A 1 kg mass is moving in a circle of radius 2 m on a flat frictionless table at
ID: 2195218 • Letter: A
Question
A 1 kg mass is moving in a circle of radius 2 m on a flat frictionless table at the end of a string. The speed of the mass is 3.8 m/s. The string routes through a hole in the center of the table and is held by you underneath the table A. What is the angular momentum of the mass? (In kgm^2/s) B. What is the tension in the string? C. If you pull on the string so that the radius of the circle decreases to one half its former value, what is the new tension in the string? D. How much work did you do to reduce the radius by a factor of one-half?Explanation / Answer
A) ang momentum = mvr = 1 * 3.8 * 2 = 7.60 kgm^2/s
B) tension = mv^2 / r = 1 * 3.8^2 / 2 = 7.22 Newtons
C) if r is half the former value, then the speed doubles so that ang mom is the same
new r is 1 new speed is 7.6
and tension = m v^2 / r = 1 * 7.6^2 / 1 = 57.76 Newtons
D) initial KE = (1/2) m v^2 = (1/2) * 1 * 3.8^2 = 7.22 Joules
final KE = (1/2) * 1 * 7.6^2 = 28.88 Joules
work = final KE - initial KE = 21.66 Joules
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