This preview shows page 1. Sign up to view the full content.
Unformatted text preview: We take our +x direction toward and perpendicular to the rod and +y direction vertically upward. The ball does not move in the y direction. Newtons 2 nd law yields 0 = T 1 sin T 2 sin mg T 1 = (T 2 sin + mg)/sin = 39.6 N In unit-vector notation, the net string force is T 1 + T 2 = (T 1 + T 2 )cos i + (T 1 T 2 )sin j = 39.4 i + 14.7 j The magnitude and direction of the net string force are | T 1 + T 2 | = (39.4 2 + 14.7 2 ) = 42.1 N, = tan-1 (14.7/39.4) = 20.5 above the horizontal in xy plane b) What is the speed of the ball? Newtons 2 nd law in the x direction yields ma = (T 1 + T 2 )cos a = (T 1 + T 2 )cos /m = 26.3 m/s 2 The ball is in a circular motion. Therefore, this acceleration is centripetal and expressed as a = v2/R. We have a = v 2 /R v = (aR) = (aLcos ) = 4.17 m/s...
View Full Document
- Summer '08