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Unformatted text preview: attached to a hanging cylinder of mass M = 4.00 kg by a cord through a hole in the table. The hanging cylinder is at rest. a) What is the tension in the cord? Taking the positive direction downward, we apply Newton’s 2 nd law to the cylinder 0 = Mg-T ⇒ T = Mg = 39.2 N b) What is the speed (in m/s) of the puck? The puck is moving in a circle. Hence it accelerates with centripetal acceleration v 2 /R toward the center. Taking our positive direction toward the center, Newton’s 2 nd law yields m(v 2 /R) = T v = √ (TR/m) = 2.80 m/s...
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This note was uploaded on 12/10/2011 for the course PHY 2048 taught by Professor Field during the Spring '08 term at University of Florida.
- Spring '08