Physics

Points Earned: | 20.0/30.0 | |

Correct Answer(s): | a) vp^2=vo^2+2*a*y y=(vp^2-vo^2)/2*a=(0-20^2)/2*(-9.8)=-400/-19.6=20.41=20.4m b) vp=vo-9.8*t 0=20-9.8*tp tp=20/9.8=2.041=2.04s c) y=vo*t+0.5*a*t^2 0=20*t-4.9*t^2 t=20/4.9=4.0816=4.08s d) v=vo+a*t=20-9.8*4.0816= -20.0m/s f) vf^2=vo^2+2*a*d=20^2+2*(-9.8)*(-98)=2321 vf= -48.174= -48.2m/s e) vf=vo+a*t t=(vf-vo)/a=(-48.17-20)/-9.8=6.96s |

Rotational Kinematics

To understand circular motion, it is first necessary to understand how it differs from motion in a straight line. When an object goes around in a circle, it subtends an arc of 360 degrees for every full revolution. This angular distance can also be measured in radians and there are 2*pi radians in one revolution. In calculations it is normally necessary to express all angular quantities, such as angles, velocities, and accelerations, in radians. Because these may be given or requested in degrees or revolutions, it is necessary to be able to convert from and to radians (rad). The relationship is

1 revolution = 360 degrees = 2*pi (6.2832) rad

It is also necessary to understand the relationship between the angular quantities and their linear equivalents. Because the perimeter of a circle in terms of its radius is 2*pi*r, it is apparent that that is the linear distance for one revolution. In general, the relationship between the angle theta (th) and the linear distance s along an arc is

s = r*th

The radius r turns out to be the translating factor for velocity and acceleration as well. Angular velocity is called omega (om) and angular acceleration is called alpha (al). Thus

v = r*om and a = r*al

In all these relationships the linear quantities should be in meters (or meters per) and the angular quantities in radians (or radians per).

Example. A car on a circular track (r=500m) accelerates uniformly from 36.0km/h to 90.0km/h while making two (2) revolutions around the track. How long does this maneuver take,...