Please Help!!!!!!!!!!!! A dirigible is flying across the Atlantic Ocean at 48.00
ID: 1836280 • Letter: P
Question
Please Help!!!!!!!!!!!!
A dirigible is flying across the Atlantic Ocean at 48.000degree N latitude when the pilot falls asleep, and the Coriolis "force" causes the dirigible to deflect. Assume: The deflection is small enough so the latitude does not change appreciably. The speed of the dirigible remains 120.00 m/s at all times. The altitude of the dirigible remains constant. Show that the motion of the dirigible is a circle. Find R, the radius of the circle, in km. A horizontal wheel (radius 2.2200 m) sits on a horizontal surface, and initially is at rest. At t = 0, the wheel begins to accelerate at a constant a = 0.64900k rad/s^2. Assume: the axes are as shown: upward is the z-direction; the x-axis is toward you. The center of mass of the wheel remains stationary. At t = 0, a bug of mass 6.3000tiems10^3 kg starts at the center of the wheel and walks along a radius at constant speed v = 0.24200 m/s^2 relative to the wheel. Assume Initially, the bug is moving along the x-axis. (Of course, as the wheel turns the bug's direction changes, but not his speed relative to the wheel.) g = 9.8030 m/s^2. When the wheel has turned one-fourth of a revolution, the bug is moving in the y-direction. At this moment, find f, the net friction felt by the bug. Assume the bug has enough friction to hold him on the wheel; express your answer as numbers in i-j-k notation. Suppose, when the wheel has turned one-fourth of a revolution, static friction is just able to keep the bug from slipping. Find ps, the coefficient of static friction.Explanation / Answer
i.
angular speed of wheel when it has turned one-forth revolutions.
using wf^2 - wi^2 = 2(alpha)(theta)
wf^2 - 0 = 2 ( 0.649) ( 2 pi / 4 rad)
wf = 1.43 rad/s
time taken to reach this speed,
wf = wi + alpha*t
1.43 = 0 + 0.649t
t = 2.2 sec
distance of bug from centre,r = 0.242t = 0.532 m
centripetal force = m w^2 r
= (6.3 x 10^-3) ( 1.43^2) ( 0.532) = 6.86 x 10^-3 N
tangential force = m a_t = m * alpha * r
= (6.3 x 10^-3 ) ( 0.649 x 0.532 ) = 2.175 x 10^-3 N
friction force = - 2.175e-3 i - 6.86e-3 j + 0k N
ii.
magnitude of friction force = sqrt(2.175^2 + 6.86^2) = 7.20 N
and N = mg = 6.3 x 9.803 x 10^-3 = 61.76 N
and f = us N
7.20 = us 61.76
us = 0.117