What Determines the Natural Period of Vibration in a Pinned Disk System?

[yoamocuy]

Homework Statement

The disk, hvng a weight of 15kN, is pinned at it center O and supports the blck A that has a weight of 3 kN. If the belt which passes over the disk is not allowed to slip at its contacting surface, determine the natural period of vibration of the system.

Homework Equations

$$\Sigma$$Mo=Mo
T=1/f
Mo=I _$$\alpha$$
I=(1/2)*M*r²

The Attempt at a Solution

M=I _$$\alpha$$
I=1/2*M*r²
I=(1/2)*(15000/9.81)*(0.75)²
I=430 kg*m²
therefore Mo=430*_$$\alpha$$

$$\Sigma$$Mo=Mo
430$$\alpha$$=(0.75)(K*x)-(0.75)*(3000)
x=0.75sin($$\vartheta$$)
sin($$\vartheta$$)=1 because $$\vartheta$$ is very small
therefore $$\vartheta$$=0.75

430_$$\alpha$$=(0.75)(80000)(0.75)$$\vartheta$$-(0.75)*(3000)

At this point I'm not sure what to do. I have 2 terms in angular components and one in rectangular components. I'm not sure if I can just convert the moment provided by the weight into an angular compononent or if I need to somehow include its moment with the momnt provided by the spring. Any advice would be greatly appreciated.

 

[ideasrule]

T=1/f

Where did you get this?

430$$\alpha$$=(0.75)(K*x)-(0.75)*(3000)

I once made the same mistake on a major exam and it cost me dearly. The weight does not exert 0.75*3000 N because it's accelerating. Draw a free-body diagram on the weight to see what it exerts.

x=0.75sin($$\vartheta$$)
sin($$\vartheta$$)=1 because $$\vartheta$$ is very small
therefore $$\vartheta$$=0.75

You don't need to assume that theta is very small; no approximations are required here. Even if you did, sin(theta) is equal to 0 when theta is small, not 1.

430_$$\alpha$$=(0.75)(80000)(0.75)$$\vartheta$$-(0.75)*(3000)

How did you get this? The only equation you need is this:

430$$\alpha$$=(0.75)(K*x)-(0.75)*(3000)

when you correct the mistake, you'll see that it takes the form d^2x/dt^2 + m^2x=0. The period is just m. (Huge hint: x=r*theta).

 

[yoamocuy]

T=1/f

Where did you get this?

period=1/frequence

I once made the same mistake on a major exam and it cost me dearly. The weight does not exert 0.75*3000 N because it's accelerating. Draw a free-body diagram on the weight to see what it exerts.

Isn't the spring holding the system in equilibrium? We are just slightly disrupting the system to provide a small vibration so we can equate for the natural vibration arent we?






430$$\alpha$$=(0.75)(80000*0.75*theta)-(0.75)*(3000)

How did you get this? The only equation you need is this:

430$$\alpha$$=(0.75)(K*x)-(0.75)*(3000)

That is the equation I used, I just plugged in the 80000 in for the value of k and 0.75*theta in for the value of x.

 

[ideasrule]

Isn't the spring holding the system in equilibrium? We are just slightly disrupting the system to provide a small vibration so we can equate for the natural vibration arent we?

I didn't understand the second sentence, but the system isn't in equilibrium because it's vibrating. It's always good to do things rigorously and only approximate if absolutely necessary.