Solving the Rotational Angle of a Strangely-Shaped Slab

[mopar969]

A thin slab of material of uniform density with a total mass 3.0 kg is cut in a strange shape: It is bounded by the curves y=4-x^2 and x=0. Initially it is lying in the xy plane. It is free to rotate about the y axi. A force of 8.0 Newtons is constantly applied perpendicular to the slab at the outer bottom corner. What angle (In radians) has it rotated through after 20 seconds?

Please help me get started with this problem.

 

[kjohnson]

The main thing you will want to do is to determine the moment of inertia from the function given. To do this you will have to do an integral of r^2dm. Then simply use M = I$$\alpha$$ to determine the angular acceleration.

 

[mopar969]

The integral of r^2 dm is 1/3r^3. Buthow did you know to integrate r^2 dm?

 

[kjohnson]

The integral of r^2 dm is 1/3r^3.

That is the integral of r^2 dr... you need to find a relationship between m & r so you can put dm in terms of dr...

The reason you do that integral is because that is the definition of the moment of inertia. And since you don't have a nice common shape like a rod, circle, etc. you must integrate.

 

[mopar969]

Now what do I do to solve the problem. Also what are my limits when I integrated?

 

[mopar969]

How do I apply the M=I to the alpha to this problem. Also, how do I figure in the time?

 

[mopar969]

My professor states that this method is incorrect and that I need to find the area of the object to solve the problem. Please help asap.

 

[tiny-tim]

My professor states that this method is incorrect and that I need to find the area of the object to solve the problem. Please help asap.

kjohnson is correct …

you need ∫ r²dm (= density times ∫∫ x²dxdy), not ∫ r²dr

please show us your calculation for ∫ r²dm

(if you don't understand how to calculate a https://www.physicsforums.com/library.php?do=view_item&itemid=31" you'd better say so now )

 

[mopar969]

Thank you for clearing that up. Here is what I have done so far:
dm=(M/A)(dx)(dy)
So, I=integral of (R^2)(M/A)(dx)(dy)

But now how do I integrate that and what are my limits.

 

[tiny-tim]

Thank you for clearing that up. Here is what I have done so far:
dm=(M/A)(dx)(dy)
So, I=integral of (R^2)(M/A)(dx)(dy)

But now how do I integrate that and what are my limits.

your limits are the area given in the question …

get on with it!​

 

[mopar969]

So my limits are from 0 to 2. But How do I finish the rest of the problem?

 

[kjohnson]

Thank you tiny-tim for helping clear some of that up for him..

Once you have properly calculated the moment of inertia as stated above the next step is to solve for alpha. This is done by summing your moments and setting that equal to moment of inertia multiplied by alpha M=I $$\alpha$$. Since you only have one force you only have one moment, then simply solve for alpha.

Once you have alpha its simply a constant angular acceleration problem where you know the acceleration and time.