Question

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.

Answer 1

Answer:

$E_L=4.667J$

Explanation:

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

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

$E_p=m*g*r$

$E_p=1.3kg*9.8m/s^2*0.45m$

$E_p=5.733 J$

Kinetic energy:

$K_{Es}=\frac{1}{2}*I*w^2$, $K_{Ep}=\frac{1}{2}*m*v^2$

$w=v^2/r$

Energy lost :

$E_p-K_{Es}-K_{Ep}=0$

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

$E_L=4.667J$