Question

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.

Answer 1

Answer:

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

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 $M_{b}$ due to the height covered= KE gained by $M_{b}$ + KE gained by pulley + KE gained by $M_{a}$ + frictional heat gained by $M_{a}$.......(1)

frictional force by $M_{a}$ = $\mu_{k} m_{a} g$

distance covered = d

KE gained by $M_{b}$ = $o.5m_{b} v^{2}$

KE gained by $M_{a}$ = $o.5m_{a} v^{2}$

Frictional heat gained by $M_{a}$ = $\mu_{k} m_{a} gd$

KE gained by the pulley = $0.5Iw^{2}$ = $0.5 I\frac{v^{2} }{R^{2} }$

Energy lost by $M_{b}$ due to height = $M_{b}gd$

Substituting these into equation (1)

$m_{b} gd = 0.5m_{b} v^{2} + 0.5m_{a} v^{2} + \mu_{k} m_{a}gd + 0.6\frac{I}{R^{2} } v^{2}$

$m_{b} gd - \mu_{k} m_{a}gd = 0.5m_{b} v^{2} + 0.5m_{a} v^{2} + 0.5\frac{I}{R^{2} } v^{2}$

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

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