# Confused with the effects of moment of inertia and torque

According to the definition of torque, if the distance from the pivot is large the force required to apply will be smaller to move with the specified angular velocity/acceleration.

And we can experience that in the movement of door about hinges and also in that of seesaw.

But consider two cylinders of two different radii but same mass. When we rotate those cylinders with the same force(torque), the cylinder with smaller radius is easier to rotate than the cylinder with larger radius. Also by formula, moment of inertia will be smaller for smaller radius cylinder and will attain greater angular velocity/acceleration with same force(torque) applied than the cylinder with larger radius.

$$τ=Iαsinθ$$, where $$I = (MR^2)/2$$ in case of solid cylinder. From this equation, for the same force applied at the tangent (i.e. θ = 90 degree) of the two cylinders' surface separately, having the same mass M but have different radius R:

Will the angular acceleration increase in cylinder having small radius (or) we need more force to apply to the cylinder with smaller R, to achieve the angular acceleration attained in the cylinder having bigger R.

I got this doubt as inertia is the resistance to change from its original state and the moment of inertia is bigger in the cylinder having bigger radius according to the formula. So will the same force make lesser angular acceleration in the cylinder having bigger radius?

EDIT:Please ignore the question and details in the text above as it may be confusing. I am rephrasing the question below clearly.

It's a common practical experience we feel, when giving some force for rotational movements, it will be easier when distance is more from the center. An example is when increasing the length of car jack lever by some extensions it will be easier to operate and lift up the car. It means we are giving less force. Am I right?

But if moment of inertia is giving resistance to move easily, by formula it is showing that increase in radius alone and mass being the same, increases the required torque/force to achieve the same angular acceleration /velocity. It is obvious through formula that increase in radius increases required linear distance to be covered to reach the same angular movement and it's consequence leads to angular velocity and acceleration.

But by practical experience, we feel if distance increases it reduces the difficulty. So what does moment of inertia actually mean?

• I'm having a really hard time understanding what you are asking in the last paragraph. Commented Dec 14, 2023 at 12:42
• Are you asking why is it more difficult to cause the same angular acceleration for a solid cylinder than it is for a solid door where the radius of the cylinder is the same as the distance between the hinge and edge of the door? Commented Dec 14, 2023 at 12:48
• $\tau=I\alpha$, what is $\theta$? Commented Dec 14, 2023 at 13:41
• Maybe the best way for me to find out what you are asking is for you to show me why I haven't answered your question. Commented Dec 14, 2023 at 13:48
• If you are confused, then work through the problem mathematically and see what the equations tell you. Commented Dec 14, 2023 at 14:25

What I am missing here/where I am wrong, to understand the difference, that in case of torque large length from pivot requires less force for door/seesaw but in case of cylinder lesser radius from pivot requires less force?

I'm having difficulty following you, but I think what you are asking is equivalent to why it requires more force to cause the same angular acceleration of a solid cylinder than a solid door of the same mass and radius of rotation. If that is the case, the short answer is the moment of inertia $$I$$, which is a measure of resistance to angular acceleration, analogous to mass being a measure of resistance to linear acceleration, is greater for the cylinder than the door.

For the solid cylinder:

$$I_{cyl}=\frac {1}{2}MR^2\tag{1}$$

For the solid door:

$$I_{door}=\frac{1}{3}MR^2\tag{2}$$

The angular acceleration of each is given by

$$\alpha=\frac{\tau}{I}= \frac{RF}{I}\tag{3}$$

$$F=\frac{\alpha I}{R}\tag{4}$$

For the same R it requires more force to cause the same angular acceleration for the cylinder than the door since $$I_{cyl}\gt I_{door}$$

Hope this helps.

