A pendulum (using massless, extensionless string and a point mass at the end) of

Question

A pendulum (using massless, extensionless string and a point mass at the end) of length 1.35 m and mass 3.0 kg is created.

(a) Determine the time it takes the pendulum to first be vertical assuming a release angle of 5° from vertical and use near the surface of the Earth.

(b) Determine the time it takes the pendulum to first be vertical assuming a release angle of 10° from vertical and use near the surface of the Earth.

(c) Determine the time it takes the pendulum to first be vertical assuming a release angle of 90° from vertical and use near the surface of the Earth.

(d) Determine the time it takes the pendulum to first be vertical assuming a release angle of 5° from vertical, use near the surface of the Earth, and double the mass of the original pendulum.

(e) Determine the time it takes the pendulum to first be vertical assuming a release angle of 5° from vertical and use on the Moon.

(f) Compare the time it takes the pendulum to first be vertical in part a, to a pendulum that is identical in all respects except the mass is spread over the length of the pendulum. You do not need to quantify this case. Just state which time is bigger, if any, and why.

Explanation / Answer

ignoring air friction, irrespective of the release angle,

the pendulum will take 1/4 th of the time period to be first vertical again.

time period of the pendulum=2*pi*sqrt(l/g)

where l=length of the pendulum,g =acceleration due to gravity

for near the surface of eart, g=9.8 m/s^2

given l=1.35 m

then time period=2*pi*sqrt(1.35/9.8)=2.332 seconds

part a:

time taken=(1/4)*time period=0.583 seconds

part b:

time taken=(1/4)*time period=0.583 seconds

part c:

time taken=(1/4)*time period=0.583 seconds

part d:

as time period does not depend upon mass, even though mass is doubled,

time period will remain the same

time taken=(1/4)*time period=0.583 seconds

part e:

on the moon, g=1.62 m/s^2

then time period=time taken=2*pi*sqrt(1.35/1.62)=5.7357 seconds

time taken=(1/4)*time period=1.4339 seconds

part f:

if mass is uniformly distributed, it will become a physical pendulum

for a physical pendulum, time period is dependent upon the distrivuted mass,

distance from the point of support to the center of mass and the moment of inertia.

so the physical pendulum will have higher time period and hence time taken to be vertical for the first time will be higher.

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