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# capa11 - 1 The corresponding expression to Newton II(Fnet =...

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1) The corresponding expression to Newton II (F net = ma) in ‘Rotation Land’ is: τ = I· α b α = τ / I For a rod pivoting around its end: I = 1/3 m L 2 (see last CAPA set) Torque: τ = mg·L/2·cos θ b α = 3·g·cos θ / (2·L) Therefore, the angular acceleration α is smaller for the long rod compared to the short one. |a y | = α ·L·cos θ =3·mg·L/2·cos θ /(mL 2 ) ·L·cos θ = 3/2·g·cos 2 θ b a y does not depend on L. Tangential acceleration at C.M. is α ·L/2 = 3/4·g·cos θ 2) Conservation of energy: mgh = ½ I ω 2 b ω = sqrt(2mgh/I) , sqrt: ‘square root’ Now be careful when you evaluate this expression! Use the ‘parallel axes theorem’ to calculate the moment of inertia ‘I’ for the axes shown in the figure! (don’t just use I C.M. ): Parallel axes theorem: I = I C.M. + mR 2 b Speed at C.M.: r C.M. · ω = r C.M. ·sqrt(2mgh/I) 3) When calculating the moment of inertia it might help to consider the two rods as one system with a mass 2M and a length 2L pivoting around one end (see problem 1 above). b

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capa11 - 1 The corresponding expression to Newton II(Fnet =...

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