Friday

February 12, 2016
Posted by **Anonymous** on Saturday, January 11, 2014 at 4:19pm.

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**, Saturday, January 11, 2014 at 5:49pmIs this an 8:01 problem ? If not I will do it.

- Is this an 8:01 problem? -
**Damon**, Saturday, January 11, 2014 at 5:50pmIf it is not I will help.

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**shaka**, Saturday, January 11, 2014 at 5:53pmno it isn't

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**Damon**, Saturday, January 11, 2014 at 5:54pmThanks

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**Damon**, Saturday, January 11, 2014 at 6:01pmLets 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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**Anonymous**, Saturday, January 11, 2014 at 6:06pmbut how do you get alpha?

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**Damon**, Saturday, January 11, 2014 at 6:17pmNow 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

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**Damon**, Saturday, January 11, 2014 at 6:19pmHey, I am kind of slow and can not do everything at once.

Check arithmetic!

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**Damon**, Saturday, January 11, 2014 at 6:31pmwhen 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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**vivipop**, Saturday, January 11, 2014 at 7:25pmHey 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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**Anonymous**, Saturday, January 11, 2014 at 7:31pmhow is the period found?

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**shaka**, Saturday, January 11, 2014 at 7:33pmthis is an exercise of the exam, somoone posted in my name. no problem though . cheaters are fooling themselves.

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**shaka**, Saturday, January 11, 2014 at 7:35pmI really pity these fools, who do not have a clue about anything...

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**shaka**, Saturday, January 11, 2014 at 7:35pmare you the real shaka? or are you a saboture

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**shaka**, Saturday, January 11, 2014 at 7:36pmthe fact that your browsing these forums means you are a cheater as well

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**Damon**, Saturday, January 11, 2014 at 8:06pmAt this point anyone who needs help finding the period is in serious trouble.

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**da**, Saturday, January 11, 2014 at 8:18pmhahaha

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**fighter**, Sunday, January 12, 2014 at 11:29amhey DAMON can u please tell about C part of the question

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**Damon**, Sunday, January 12, 2014 at 12:31pmI did.

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**C++MeetsJava**, Monday, January 13, 2014 at 10:31amThis 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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**C++MeetsJava**, Monday, January 13, 2014 at 10:33amthis will surely hurt more than help but here you go

Period = 2*pi sqrt([kl^2+mgl]/[(2mr^2)/3+ ml^2)