# PHYSICS(HELP)

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 small-angle 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=

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1. Is this an 8:01 problem ? If not I will do it.

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posted by Damon
2. If it is not I will help.

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posted by Damon
3. no it isn't

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posted by shaka
4. Thanks

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posted by Damon
5. 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.phy-astr.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

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posted by Damon
6. but how do you get alpha?

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7. Now let's do Moments about the pivot point
let angle theta = T
restoring Force = - k x - m g sin T
so
F = -k x - m g 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

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posted by Damon
8. Hey, I am kind of slow and can not do everything at once.
Check arithmetic!

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posted by Damon
9. 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|=

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posted by Damon
10. 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.

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posted by vivipop
11. how is the period found?

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12. this is an exercise of the exam, somoone posted in my name. no problem though . cheaters are fooling themselves.

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posted by shaka
13. I really pity these fools, who do not have a clue about anything...

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posted by shaka
14. are you the real shaka? or are you a saboture

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posted by shaka
15. the fact that your browsing these forums means you are a cheater as well

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posted by shaka
16. At this point anyone who needs help finding the period is in serious trouble.

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posted by Damon
17. hahaha

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posted by da
18. hey DAMON can u please tell about C part of the question

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posted by fighter
19. I did.

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posted by Damon
20. 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

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21. 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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