• My doubt is, as the distance from the pivot is greater the force required will be smaller to achieve certain angular velocity/acceleration. But τ=Iαsinθ, where I = (MR^2)/2 in case of solid cylinder. From this equation, for the same force applied at the tangent (i.e θ = 90 degree) of the two cylinders surface separately which have the same mass M but have different radius R: will the angular acceleration increases in cylinder having small radius (or) we need more force to apply to the cylinder with smaller R, to achieve the torque/angular acceleration attained in the cylinder having bigger R. Commented Dec 14, 2023 at 14:02
• I got this doubt as Inertia is the resistance to change from its original state and the moment of inertia is bigger in the cylinder having bigger radius according to the formula. So will the same force make lesser angular acceleration in the cylinder having bigger radius? Commented Dec 14, 2023 at 14:09
• @ayyappanmuthukrishnan “My doubt is, as the distance from the pivot is greater the force required will be smaller to achieve certain angular velocity/acceleration” That is correct, but only for a given moment of inertia. If the moment of inertia is different, the force will be different. Commented Dec 14, 2023 at 14:54
• “But $τ=Iαsinθ$, where $I = (MR^2)/2$ in case of solid cylinder”. That’s correct. Commented Dec 14, 2023 at 14:54
• From this equation, for the same force applied at the tangent of the two cylinders surface separately which have the same mass M but have different radius R: will the angular acceleration increases in cylinder having small radius…” Yes, because the moment of inertia, which is the resistance to angular acceleration, is less for the smaller radius cylinder. That means if the same force is applied to the two cylinders, the angular acceleration of the smaller radius cylinder will be greater than the larger radius cylinder. Commented Dec 14, 2023 at 14:54

We have

$$Fd = I \alpha$$

where $$F$$ is the applied force, $$d$$ is the perpendicular distance between the line of this force and the axis (this is simpler than introducing a factor of $$\sin \theta$$), $$I$$ is the moment of inertia and $$\alpha$$ is the angular acceleration.

For the door, if we keep $$I$$ and $$\alpha$$ constant, then we have

$$\displaystyle F \propto \frac 1 d$$

so as we make $$d$$ larger then the force $$F$$ required to achieve a given angular acceleration $$\alpha$$ becomes smaller.

For the cylinders we have $$d=R$$ and $$I \propto R^2$$ (if the cylinders have the same mass $$M$$). So if $$\alpha$$ is constant we have

$$FR \propto R^2 \\ \Rightarrow F \propto R$$

so as we make $$R$$ larger the force $$F$$ required to achieve a given angular acceleration $$\alpha$$ now increases. This is because although a larger $$R$$ increases the torque, it increases the moment of inertia $$I$$ by a larger amount since $$I \propto R^2$$.

• Okay, But what is the formula for moment of inertia in case of door? Bob D has mentioned it as (MR^2)/3. So it's moment of inertia is also proportional to R^2. Is it due to that in cylinder the force has to push the part in the other half of pivot. And in case of door we apply force only to the one side of pivot? Commented Dec 14, 2023 at 15:39
• @ayyappanmuthukrishnan In my answer there is only one door, so in the case of the door $I$ is constant. We can, however, vary $d$, which is the distance between the pivot of the door and the line of the force $F$. Bob thinks you are comparing a cylinder with a door of similar dimensions. I think you are comparing one cylinder with another with different values of $R$. Commented Dec 14, 2023 at 15:49

Mass moment of inertia is not the only consideration when feeling resistance to motion. You have to separate the idea of mass moment of inertia which indicates how far from a pivot the mass is mostly concentrated, from the idea of a moment-arm which is how far from a pivot are you applying a force.

There are two scenarios to consider.

1. Free Floating Body - Imagine an object in space at rest. This body has mass $$m$$ and a radius of gyration $$\kappa$$, such that the mass moment of inertia value about the center of mass is $$I= m \kappa^2$$

At some instant an impulse $$J$$ is applied at a distance $$d$$ (the moment arm) from the center of mass, as shown below:

As a response to the impulse, the body is set in motion with a rotation about some point a distance $$c$$ from the center of mass on the other side as the impulse was applied. The resulting rotational speed $$\Omega$$ means that the point of application of the impulse is moving with speed

$$v = (c+d) \Omega$$

and the effective mass the impulse feels (call it $$m_{\rm eff}$$) can be found from the relationship

$$J = m_{\rm eff}\; v = m_{\rm eff}\; (c+d) \Omega$$

and for this situation the momentum linear and angular momentum transfer is

\left. \begin{aligned} J & = m c \Omega \\ d J & = m \kappa^2 \Omega \end{aligned} \right\} \; \begin{aligned} c & = \frac{\kappa^2}{d} \\ \Omega & = \frac{d J}{m \kappa^2} \end{aligned}

which gives us the location and magnitude of the resulting motion. Now use this result in the equation for the effective mass to get

$$m_{\rm eff} = \frac{m \kappa^2}{\kappa^2 + d^2} = \frac{1}{ \frac{1}{m} + \frac{d^2}{I}}$$

