Return to Answer

Let's derive $\tau = I\alpha$ to see the conditions of its validity.

At any instant, we may describe the motion of a rigid body as a rotation about an arbitrary point, on top of a translation. Specifically, the velocity of any point on the rigid body may be expressed as $$\vec v = \vec v_0 +\vec \omega\times\vec r,$$ where $\vec v_0$ is the velocity of the center of rotation you have chosen, $\vec r$ is the relative position with respect to this point, and $\vec \omega$ is the angular velocity. To simplify things a bit, we can break down the position vector into two components: $\vec r = \vec R + \vec r_\parallel$, where $\vec R$ is perpendicular to $\vec \omega$ (and hence the axis of rotation), and $\vec r_\parallel$ is parallel to it. Then, we can write $$\vec v = \vec v_0 +\vec \omega\times\vec R.$$

Taking the derivative with respect to time, $$\vec a = \vec a_0 + \vec \alpha \times \vec R + \vec \omega\times (\vec v - \vec v_0)$$ $$=\vec a_0 + \vec \alpha\times \vec R + \vec \omega\times (\vec \omega \times \vec R).$$ The second term on the right hand side is the tangential acceleration, while the third term is the centripetal acceleration. Using the vector identity $\vec A \times (\vec B \times \vec C) = (\vec A \cdot \vec C)\vec B-(\vec A \cdot \vec B)\vec C$, we may simplify the centripetal acceleration term to obtain $$\vec a=\vec a_0 + \vec \alpha\times \vec R-\omega ^2\vec R.$$

Now to get the out-of-plane component of the torque, $$\require{cancel}\vec \tau =\int\rho(\vec r)\vec R\times \vec a \ dV$$ $$=\left(\int\rho\vec R\ dV\right)\times \vec a_0 + \int\rho\vec R\times (\vec \alpha \times \vec R) \ dV - \int\rho\omega^2\cancelto{0}{(\vec R \times \vec R)}\ dV$$ $$=\left(\int\rho\vec R\ dV\right)\times \vec a_0 + \left(\int\rho R^2\ dV\right)\vec\alpha$$ $$=M\vec R_\text{CM}\times\vec a_0 + I\vec\alpha$$ or $$\tau - M\vec R_\text{CM}\times\vec a_0 = I\vec\alpha.$$

We see that $\vec\tau = I\vec \alpha$ holds if only if $- M\vec R_\text{CM}\times\vec a_0 = 0$, where $\vec R_\text{CM}$ is the position of the center of mass relative to the reference point you pick. $- M\vec R_\text{CM}\times\vec a_0$ is the fictitious torque term that you need to account for to be able to use $\vec\tau = I\vec\alpha$ around an arbitrary point.

If you pick $A$, this point has no downward acceleration, so $-M\vec R_\text{CM}\times\vec a_0 = 0.$ If you pick $B$ (the center of mass), $\vec R_\text{CM} = 0$, so $-M\vec R_\text{CM}\times\vec a_0 = 0.$ Finally, for point $C$, the downward (tangential) component of $\vec a_0$ is $-2\alpha R$, so $-M\vec R_\text{CM}\times\vec a_0$ is $2MR^2\alpha$, out-of-plane (counter-clockwise). We can write in the counter-clockwise direction $$\tau_C + 2MR^2\alpha = I_C \alpha.$$ Plugging in $\tau_C = \frac13MRg$ and $I_C=\frac32MR^2$, we find the correct result $\alpha = -\frac{2g}{3R}$.

Let's derive $\tau = I\alpha$ to see the conditions of its validity.

At any instant, we may describe the motion of a rigid body as a rotation about an arbitrary point, on top of a translation. Specifically, the velocity of any point on the rigid body may be expressed as $$\vec v = \vec v_0 +\vec \omega\times\vec r,$$ where $\vec v_0$ is the velocity of the center of rotation you have chosen, $\vec r$ is the relative position with respect to this point, and $\vec \omega$ is the angular velocity. Taking the derivative with respect to time, $$\vec a = \vec a_0 + \vec \alpha \times \vec r + \vec \omega\times (\vec v - \vec v_0)$$ $$=\vec a_0 + \vec \alpha\times \vec r + \vec \omega\times (\vec \omega \times \vec r).$$ The second term on the right hand side is the tangential acceleration, while the third term is the centripetal acceleration.

