Answer to Question #200644 in Mechanics | Relativity for Master j

Question

A crate is lowered at a constant velocity of 2.5 ms^-1 by means of a cable wrapped round a pulley. The effective diameter of the pulley 0.20m. Calculate the:

i. angular velocity of the pulley

ii. number of revolutions it makes when the crate descends 15 m.

Answer 1

Solution.

$v=2.5m/s;$

$d=0.20m;$

$i. v=\omega r\implies \omega=\dfrac{v}{r}; \omega=\dfrac{2.5m/s}{0.10m}=25s^{-1};$

$ii. l=15m;$

$C=2\pi r;$

$C=2\sdot3.14\sdot0.1m=0.628m;$

$N=\dfrac{l}{C};$

$N=\dfrac{15m}{0.628m}=23;$

Answer: $i. 25 s^{-1};$

$ii. 23.$