Question

A light 300mm diamter pulley is held stationairy and carries a 10kg mass haning on a light cord passing around the pulley. Find the vertical acceleration of the mass when the pulley is released. Frictional torque is the pulley bearings is 3N.m.

Clearly show the following value

Ek=?

Ep=?

E=?

Energy equation=??

v=??

Answer 1

$g = 9.8\frac{m}{s^2}$

$d = 300mm$

$r = \frac{d}{2}= 150mm= 0.15m$

$I_{fr}= 3N*m$

$I_{fr}= F_{fr}*r$

$F_{fr}= \frac{I_{fr}}{r}=\frac{3}{0.15}=20N$

$\text{Let the load descend a distance }h$

$E_k =\frac{mv^2}{2}$

$E_p= -mgh$

$E_{fr}=F_{fr}h$

$E_k+E_p+E_{fr}=0$

$\frac{mv^2}{2} - mgh +F_{fr}h=0$

$v^2= \frac{2h*(mg-F_{fr})}{m}(1)$

$v=\sqrt{ \frac{2h*(mg-F_{fr})}{m}}$

$v^2 = 2ah$

$\text{From (1)}$

$2ah = \frac{2h*(mg-F_{fr})}{m}$

$a = \frac{(mg-F_{fr})}{m}=\frac{10*9.8 -20}{10}=7.8\frac{m}{s^2}$

$\text{Answer: }a =7.8\frac{m}{s^2}$