To simplify things a bit, since this is a 2D problem with the axis of rotation perpendicular to the page, I will assume that $\omega$ and $\alpha$ are both perpendicular to the page, i.e. (anti-)parallel to each other. We can break down the position vector into two components: $\vec r = \vec R + \vec r_\parallel$, where $\vec R$ is perpendicular to $\vec \omega$ (and hence the axis of rotation), and $\vec r_\parallel$ is parallel to it. Then, $\vec\alpha\times\vec r = \vec\alpha\times\vec R$ and $\vec\omega\times\vec r = \vec\omega\times\vec R$. Using the vector identity $\vec A \times (\vec B \times \vec C) = (\vec A \cdot \vec C)\vec B-(\vec A \cdot \vec B)\vec C$, we may simplify the centripetal acceleration term to obtain $$\vec a=\vec a_0 + \vec \alpha\times \vec R-\omega ^2\vec R.$$

Now to get the out-of-plane component of the torque, $$\require{cancel}\vec \tau =\int\rho(\vec r)\vec R\times \vec a \ dV$$ $$=\left(\int\rho\vec R\ dV\right)\times \vec a_0 + \int\rho\vec R\times (\vec \alpha \times \vec R) \ dV - \int\rho\omega^2\cancelto{0}{(\vec R \times \vec R)}\ dV$$ $$=\left(\int\rho\vec R\ dV\right)\times \vec a_0 + \left(\int\rho R^2\ dV\right)\vec\alpha$$ $$=M\vec R_\text{CM}\times\vec a_0 + I\vec\alpha$$ or $$\tau - M\vec R_\text{CM}\times\vec a_0 = I\vec\alpha.$$

We see that $\vec\tau = I\vec \alpha$ holds if only if $- M\vec R_\text{CM}\times\vec a_0 = 0$, where $\vec R_\text{CM}$ is the position of the center of mass relative to the reference point you pick. $- M\vec R_\text{CM}\times\vec a_0$ is the fictitious torque term that you need to account for to be able to use $\vec\tau = I\vec\alpha$ around an arbitrary point.

If you pick $A$, this point has no downward acceleration, so $-M\vec R_\text{CM}\times\vec a_0 = 0.$ If you pick $B$ (the center of mass), $\vec R_\text{CM} = 0$, so $-M\vec R_\text{CM}\times\vec a_0 = 0.$ Finally, for point $C$, the downward (tangential) component of $\vec a_0$ is $-2\alpha R$, so $-M\vec R_\text{CM}\times\vec a_0$ is $2MR^2\alpha$, out-of-plane (counter-clockwise). We can write in the counter-clockwise direction $$\tau_C + 2MR^2\alpha = I_C \alpha.$$ Plugging in $\tau_C = \frac13MRg$ and $I_C=\frac32MR^2$, we find the correct result $\alpha = -\frac{2g}{3R}$.

Let's derive $\tau = I\alpha$ to see the conditions of its validity.

At any instant, we may describe the motion of a rigid body as a rotation about an arbitrary point, on top of a translation. Specifically, the velocity of any point on the rigid body may be expressed as $$\vec v = \vec v_0 +\vec \omega\times\vec r,$$ where $\vec v_0$ is the velocity of the center of rotation you have chosen, $\vec r$ is the relative position with respect to this point, and $\vec \omega$ is the angular velocity. To simplify things a bit, we can break down the position vector into two components: $\vec r = \vec R + \vec r_\parallel$, where $\vec R$ is perpendicular to $\vec \omega$ (and hence the axis of rotation), and $\vec r_\parallel$ is parallel to it. Then, we can write $$\vec v = \vec v_0 +\vec \omega\times\vec R.$$

Taking the derivative with respect to time, $$\vec a = \vec a_0 + \vec \alpha \times \vec R + \vec \omega\times (\vec v - \vec v_0)$$ $$=\vec a_0 + \vec \alpha\times \vec R + \vec \omega\times (\vec \omega \times \vec R).$$ The second term on the right hand side is the tangential acceleration, while the third term is the centripetal acceleration. Using the vector identity $\vec A \times (\vec B \times \vec C) = (\vec A \cdot \vec C)\vec B-(\vec A \cdot \vec B)\vec C$, we may simplify the centripetal acceleration term to obtain $$\vec a=\vec a_0 + \vec \alpha\times \vec R-\omega ^2\vec R.$$

Now to get the out-of-plane component of the torque, $$\require{cancel}\vec \tau =\int\rho(\vec r)\vec R\times \vec a \ dV$$ $$=\left(\int\rho\vec R\ dV\right)\times \vec a_0 + \int\rho\vec R\times (\vec \alpha \times \vec R) \ dV - \int\rho\omega^2\cancelto{0}{(\vec R \times \vec R)}\ dV$$ $$=\left(\int\rho\vec R\ dV\right)\times \vec a_0 + \left(\int\rho R^2\ dV\right)\vec\alpha$$ $$=M\vec R_\text{CM}\times\vec a_0 + I\vec\alpha$$ or $$\tau - M\vec R_\text{CM}\times\vec a_0 = I\vec\alpha.$$

We see that $\vec\tau = I\vec \alpha$ holds if only if $- M\vec R_\text{CM}\times\vec a_0$, where $\vec R_\text{CM}$ is the position of the center of mass relative to the reference point you pick. $- M\vec R_\text{CM}\times\vec a_0$ is the fictitious torque term that you need to account for to be able to use $\vec\tau = I\vec\alpha$ around an arbitrary point.

