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Need help calculating vector angular momentum and torque due to an external force?
I'm not exactly sure what formulas to use here. Any help would be appreciated.
1 Answer
- EMLv 77 years ago
(a)
L = r x p
L = m(r x v)
v = r'(t) = <-40sin4t, 40cos4t>
L(t) = (0.5)(<10cos4t + 6, 10sin4t + 6> x <-40sin4t, 40cos4t>)
L(t) = (0.5)[(10cos4t + 6)(40cos4t) - (10sin4t + 6)(-40sin4t)]
L(t) = (0.5)(400cos²4t + 240cos4t + 400sin²4t + 240sin4t)
L(t) = 200 + 120(cos4t + sin4t)
L(π/8) = 200 + 120(cos(π/2) + sin(π/2))
L(π/8) = 200 + 120
L(π/8) = 320 kg-m²/s (+z direction)
(b)
F(t) = dp/dt
F(t) = m(dv/dt)
F(t) = (0.5)(d/dt[<-40sin4t, 40cos4t>])
F(t) = (0.5)<-160cos4t, -160sin4t>
F(π/4) = (0.5)<-160cosπ, -160sinπ>
F(π/4) = <80, 0>
|| F(π/4) || = 80 N
(c)
τ = r x F
τ(t) = <10cos4t + 6, 10sin4t + 6, 0> x <2, 4, -1>
τ(t) = <-10sin4t - 6, 10cos4t + 6, 4(10cos4t + 6) - 2(10sin4t + 6)>
τ(t) = <-10sin4t - 6, 10cos4t + 6, 40cos4t - 20sin4t + 12>
τ(π/24) = <-10sin(π/6) - 6, 10cos(π/6) + 6, 40cos(π/6) - 20sin(π/6) + 12>
τ(π/24) = <-5 - 6, 5sqrt(3) + 6, 20sqrt(3) - 10 + 12>
τ(π/24) = <-11, 5sqrt(3) + 6, 20sqrt(3) + 2> N-m
