Question

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=

Answer 1

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

Answer 2

If it is not I will help.

Answer 3

no it isn't

Answer 4

Thanks

Answer 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

Answer 6

but how do you get alpha?

Answer 7

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

Answer 8

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

Answer 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|=

Answer 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.

Answer 11

how is the period found?

Answer 12

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

Answer 13

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

Answer 14

are you the real shaka? or are you a saboture

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

I did.

Answer 19

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

Answer 20

this will surely hurt more than help but here you go

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