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A yo-yo has mass M = 0.550 kg and rotational inertia I = 3.40 xx 10^(-4) kg m^(2)?. It rolls from rest down a string of length L = 17.0 cm. What is its angular speed at the bottom? |
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Answer» Solution :The mechanical energy of the yo-yo is conserved during the descent. Calculations: Let the bottom position be the reference LEVEL for zero gravitational potential energy. Then at the initial POINT (at height y = L) the mechanical energy is `E_(i) = K_(i) + U_(i) = 0+ MgL`. The symbol `K_(i)` is the sum of the translational kinetic energy and ROTATIONAL kinetic energy, both of which are zero. At the bottom (at height y= 0), the translational kinetic energy and gravitational potential energy are zero. So, the mechanical energy is just the rotational kinetic energy `1//2Iomega^(2)` and we have `E_(f) = K_(f) +U_(f) =(1)/(2)Iomega^(2)` Because the mechanical energy does not change, `E_(f)=E_(i)` `(1)/(2)Iomega^(2) = MgL` Solving for omega and SUBSTITUTING the given values LEAD to `omega= 73.4 "rad"//s` |
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