You can interpret the above geometrically, as you see that $$m_{\rm eff} \leq m$$ in general. For the special case where the moment-arm $$d$$ is zero (or the mass moment of inertia is infinite) then $$m_{\rm eff} = m$$. For all other cases, the effective mass is less than the actual mass.

The longer the moment arm $$d$$ the less the effective mass is, and this is what you have described in your question. It is possible with a really large moment arm to get the effective mass to approach zero given the $$d^2$$ in the denominator.

The lesser the mass moment of inertia $$I$$ is the less is the effective mass since the denominator becomes larger. This is because it is easier to turn something with less mass moment of inertia.

So these results are consistent with observation and intuition.

2. Constrained Body - This is the case where the body is pivoted and we know the pivot distance $$c$$, but there is an additional reaction impulse $$R$$ applied at the pivot as a result of the applied impulse $$J$$ with moment-arm distance $$d$$.

The equation of momentum transfer are now:

\left. \begin{aligned} J + R & = m c \Omega \\ d J - c R & = m \kappa^2 \Omega \end{aligned} \right\} \; \begin{aligned} \Omega & = \frac{J ( c+ d) }{m ( c^2+\kappa^2)} \\ R & = \frac{J ( c d - \kappa^2 )}{c^2 + \kappa^2} \end{aligned}

Now use the results in the equation for effective mass to get

$$m_{\rm eff} = \frac{m ( c^2 + \kappa^2)}{(c+d)^2} = \frac{I + m c^2}{(c+d)^2}$$

The interpretation of this result is the ratio of the mass moment of inertia about the pivot ($$I+m c^2$$), to the distance of the applied impulse (from the pivot) squared ($$(c+d)^2$$). This can either be larger or lesser than the mass of the object.

In fact, the case where the mass moment of inertia becomes exceedingly large is the case where the effective mass also becomes exceedingly large. I think this corresponds to the situation in your question where large MMOI results in difficulting moving an object.

The case where the effective mass approaches zero for a constrained body does not exist unless the body is pivoted about the center of mass ($$c=0$$) and it is a point mass ($$\kappa=0$$).

The special case where $$m_{\rm eff} = m$$ occurs when the moment arm is $$d = \sqrt{c^2 + \kappa^2}-c$$ which is interpreted as the radius of gyration about the pivot minus the center of mass distance. It is not any noteworthy situation.

For the constrained body the larger the moment arm the less the effective mass (just like the free body), and the more the mass moment of inertia is the more the effective mass (like the free body also), but both with different "strengths" for these effects.

• I could not understand the formulae used. Is there any key references I can look to understand how it is derived. Also in the below link I see physics.stackexchange.com/questions/691097/… that it was mentioned as "Once object is rotating at certain angular velocity ω, if there is no (net) torque the object will keep rotating at the same angular velocity". If so I have a doubt that MOI creates difficulty in rotating, then if torque applied is removed it should stop rotating quickly for large MOI right? Commented Jan 26 at 11:18
• @ayyappanmuthukrishnan - maybe read this answer (physics.stackexchange.com/a/783565/392) that derives the concept of effective mass as seen by an impulse or collision. Commented Jan 26 at 14:22
• @ayyappanmuthukrishnan - MMOI describes the difficulty in changing rotation. Commented Jan 26 at 14:26
• If two cylinders with same mass but different radius is rotated by applying same quantum of force/torque, I am clear now that cylinder with greater radius will have lesser angular velocity. Now while rotating if the applied force to rotate is removed, which cylinder will stop rotating first, assuming that other environment conditions are same for both. Commented Feb 2 at 13:37
• @ayyappanmuthukrishnan - neither. They will keep spinning maintain their angular momentum when no net torque is applied. Commented Feb 2 at 14:05