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CH10.Problems

CH10.Problems. JH. 132. P10.4: The angular position of a point on a rotating wheel is given by: θ (t)= 2.0 + 4.0t 2 + 2.0t 3 , where θ is in radians and t is in seconds. At t 0, what are ( a) the point’s angular position and (b) its angular velocity ?

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CH10.Problems

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  1. CH10.Problems JH.132

  2. P10.4: The angular position of a point on a rotating wheel is given by: θ(t)= 2.0 + 4.0t2 + 2.0t3, where θis in radians and t is in seconds. At t 0, what are (a) the point’s angular position and (b) its angular velocity? (c) What is its angular velocity at t 4.0 s? (d) Calculate its angular acceleration at t 2.0 s. (e) Is its angular acceleration constant? f) Average angular velocity between t=0 and t=3 θ=2.0 rad, ω = 0; Ω=8t+6t^2=128 rad/sec α= 8+12t= 32 rad/sec^2 No, Avgω = Δθ/Δt= ((2+4*3^2+2*3^3)-(2))/(3-0) = (92-2)/3 = 30rad/sec

  3. a) 10-15, answer t = 335s b) By selecting 10-16 40 = 0- 0.5* alpha * 335^2 Answer = -4.5*10^-3 rad/sec c) Use 10-13 and calculator 20 = 1.5 t+0.5*(-4.5*10^-3)t^2 0= 0.5*(-4.5*10^-3)t^2 +1.5t -20*2Pi T=568 or 98s Since answer in part a is only 335 we cannot take 568 so t=98s

  4. a)if ws = 6 then length of y axis=9 Then alpha = 9/6=1.5rad/s^2 b) At t = 4, w = 4, K =1/2 I 4^2 At t= 0 , w = -2, K = ½ I 2^2 = 1.6/4=0.4

  5. Mg h = ½ I w^2 M g (L/2* Sin(45) ) = ½ (1/3 M L^2) w^2 W^2 = 3 g Sin(45) / L W = 3.2 rad

  6. K = KL+ KP + KR KL = ½ (2kg) * v^2 KR = ½ (4kg) * v^2 KP = ½ (I) * w^2 V=R W, W= 2.0m/s / 0.030 m= 66.7rad/s K = ½ (2+4)* 2^2 + ½ (0.0045)* 66.7^2 = 22 J

  7. K = ½ I w^2 I= sum (m r^2) For all the point masses r is the same: r^2 = 2^2+3^2= 4+9= 13 I = (3+2+2+4)*13= 11*13 K = ½ (11*13) 6.0^2 = (2574) = 2.6 kJ

  8. Use parallel axis theorem Rod Icom+ L/2 shift Sphere Icom + R shift Rod: I = 1/12 M L^2 + M (L/2)^2 Sphere I= 2/5 M R^2 + M (R)^2

  9. TL, TR, ml g=0.5g, mrg = 0.4 g, Using a as +ve direction and CCW as +ve Left: ml g – Tl = ml a Right: Tr - mr g = mr a Pulley (Tl – Tr )R = (1/2 M R^2) alpha alpha= + a/R Add eqns: (ml- mr)g = (ml+mr)a + M/2 a a= g (ml-mr)/(ml+mr+M/2) = 0.59 m/s^2

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