# Why are the two torque equations wrong together?

There are two torque equations :

$$\tau = I \alpha \tag{1}$$ $$\tau = F r \tag{2}$$

In the second equation, we imagine that $$F$$ is applied at $$90°$$, and that $$I = k m r^2$$. ($$k$$ is the constant responsible for the shape and the axis)

Then by equating those two formulas, we can get that: $$\alpha = \frac{k \omega}{t}.$$ But change in $$\omega$$ over time without any constant is $$\alpha$$ already. Why these formulas do not work together?

• Hi, I got $\vec{F} \times \vec{r} = km | \vec{r} |^2 \vec{\alpha}$ from your explaination. Where did $\omega, t$ come from? – Thormund May 16 at 10:35
• Because torque is the change in angular momentum over time. or the alpha is omega over time. so T = (I * omega) / time = I * alpha. – GameOver May 16 at 12:00
• Did you mean the time derivative of the angular velocity vector? – Thormund May 16 at 12:23
• Of angular momentum vector – GameOver May 17 at 8:07
• Even so, I'm still getting $| \vec{\alpha} | = \frac{1}{kmr^2} | \vec{F} \times \vec{r} | \not\Rightarrow \alpha = \frac{k\omega}{t}$. Can you please show your derivation properly. – Thormund May 17 at 11:12

Both the equations are completely consistent with each other. Equating both the equations (assuming that $$F$$ is the only force acting on the body, thus we can use $$F=ma$$)
$$Fr=I\alpha\Longrightarrow mar=kmr^2 \alpha\Longrightarrow \boxed{a=kr\alpha}$$
This equation is true for any general body as long as there's only one force acting at a distance $$r$$ from the center of mass and acting perpendicular to the line joining the center of mass and the point of application of the force.
• @GameOver In a rolling cylinder, there is also friction and $F$ isn't the only force. If you would have pulled a cylinder in space (where there would be no other force), then my $kr\alpha$ will be definitely equal to $a$. – user258881 May 16 at 14:57
• @GameOver $k$ will be the same in both the places, however the relation between $a$ and $\alpha$ will be different depending on the friction acting on the body. In the ground case (with friction), $a=r\alpha$. In the space case, $a=kr\alpha$. – user258881 May 17 at 8:36