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A spool of thin wire (with inner radius r = 0.45 m, outer radius R = 0.60 m, and moment of inertia Icm = 0.8208 kg·m2) pivots on a shaft. The wire is pulled down by a mass M = 1.300 kg. After falling a distance D = 0.500 m, starting from rest, the mass has a speed of v = 63.1 cm/s. Calculate the energy lost to friction during that time.

Respuesta :

davidlcastiblanco

Answer:

[tex]E_L=4.667J[/tex]

Explanation:

The energy lost can be model by the energy equation in potential and kinetic energy in each step of the motion

Given: [tex]r=0.45m[/tex],[tex]R=0.60m[/tex],[tex]I_m=0.8208 kg*m^2[/tex],[tex]v=63.1cm/s[/tex], [tex]m=1.3kg[/tex]

[tex]E_p=m*g*r[/tex]

[tex]E_p=1.3kg*9.8m/s^2*0.45m[/tex]

[tex]E_p=5.733 J[/tex]

Kinetic energy:

[tex]K_{Es}=\frac{1}{2}*I*w^2[/tex], [tex]K_{Ep}=\frac{1}{2}*m*v^2[/tex]

[tex]w=v^2/r[/tex]

Energy lost :

[tex]E_p-K_{Es}-K_{Ep}=0[/tex]

[tex]5.733J-\frac{1}{2}*0.8208kg*(\frac{(0.631m/s)^2}{0.45m})^2-\frac{1}{2}*1.3kg*(0.631m/s)^2=0[/tex]

[tex]E_L=4.667J[/tex]

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