The pulley has radius R and moment of inertia I. The rope does not slip over the pulley. The coefficient of friction between block A and the tabletop is 4k. the system is released from rest, and block B descends. a) Use energy methods to calculate the speed of block B as a function of the distance d that it has descended. b) Use forces and torques to calculate the speed of block B as a function of the distance d that it has descended.

Respuesta :

Answer:

[tex]v = \sqrt{\frac{2(m_{b} gd - \mu_{k} m_{a}gd)}{m_{b} +m_{a} + \frac{I}{R^{2} } } }[/tex]

Explanation:

Energy = force * distance

The heat energy lost by the block B equals the sum of the energy gained.

Energy lost = sum of energy gained

Energy lost by [tex]M_{b}[/tex] due to the height covered= KE gained by [tex]M_{b}[/tex] + KE gained by pulley + KE gained by [tex]M_{a}[/tex] + frictional heat gained by [tex]M_{a}[/tex].......(1)

frictional force by [tex]M_{a}[/tex] = [tex]\mu_{k} m_{a} g[/tex]

distance covered = d

KE gained by [tex]M_{b}[/tex] = [tex]o.5m_{b} v^{2}[/tex]

KE gained by [tex]M_{a}[/tex] = [tex]o.5m_{a} v^{2}[/tex]

Frictional heat gained by [tex]M_{a}[/tex] = [tex]\mu_{k} m_{a} gd[/tex]

KE gained by the pulley = [tex]0.5Iw^{2}[/tex] = [tex]0.5 I\frac{v^{2} }{R^{2} }[/tex]

Energy lost by [tex]M_{b}[/tex] due to height = [tex]M_{b}gd[/tex]

Substituting these into equation (1)

[tex]m_{b} gd = 0.5m_{b} v^{2} + 0.5m_{a} v^{2} + \mu_{k} m_{a}gd + 0.6\frac{I}{R^{2} } v^{2}[/tex]

[tex]m_{b} gd - \mu_{k} m_{a}gd = 0.5m_{b} v^{2} + 0.5m_{a} v^{2} + 0.5\frac{I}{R^{2} } v^{2}[/tex]

[tex]m_{b} gd - \mu_{k} m_{a}gd = 0.5m_{b} v^{2} + 0.5m_{a} v^{2} + 0.5\frac{I}{R^{2} } v^{2}\\v^{2} = \frac{m_{b} gd - \mu_{k} m_{a}gd}{0.5m_{b} +0.5m_{a} + \frac{0.5I}{R^{2} } } \\v = \sqrt{\frac{m_{b} gd - \mu_{k} m_{a}gd}{0.5m_{b} +0.5m_{a} + \frac{0.5I}{R^{2} } } }[/tex]

[tex]v = \sqrt{\frac{2(m_{b} gd - \mu_{k} m_{a}gd)}{m_{b} +m_{a} + \frac{I}{R^{2} } } }[/tex]

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