Emergency, derive or derivative of moment of intertia

[Clairepie]

Homework Statement

Needs to be in TODAY (yeah I know I am cutting it close!)
Derive an expression for the moment of inertia of the masses at the ends of the arms in terms of l, θ and m
Is the example question asking for the derivative or is it asking to use the terms above in an equation.

Homework Equations

I=MR^2
I=$$\Gamma$$$$\alpha$$
R=l sin θ

The Attempt at a Solution

I=(mgl)/(d^2θ/d^2t) or I=2m(l sinθ)^2
It's more the term Derive that is bugging me.

Thanks guys

 

[vela]

Derive doesn't mean differentiate. The problem is simply asking you to calculate I.

 

[Doc Al]

Derive an expression for the moment of inertia of the masses at the ends of the arms in terms of l, θ and m

The moment of inertia of what?

Is the example question asking for the derivative or is it asking to use the terms above in an equation.

I'm guessing (since you didn't give the full problem) that they just want you to express the moment of inertia of some body in terms of those quantities. Not take a derivative! Derive = 'come up with'.

 

[Clairepie]

Thanks, it seems so obvious now! Having a really bad brain fog day & you've both saved my bacon! I have the rest I think,

Thank you again with extra karma your way

(Cognitive dysfunctional) Claire

 

[rwisz]

I would first clarify the question by asking for more information or context. Is this a physics or engineering problem? What is the system being studied? What are the assumptions and simplifications being made? This will help in determining the appropriate approach for solving the problem.

Assuming this is a classical mechanics problem, the moment of inertia is a measure of an object's resistance to rotational motion and is defined as the sum of the products of each mass element in the object with the square of its distance from the axis of rotation. In the case of the example question, we are dealing with a system of masses at the ends of arms, which can be thought of as a simple pendulum.

To derive an expression for the moment of inertia of this system, we need to consider the rotational motion of each mass separately. We can start by considering the moment of inertia of a single mass m at a distance l from the axis of rotation, which is given by I = ml^2. However, since the masses are not rotating about their own center of mass, we need to take into account the rotational motion of the entire system. This can be done by using the parallel axis theorem, which states that the moment of inertia of a system about an axis parallel to the original axis is equal to the sum of the moment of inertia about the original axis and the product of the total mass and the square of the distance between the axes.

In this case, the original axis of rotation is at the center of mass of the system, and the parallel axis is at the end of the arm. Therefore, the moment of inertia of the system can be expressed as I = 2ml^2 + 2m(l sinθ)^2 = 2ml^2(1 + sin^2θ). This is the desired expression for the moment of inertia in terms of l, θ, and m.

In summary, the example question is asking for the derivation of an expression for the moment of inertia of the masses at the ends of the arms, taking into account the rotational motion of the entire system. By considering the individual moments of inertia and using the parallel axis theorem, we can derive the desired expression.