A pendulum consists of a bob of mass of 0.34 kg, attached to a string of length
ID: 1630149 • Letter: A
Question
A pendulum consists of a bob of mass of 0.34 kg, attached to a string of length L = 1 meters. It is moving to the right with a speed of 4.49 m/sec at point A. What we'd like to know is what the the velocity of the bob when it reaches point B which is 37 degrees from vertical. There are two different ways to look at this problem, each of which will give us the same answer. Let's look at each separately. (For all of the questions below, let's start by defining the potential energy of the bob to be zero at point A.) First, lets look at the question in terms of the work energy theorem. (WTOT=KE)
a: Now, lets look at the problem in terms of mechanical energy. That is,ME=U+KE and ME=Wnon-con. (Recall we are defining the potential energy to be zero at point A.) What is the kinetic energy of the bob at point A?
b: What is the potential energy of the bob at point A?
c: What is the mechanical energy of the bob at point A?
d: What is work done by non-conservative forces as the bob moves from point A to point B?
e: What is the mechanical energy of the bob at point B?
f: What is the potential energy of the bob when it reaches point B?
g: What is the kinetic energy of the bob when it reaches point B?
h: What is the speed of the bob at point B?
Explanation / Answer
a: What is the kinetic energy of the bob at point A?
Kinetic Energy at A = ½ * 0.34 * 4.492 = 3.4272 J
b: What is the potential energy of the bob at point A?
PE = mgh = 0 J
c: What is the mechanical energy of the bob at point A?
mechanical energy of the bob at point A = kinetic energy of the bob at point A = 3.4272 J
d: What is work done by non-conservative forces as the bob moves from point A to point B?
work done by non-conservative force = Increase in PE = m*g*(L – Lcos )
Increase of PE = 0.34 * 9.81 * (1 – 1*cos 37) = 0.6716 J
e: What is the mechanical energy of the bob at point B?
the mechanical energy of the bob at point B = PE + KE = 0.6716 + 3.4272
= 4.0988 J
f: What is the potential energy of the bob when it reaches point B?
potential energy of the bob when it reaches point B = same as PE = 0.34 * 9.81 * (1*cos 37) = 0.6716 J
g: What is the kinetic energy of the bob when it reaches point B?
kinetic energy of the bob when it reaches point B = (KEA – Increase of PE) = 3.4272 J – 0.6716 J = 2.7556 J
h: What is the speed of the bob at point B?
½ * 0.34 * v2 = 2.7556
v = (16.20941176)
v = 4.026 m/s
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