qz4sol_3885f10

# qz4sol_3885f10 - We take our x direction toward and...

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TA: Tomoyuki Nakayama Monday, September 27, 2010 PHY 2048: Physic 1, Discussion Section 3885 Quiz 4 (Homework Set #5) Name: UFID: Formula sheets are not allowed. Calculators are allowed. Do not store equations in your calculator. You need to show all of your work for full credit. ________________________________________________________________________________ In the figure below right, a 1.50 kg ball is connected by means of two massless strings, each of length L = 1.00 m, to a vertical rotating rod. The strings are tied to the rod with separation d = 1.50 m and are taut. The tension in the lower string is 20.0 N. a) What are the magnitude and direction of the net string force? Three forces are exerted on the ball: tension in the upper string T 1 , tension in the lower string T 2 and the gravitational force. The angle between the upper string and the horiz4ontal is θ = sin -1 ((d/2)/L) = 48.6º
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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 unit-vector 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 = tan-1 (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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