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?

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    $\begingroup$ Hi, I got $\vec{F} \times \vec{r} = km | \vec{r} |^2 \vec{\alpha}$ from your explaination. Where did $\omega, t$ come from? $\endgroup$ – Thormund May 16 at 10:35
  • $\begingroup$ Because torque is the change in angular momentum over time. or the alpha is omega over time. so T = (I * omega) / time = I * alpha. $\endgroup$ – GameOver May 16 at 12:00
  • $\begingroup$ Did you mean the time derivative of the angular velocity vector? $\endgroup$ – Thormund May 16 at 12:23
  • $\begingroup$ Of angular momentum vector $\endgroup$ – GameOver May 17 at 8:07
  • $\begingroup$ 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. $\endgroup$ – 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.

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  • $\begingroup$ How can it be k * r * alpha? If this is a cylinder spinning around its axis then k = 0.5. So how can acceleration be the half of r *alpha? $\endgroup$ – GameOver May 16 at 12:04
  • $\begingroup$ @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 $a$. $\endgroup$ – user258881 May 16 at 14:57
  • $\begingroup$ But k depends on the shape and the axis of rotation. k is the constant in moment of inertia. I do not think it is affected by weather you rotate it in vacuum or on earth. So k of any cylinder rotating around its axis will be 0.5, but the acceleration of that cylinder is not the half of its rotational acceleration times radius. $\endgroup$ – GameOver May 17 at 8:12
  • $\begingroup$ @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, $a=kr\alpha$. $\endgroup$ – user258881 May 17 at 8:36
  • $\begingroup$ Are we referring to an acceleration of a point of the object? If so, then traveled circumference is 2piR, and by doing some math we get that omega * R = velocity, thus the same with acceleration. So there is no k in there. If we are referring to the acceleration of a body as a whole, then the linear body acceleration in space or in air of the body is zero, so it is also not kR*alpha. What do I keep missing in there? $\endgroup$ – GameOver May 17 at 13:57

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