PHYSICS(HELP)
posted by Anonymous .
A pendulum of mass m= 0.8 kg and length l=1 m is hanging from the ceiling. The massless string of the pendulum is attached at point P. The bob of the pendulum is a uniform shell (very thin hollow sphere) of radius r=0.4 m, and the length l of the pendulum is measured from the center of the bob. A spring with spring constant k= 7 N/m is attached to the bob (center). The spring is relaxed when the bob is at its lowest point (θ=0). In this problem, we can use the smallangle approximation sinθ≃θ and cosθ≃1. Note that the direction of the spring force on the pendulum is horizontal to a very good approximation for small angles θ. (See figure)
Take g= 10 m/s2
(a) Calculate the magnitude of the net torque on the pendulum with respect to the point P when θ=5∘. (magnitude; in Nm)
τP=
(b) What is the magnitude of the angular acceleration α=θ¨ of the pendulum when θ=5∘? (magnitude; in radians/s2)
α=
(c) What is the period of oscillation T of the pendulum? (in seconds)
T=

PHYSICS(HELP) 
Damon
Is this an 8:01 problem ? If not I will do it.

Is this an 8:01 problem? 
Damon
If it is not I will help.

PHYSICS(HELP) 
shaka
no it isn't

PHYSICS(HELP) 
Damon
Thanks

PHYSICS(HELP) 
Damon
Lets do moments of inertia about the pivot point
First the moment of inertia of this thin shell about a horizontal line through its center is (2/3) mR^2
for derivation see http://hyperphysics.phyastr.gsu.edu/hbase/isph.html
This has to be transferred up to the pivot point using the parallel axis theorem
I = m L^2 + (2/3) m R^2
here
m = .8
L = 1
R = .4
so
I = .8 + .085 = .885 
PHYSICS(HELP) 
Anonymous
but how do you get alpha?

PHYSICS(HELP) 
Damon
Now let's do Moments about the pivot point
let angle theta = T
restoring Force =  k x  m g sin T
but sin t is about T in radians for small T
so
F = k x  m g T
x is about L T
so
F = k L T  m g T
moment = M = F L = (k L^2 + mgL)T
moment = I alpha
(k L^2 + mgL)T = .885 d^2T/dt^2
well calculate the coefficient of T on the left
(7 + 8) T = .885 d^2T/dt^2
.885 d^2T/dt^2 =  15 T
let T = A sin ( w t)
then d^2T/dt^2 =  w^2 T
so
.885 w^2 = 15
w = 2 pi f = 2 pi/period = 4.11 
PHYSICS(HELP) 
Damon
Hey, I am kind of slow and can not do everything at once.
Check arithmetic! 
PHYSICS(HELP) 
Damon
when you do these, use T in RADIANS
5 degrees *pi/180
(a) Calculate the magnitude of the net torque on the pendulum with respect to the point P when θ=5∘. (magnitude; in Nm)
Torque = 15T where T = 5 * pi/180
τP=
(b) What is the magnitude of the angular acceleration α=θ¨ of the pendulum when θ=5∘? (magnitude; in radians/s2)
alpha = d^2T/dt^2 =  15 T /.885
α= 
PHYSICS(HELP) 
vivipop
Hey Damon, i just want to say thank you for making me believe in myself one more time. i got this but i didn't think my workings were right,so i was scared to key in my answers. Now,i am sooo confident in myself. thank you.

PHYSICS(HELP) 
Anonymous
how is the period found?

PHYSICS(HELP) 
shaka
this is an exercise of the exam, somoone posted in my name. no problem though . cheaters are fooling themselves.

PHYSICS(HELP) 
shaka
I really pity these fools, who do not have a clue about anything...

PHYSICS(HELP) 
shaka
are you the real shaka? or are you a saboture

PHYSICS(HELP) 
shaka
the fact that your browsing these forums means you are a cheater as well

PHYSICS(HELP) 
Damon
At this point anyone who needs help finding the period is in serious trouble.

PHYSICS(HELP) 
da
hahaha

PHYSICS(HELP) 
fighter
hey DAMON can u please tell about C part of the question

PHYSICS(HELP) 
Damon
I did.

PHYSICS(HELP) 
C++MeetsJava
This problem is really all about an expression summarizing harmonic motion.
So your period for this problem is oddly similar to
T = 2pi * sqrt (m/k)
but now m = moment of inertia
and k = expanded coefficient, that was given above by Mr. Damon
Also make sure that you end up with seconds as units 
PHYSICS(HELP) 
C++MeetsJava
this will surely hurt more than help but here you go
Period = 2*pi sqrt([kl^2+mgl]/[(2mr^2)/3+ ml^2)
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