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Physics Lesson 10.2.5 - Energy in a Simple Pendulum
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Welcome to our Physics lesson on Energy in a Simple Pendulum, this is the fifth lesson of our suite of physics lessons covering the topic of Pendulums. Energy in Simple Harmonic Motion, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.
Energy in a Simple Pendulum
To know the maximum speed and the highest position of the bob, we use the law of energy conservation. Thus, if considering the bob at position 1 at rest, it is obvious that its kinetic energy is zero. Therefore, since h = L - L × cos θ, we have
= Gravitational Potential Energy (1) + Kinetic Energy (1)
= Gravitational Potential Energy (1) + 0
= Gravitational Potential Energy (1)
= m × g × h
= m × g × (L - L × cos θ)
= m × g × L (1 - cos θ)
On the other hand, the kinetic energy is at maximum at the lowest point (position 2 or vertical position). In this position, gravitational potential energy is zero and all mechanical energy is due to kinetic energy. Thus, we can write
= Gravitational Potential Energy (2) + Kinetic Energy (2)
= 0 + Kinetic Energy (2)
= m × v2/2
These two energies are equal from the law of mechanical energy conservation. Thus, we can write
Simplifying mass we obtain,
Thus,
Or
After calculating v for known values of L and θ, we can find h as well, giving that
Example 2
A 2 kg bob is attached at the end of a 1 m thread as shown in the figure.
Calculate:
- Initial height h0 of the bob
- Period of this simple pendulum
- Height of the bob at t = 3.2 s
- Speed of the bob at t = 3.2 s
Take cos 200 = 0.94, sin 200 = 0.34 and g = 10 m/s2.
Solution 2
a The initial height h0 of the bob is
= L(1 - cos θ )
= 1 × (1-cos 200)
= 1 - 0.94
= 0.06 m
b The period of this pendulum is
= 2 × 3.14 × √1/10
≈ 2 s
c To calculate the height of the bob at a given time we must calculate the angle formed by the thread to the vertical at that specific time. for this, we use the equation of SHM for pendulums,
where t = 3.2 s, θ0 = 200 = π/9rad and ω = 2π/T = 2π/2 = π rad/s. Thus, we obtain
= π/9 × (-0.81)
= -0.283 rad
Thus, at t = 3.2 s, the object is at
= 1 - 1 × 0.96
= 0.04 m
d The speed of the bob at t = 3.2 s is calculated through the energetic method. Thus, since initially (at h0 = 0.06 m) the bob had only gravitational potential energy, the amount of mechanical energy of the bob during the entire motion is
m × g × h0 + 0 = m × g × h + m × v2/2
g × h0 = g × h + v2/2
10 × 0.06 = 10 × 0.04 + v2/2
0.6 = 0.4 + v2/2
v = √2 × (0.6 - 0.4)
= √0.4
= 0.63 m/s
You have reached the end of Physics lesson 10.2.5 Energy in a Simple Pendulum. There are 5 lessons in this physics tutorial covering Pendulums. Energy in Simple Harmonic Motion, you can access all the lessons from this tutorial below.
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