TA: Tomoyuki Nakayama
Monday, January 24, 2011
PHY 2048: Physic 1, Discussion Section 3705
Quiz 1 (Homework Set #2)
Name:
UFID:
Formula sheets are not allowed. Do not store equations in your calculator. You have to solve
problems on your own; memorizing final algebraic expressions from homework assignments and
just plugging numbers into them will not give you full credit. Leave all your work.
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The figure below right shows a pendulum of length L = 1.20 m. Its bob (which effectively has all the
mass) has speed v0 when the cord makes an angle of θ0 = 45.0 with the vertical.
a) What is the least value that v0 can have if the
pendulum is to swing down and then up to a horizontal
position?
Two forces are exerted on the pendulum bob: the
tension force by the string and the gravitational force.
Only the gravitational force, which is conservative,
does work. Therefore, the mechanical energy of the
pendulum is conserved. The energy conservation
equation leads to
Ei = Ef ⇒ (1/2)mv02 + mgL(1 – cosθ0) = 0 + mgL
v0 = √(2gL cosθ0) = 4.08 m/s
b) What is the least value that v0 can have if the pendulum is to swing down and then up to a vertical
position with the cord remaining straight?
Since the cord is remaining straight, the bob is in a circular motion and the tension in the string is
always equal to or larger than 0. If v0 has the minimum value, then the tension becomes at the top.
Therefore, the velocity of the bob at the top is
m(v2/L) = mg ⇒ v = √(gL)
The work-energy theorem yields
Ei = Ef ⇒ (1/2)mv02 + mgL(1 – cosθ0) = (1/2)mv2 + mg(2L) ⇒ (1/2)mv02 + mgL(1 – cosθ0) =
(1/2)mgL + mg(2L) ⇒ v0 = √(3gL + 2gL cosθ0) = 7.20 m/s