Physics
posted by Sarah on .
A bowling ball weighing 71.2 N is attached to the ceiling by a 3.80 m rope. The ball is pulled to one side and released; it then swings back and forth like a pedulum. as the rope swings through its lowest point, the speed of the bowling ball is measured at 4.20 m/s. At that instant, find a) the magnitude and direction of the acceleration of the bowling ball and b) the tension in the rope.

a) The magnitude of the acceleration is V^2/R and it is directed along the rope (i.e. vertical), perpendicular to the direction of motion.
b) If the rope tension is T, Newton's Second Law tells you that
F = T  Mg = Ma = M V^2/R
Solve for T
T = M*(g + V*2/R)
R is the rope length. 
So for a I got 17.64 / 3.80 = 4.64 sec is this correct
and then for b i got 71.2( 9.8+ 4.20)*2 / 3.80 and got 524.63 
hell naw

LMBO!!!