The pendulum shown to the left consists of a bob of mass of 0.41 kg, attached to
ID: 2142485 • Letter: T
Question
The pendulum shown to the left consists of a bob of mass of 0.41 kg, attached to a string of length L = 1.2 meters. It is moving to the right with a speed of 4.51 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 34 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)
5)
What is change in kinetic energy of the bob as it moves from point A to point B?
?KE =
6)
What is kinetic energy of the bob at point B?
KEB =
7)
What is velocity of the bob at point B?
vB =
8)
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?
KEA =
9)
What is the potential energy of the bob at point A?
UA=
10)
What is the mechanical energy of the bob at point A?
METOT-A=
11)
What is work done by non-conservative forces as the bob moves from point A to point B?
Wnon-con=
12)
What is the mechanical energy of the bob at point B?
METOT-B=
13)
What is the potential energy of the bob when it reaches point B?
UB =
14)
What is the kinetic energy of the bob when it reaches point B?
KEB =
15)
What is the speed of the bob at point B?
v=
Explanation / Answer
work done total = change in kinetic energy
change in kinetic energy = -0.824 J
KE at B = 3.346 J
velocity at B = 4.04 m/s
KE at A = 4.17J
PE at A = 0
ME total at A = 4.17J
work non conservative = 0
ME total at B = 4.17 J
PE at B = 0.824J
KE at B = 3.346J
velocity at B = 4.04m/s
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