Saved Bookmarks
| 1. |
A pilot of mass 70 kg in a jet aircraft moves in a vertical circle while executing the loop-the-loop at a constant speed of 200 m/s. If the radius of the circle is 2.5 km what is the force exerted by the seat on the pilot at (a) the top of the loop and (b) the bottom of the loop. |
Answer» Solution : Speed = v=200 m/s Radius = `r= 2.5 KM = 2.5 xx 10^(3)` m At the top [The pilot will be UPSIDE down in his seat] Centripetal FORCE = `(mv^(2))/r = N_(1) + mg` `N_(1) = (mv^(2))/r - mg` `N_(1) = mg(v^(2)/(rg)-1) = 70 xx 9.8 ((200 xx 200)/(2.5 xx 10^(3) xx 9.8)-1)` `=70 xx 9.8 xx 0.63` =432.18 N At the bottom Centripetal force = `(mv^(2))/r = N_(2)-mg` `N_(2) = (mv^(2))/r + mg = mg (v^(2)/(rg) +1)` `=70 xx 9.8(1.632 +1)` `=1805.6` N |
|