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. Newton’s 2 nd law yields 0 = T 1 sin θ – T 2 sin θ – mg ⇒ T 1 = (T 2 sin θ + mg)/sin θ = 39.6 N In unitvector notation, the net string force is T 1 + T 2 = (T 1 + T 2 )cos θ i + (T 1 – T 2 )sin θ 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, φ = tan1 (14.7/39.4) = 20.5º above the horizontal in xy plane b) What is the speed of the ball? Newton’s 2 nd law in the x direction yields ma = (T 1 + T 2 )cos θ a = (T 1 + T 2 )cos θ /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 θ ) = 4.17 m/s...
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 Summer '08
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 Force, Work, General Relativity, net string force

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