Is a shorter lever arm more effective in generating angular acceleration?

[Ibraheem]

Hello,
I've recently tried to come up with my own physics problems to clarify the torque equation(T=αI). I've made up two problems with the same force magnitude and mass but different lever length.It turned out to me that the shorter the lever the higher the angular acceleration. Is that possible? or have I done it wrong? and does that mean the closer the force to the axis the more effective it is ?!

all the bodies in the two problems are uniformed and the forces are perpendicular to the lever r
Problem one: problem two:
givens: givens:
F=5N F=5N
m=0.3kg m=0.3kg
r=0.2m r= 0.8m
results: results :
α=83.3 rad/s^2 α=20.83 rad/s^2
T=1 N.m T=4 N.m


I would be grateful if someone could help me with this.

 

[Dale]

Your expressions for torque are fine, but you cannot determine angular acceleration without the moment of inertia which is not shown nor described in a way that would allow its calculation.

 

[Doc Al]

I assumed that the moment of inertia is I=MR^2 = (0.3)(2)^2=1.2

That expression gives the moment of inertia of a point mass a distance R from the axis. (What value of R are you using?)

 

[Ibraheem]

Ignore my last comment I have mistaken the values of r with a different question I'm working on, sorry about that.
the values of moment of inertia are as following:

Problem 1: I=mr^2=0.3*0.2^2= 0.012
problem2 : I=mr^2=0.3*0.8^2=0.192

Again, sorry about the mistake

 

[Doc Al]

the values of moment of inertia are as following:

Problem 1: I=mr^2=0.3*0.2^2= 0.012
problem2 : I=mr^2=0.3*0.8^2=0.192

OK.

It turned out to me that the shorter the lever the higher the angular acceleration. Is that possible?

Only because you are changing the moment of inertia when you change the lever, since your masses are at that distance from the axis. If you kept the masses in the same place, thus keeping the moment of inertia the same, you'd find that shortening the lever reduces the angular acceleration.

and does that mean the closer the force to the axis the more effective it is ?!

No.

 

[Doc Al]

Key point: Don't confuse the distance the mass is from the axis, which determines the moment of inertia, with the lever arm, which determines the torque from the applied force.

Also realize that the moment of inertia of a point mass is a special case, not a general result for an arbitrary object.

 

[Ibraheem]

OK.Only because you are changing the moment of inertia when you change the lever, since your masses are at that distance from the axis. If you kept the masses in the same place, thus keeping the moment of inertia the same, you'd find that shortening the lever reduces the angular acceleration.

In both of the problems, I assumed that the forces are all perpendicular to R which makes the lever= R, T=F.r sin90 . So how can I shorten the lever without changing the moment of inertia?

 

[Doc Al]

In both of the problems, I assumed that the forces are all perpendicular to R which makes the lever= R, T=F.r sin90 . So how can I shorten the lever without changing the moment of inertia?

Realize that the lever arm has nothing to do with the distance the mass is from the axis. Unless you choose to use the same distance for both, which is what you did.

If you want to keep using a point mass, just fix its distance from the axis. Keep it the same for both calculations, so they have the same moment of inertia. Then, when finding the torque, use a different lever arm for each calculation.

 

[Ibraheem]

I still don't get the difference between the distance and the lever arm.I know that the lever arm is a vector, but wouldn't the lever arm have a magnitude equal to the distance if the force is perpendicular (sin(90)) ? Also, I wanted to note that I have used different distances for each problem.

 

[Doc Al]

I still don't get the difference between the distance and the lever arm.

There are two distances involved here. One is the distance the mass is from the axis. That distance is used to calculate the moment of inertia.

The other distance is the lever arm, the distance from the point of application of the force and the axis. Two very different things.

I think you are confusing yourself by using a point mass. Try this: Imagine a thin rod of mass M and length L. What's the moment of inertia of that rod, assuming the axis of rotation is at one end and perpendicular to the rod? Then apply the force at two points: At the middle of the rod (lever arm = L/2) and at the end of the rod (lever arm = L). Compare the torque and resulting angular accelerations for both cases.

 

[Ibraheem]

Would changing the point of application of the force change the distance of mass?

So if the lever arm= L/2 will the moment of inertia be I=M(L/2)^2 or if the lever arm= L will the moment of inertia be I=ML^2 ?

 

[Doc Al]

Would changing the point of application of the force change the distance of mass?

So if the lever arm= L/2 will the moment of inertia be I=M(L/2)^2 or if the lever arm= L will the moment of inertia be I=ML^2 ?

Once again, the lever arm has nothing to do with the moment of inertia. The lever arm just describes where you are applying the force and thus determines the resulting torque created by that force.

The moment of inertia depends on how the mass is distributed, not where the force is applied.

Look up the moment of inertia of a thin rod.

 

[Ibraheem]

Oh I see the difference now.The moment of inertia is constant for a rigid object so the moment of inertia of a rigid object is the proportionality constant
of T/α , and the lever arm is the line from the axis to point where the force applied and has nothing to do with the lever arm.

Thank you sir for the help