According to Fundamentals of Physics Extended 10th Edition by Halliday, Walker & Resnick, the relation between torque $(\tau)$ and moment of inertia $(I)$ and angular acceleration $(\alpha)$ is $\tau = I\alpha$, and I don't have a problem with this relation.

I think an alternate correct relation can be $\tau = I\alpha \sin\theta$. As $\tau=Fr$ was used in the deriving the formula, we get $\tau = I\alpha$. Instead, if $\tau = Fr\sin\theta$ was used in the derivation, we would've gotten $\tau = I\alpha \sin\theta$.

If we want to use $Fr\sin\theta$ for deriving the relation between torque, moment of inertia, and angular acceleration, we can see the derivation below:


$\implies \tau=m_1a_1r_1\sin\theta$

$\implies \tau=m_1\alpha r_1 r_1\sin\theta$

$\implies \tau=\alpha m_1\sin\theta(r_1)^2$

$\implies \tau_\text{net}=\alpha \sin\theta[m_1r^2_1+m_2r^2_2+m_3r^2_3+...]$

$\implies \tau_\text{net}=mr^2\alpha \sin\theta$

$\implies \tau_\text{net}= I\alpha \sin\theta$

To prove that my formula is correct, I will do a math with both my formula and $\tau=I\alpha$:


A point mass ($m = 3 \;\text{kg}$) is revolving in a circle of radius $2 \;\text{m}$. If a force $\vec F \ (|\vec F|=5 \;\text{N})$ that is not perpendicular to $\vec r$ acts on the point mass, what will be the applied torque, $\tau$?

Using $\tau=I\alpha$:


$\implies \tau=mr^2\times\frac{a}{r}$

$\implies \tau=3\times2^2\times\frac{\frac{5\times\sin(30)}{3}}{2}$

$\implies \tau=5 \;\text{N}\,\text{m}$

Using $\tau= I\alpha \sin\theta$:

$\tau= I\alpha \sin\theta$

$\implies \tau=mr^2\times\frac{a}{r}\times \sin\theta$

$\implies \tau=3\times2^2\times\frac{\frac{5}{3}}{2}\times\sin 30°$

$\implies \tau=5 \;\text{N}\,\text{m}$

So, we can see that my formula and $\tau=I\alpha$ are the same formulae. When $\theta=90°$, that is, when $\vec F$ & $\vec r$ are perpendicular, then my formula also becomes $\tau=I\alpha\times\sin 90°$$\implies \tau=I\alpha$. So, my derivation and formula are legitimate for the above reasons. How am I wrong?

  • 1
    $\begingroup$ What they call $F_t$ (and correctly relate to angular acceleration) already has the $\sin \theta$ absorbed. $\endgroup$ Jul 2, 2021 at 16:59
  • $\begingroup$ @ConnorBehan So, I am correct that the formula is $\tau = I\alpha sin\theta$? $\endgroup$ Jul 3, 2021 at 5:30
  • $\begingroup$ No. When you see a $\sin \theta$, that is a telltale sign that the equation is really a cross product of two vectors. But $I$ is not a vector. $\endgroup$ Jul 3, 2021 at 11:43
  • $\begingroup$ @user545735 (don't know if that will ping correctly), to help make sure you get what you're looking for from your bounty, can you clarify what you're looking for in an answer? Should it be only algebraic, or are vector-valued operations (dot products, cross-products, etc.) acceptable? Etc? $\endgroup$ Jul 4, 2021 at 19:12
  • 1
    $\begingroup$ For readability use \sin and \cos in the math expressions. Do you notice the difference $\tau = I \alpha \sin \theta$ ? $\endgroup$
    – JAlex
    Jul 4, 2021 at 19:49

3 Answers 3


The thing is that the relation $a_t=\alpha r$ gives the tangential component of the acceleration $a$, i.e. $a_t=a\sin{\theta}$. You can see this by differentiating $\vec{v}=\vec{\omega}\times \vec{r}$. You'd get $\vec{a}=\vec{\alpha}\times \vec{r} + \vec{\omega} \times \vec{v}$. The second term is directed along $\vec{r}$ and is called radial acceleration. The first term $\vec{\alpha}\times \vec{r}$ is perpendicular to $\vec{r}$ and is called tangential acceleration. So the tangential acceleration is only a part of the total acceleration $\vec{a}$

Even $\vec{\omega} \times \vec{r}$ only gives you the tangential velocity. Since this cross product is perpendicular to $\vec{r}$, it can't have any radial component. But the thing is, the radial component is 0. As all the particles are going in circles, the tangential velocity is equal to the total velocity $\vec{v}$. Things change when we talk about total acceleration $\vec{a}$ because, for any particle to go in a circle, it must experience a centripetal acceleration which is directed along the radius.

$$\tau=Fr\sin{\theta}$$ $$=mra\sin{\theta}$$ $$=mra_t$$ $$=mr^2\alpha$$ $$=I\alpha$$


The relationship $\tau = I \alpha$ is the correct one. It relates torque about an axis to the angular acceleration about the same axis, and the mass moment of inertia also about this axis.

It is true that if $\tau$ is a result of an offset force $F$ at a radius distance $r$, with angle $\theta$ between the force direction and the radial direction, then

$$ \tau = (r \sin \theta) F $$

But that just described the left hand side of $\tau = I \alpha$ to produce

$$ (r \sin \theta) F = I \alpha $$

Here $r \sin \theta$ is the moment arm of $F$ about the axis.

In the question you blindly substitute $F = m a$ without consideration of all possible forces acting on the body (such as pin reaction forces) and no consideration of which direction is the force $F$ acting and what direction is the acceleration $a$ possible.

This is where your equation progression fell apart.

To derive $\tau = I \alpha$ you sum up all the individual angular momentum of each particle in the body to show that $H = I \omega$. Then use Newton's second law relating force to derivative of translational momentum, for each particle of the body to find that $$\tau = \frac{{\rm d}}{{\rm d}t} H = \frac{{\rm d}}{{\rm d}t} (I \omega) = I \alpha$$

Better yet read-up on the derivation of the Newton-Euler equations of motion in vector from

$$ \begin{aligned} \sum \boldsymbol{F} & = m \boldsymbol{a} \\ \sum \boldsymbol{\tau} & = \mathrm{I} \boldsymbol{\alpha} + \boldsymbol{\omega} \times \mathrm{I} \boldsymbol{\omega} \end{aligned}$$

  • $\begingroup$ Thanks for your response! I had a question about your answer. You say that my substituting $F=ma$ was incorrect, but Halliday in his derivation also used $F=ma$ to derive $\tau=I\alpha$. What's the difference between my and his derivation? $\endgroup$ Jul 5, 2021 at 4:26
  • $\begingroup$ @AbuSafwan - I cannot comment on Halliday as I haven't read it. I do know that to complete the derivation relating torque and force, you also need the kinematics relating angular to linear motion. I also think you need it edit your question to add details such as is this a free body, or a pinned body? Also the information provided in your "answer" really belongs inside the question and the answer should be deleted. $\endgroup$
    – JAlex
    Jul 5, 2021 at 14:08
  • $\begingroup$ Could you please see my edited question? Thanks in advance! $\endgroup$ Jul 5, 2021 at 14:32

The $\sin(\theta)$ is included in the definition of the torque, which uses only the component of the force which is doing work in the tangential direction.


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