 0025 ˆ 0. 0068ˆ 0. 2 2 −3 vave = i + j = + = × v 2. Use Equation 4.28 for r(t) r : r t r t i r t j ˆ [ sin ( )] ˆ ( ) = [ cosθ ( )] + θ r and Equation 4.26 for ω(t) r : t t k z ˆ ω( ) = ω ( ) r to evaluate the expression r(t)⋅[ (t)× r(t)] = r r r ω 0 . Explain why this result is another way to see that v(t) r is always perpendicular to r(t) r for circular motion: solution(2) . Solution: (1) t r t t k { } r t i r t j r t t j r t t i z z z ˆ sin ( ) ( ) ˆ cos ( ) ( ) ˆ [ sin ( )] ˆ [ cos ( )] ˆ ω( ) × ( ) = ω ( ) × θ + θ = θ ω − θ ω v r [ ] [ ] [ ] sin ( ) cos ( ) ( ) sin ( ) cos ( ) ( ) 0 ˆ sin ( ) ( ) ˆ cos ( ) ( ) ˆ sin ( ) ˆ ( ) ( ) ( ) cos ( ) 2 2 = − = ⇒ ⋅ × = + ⋅ − r t t t r t t t r t t r t r t i r t j r t t j r t t i z z z z θ θ ω θ θ ω ω θ θ θ ω θ ω v v r 15cm 30cm Branch Tree trunk Fig.1 o 30 x y 0