# PHYSICS(HELP)

posted by on .

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=

• PHYSICS(HELP) - ,

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

• Is this an 8:01 problem? - ,

If it is not I will help.

• PHYSICS(HELP) - ,

no it isn't

• PHYSICS(HELP) - ,

Thanks

• PHYSICS(HELP) - ,

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

• PHYSICS(HELP) - ,

but how do you get alpha?

• PHYSICS(HELP) - ,

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

• PHYSICS(HELP) - ,

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

• PHYSICS(HELP) - ,

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) - ,

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) - ,

how is the period found?

• PHYSICS(HELP) - ,

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

• PHYSICS(HELP) - ,

I really pity these fools, who do not have a clue about anything...

• PHYSICS(HELP) - ,

are you the real shaka? or are you a saboture

• PHYSICS(HELP) - ,

the fact that your browsing these forums means you are a cheater as well

• PHYSICS(HELP) - ,

At this point anyone who needs help finding the period is in serious trouble.

• PHYSICS(HELP) - ,

hahaha

• PHYSICS(HELP) - ,

hey DAMON can u please tell about C part of the question

• PHYSICS(HELP) - ,

I did.

• PHYSICS(HELP) - ,

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) - ,

this will surely hurt more than help but here you go

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