# PHYSICS(HELP)

posted by .

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)

## Similar Questions

1. ### physics

A pendulum is constructed from a thin, rigid, and uniform rod with a small sphere attached to the end opposite the pivot. This arrangement is a good approximation to a simple pendulum (period = 5.97 s), because the mass of the sphere …
2. ### physics

A pendulum consists of a mass m = 0.12 kg hanging from a flexible string of length L. The string is very thin, very light, and does not stretch. It makes small oscillations, with a period of 0.663 s. What is the oscillation frequency …
3. ### physics

A simple pendulum consists of a ball of mass 5.39 kg hanging from a uniform string of mass 0.0789 g and length L. The period of oscilla- tion for the pendulum is 2.51 s. Determine the speed of a transverse wave in the string when the …
4. ### physics

A 0.33- kg pendulum bob is attached to a string 1.2 m long and hangs vertically as shown in Figure 1. What is the change in the pendulum bob’s gravitational potential energy as it swings from the lowest point of the pendulum to a …
5. ### Physics Classical Mechanics Help ASAP

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 …
6. ### Classical mechanics

A pendulum of mass m= 0.9 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 …
7. ### 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 …
8. ### 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 …
9. ### physics (Urgent)!!!!!!!!!!!!!!!!!!!!!!

A pendulum of mass m= 0.9 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 …
10. ### physics Classical Mechanics

A pendulum of mass m= 0.9 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 …

More Similar Questions

Post a New Question