If you pick $A$, this point has no downward acceleration, so $-M\vec R_\text{CM}\times\vec a_0 = 0.$ If you pick $B$ (the center of mass), $\vec R_\text{CM} = 0$, so $-M\vec R_\text{CM}\times\vec a_0 = 0.$ Finally, for point $C$, the downward (tangential) component of $\vec a_0$ is $2\alpha R$, so $-M\vec R_\text{CM}\times\vec a_0$ is $2MR^2\alpha$, out-of-plane (counter-clockwise). We can write in the counter-clockwise direction $$\tau_C + 2MR^2\alpha = I_C \alpha.$$ Plugging in $\tau_C = \frac13MRg$ and $I_C=\frac32MR^2$, we find the correct result $\alpha = -\frac{2g}{3R}$.

Let's derive $\tau = I\alpha$ to see the conditions of its validity.

At any instant, we may describe the motion of a rigid body as a rotation about an arbitrary point, on top of a translation. Specifically, the velocity of any point on the rigid body may be expressed as $$\vec v = \vec v_0 +\vec \omega\times\vec r,$$ where $\vec v_0$ is the velocity of the center of rotation you have chosen, $\vec r$ is the relative position with respect to this point, and $\vec \omega$ is the angular velocity. To simplify things a bit, we can break down the position vector into two components: $\vec r = \vec R + \vec r_\parallel$, where $\vec R$ is perpendicular to $\vec \omega$ (and hence the axis of rotation), and $\vec r_\parallel$ is parallel to it. Then, we can write $$\vec v = \vec v_0 +\vec \omega\times\vec R.$$

Taking the derivative with respect to time, $$\vec a = \vec a_0 + \vec \alpha \times \vec R + \vec \omega\times (\vec v - \vec v_0)$$ $$=\vec a_0 + \vec \alpha\times \vec R + \vec \omega\times (\vec \omega \times \vec R).$$ The second term on the right hand side is the tangential acceleration, while the third term is the centripetal acceleration. Using the vector identity $\vec A \times (\vec B \times \vec C) = (\vec A \cdot \vec C)\vec B-(\vec A \cdot \vec B)\vec C$, we may simplify the centripetal acceleration term to obtain $$\vec a=\vec a_0 + \vec \alpha\times \vec R-\omega ^2\vec R.$$

Now to get the out-of-plane component of the torque, $$\require{cancel}\vec \tau =\int\rho(\vec r)\vec R\times \vec a \ dV$$ $$=\left(\int\rho\vec R\ dV\right)\times \vec a_0 + \int\rho\vec R\times (\vec \alpha \times \vec R) \ dV - \int\rho\omega^2\cancelto{0}{(\vec R \times \vec R)}\ dV$$ $$=\left(\int\rho\vec R\ dV\right)\times \vec a_0 + \left(\int\rho R^2\ dV\right)\vec\alpha$$ $$=M\vec R_\text{CM}\times\vec a_0 + I\vec\alpha$$ or $$\tau - M\vec R_\text{CM}\times\vec a_0 = I\vec\alpha.$$

We see that $\vec\tau = I\vec \alpha$ holds if only if $- M\vec R_\text{CM}\times\vec a_0 = 0$, where $\vec R_\text{CM}$ is the position of the center of mass relative to the reference point you pick. $- M\vec R_\text{CM}\times\vec a_0$ is the fictitious torque term that you need to account for to be able to use $\vec\tau = I\vec\alpha$ around an arbitrary point.

If you pick $A$, this point has no downward acceleration, so $-M\vec R_\text{CM}\times\vec a_0 = 0.$ If you pick $B$ (the center of mass), $\vec R_\text{CM} = 0$, so $-M\vec R_\text{CM}\times\vec a_0 = 0.$ Finally, for point $C$, the downward (tangential) component of $\vec a_0$ is $-2\alpha R$, so $-M\vec R_\text{CM}\times\vec a_0$ is $2MR^2\alpha$, out-of-plane (counter-clockwise). We can write in the counter-clockwise direction $$\tau_C + 2MR^2\alpha = I_C \alpha.$$ Plugging in $\tau_C = \frac13MRg$ and $I_C=\frac32MR^2$, we find the correct result $\alpha = -\frac{2g}{3R}$.

Let's derive $\tau = I\alpha$ to see the conditions of its validity.

At any instant, we may describe the motion of a rigid body as a rotation about an arbitrary point, on top of a translation. Specifically, the velocity of any point on the rigid body may be expressed as $$\vec v = \vec v_0 +\vec \omega\times\vec r,$$ where $\vec v_0$ is the velocity of the center of rotation you have chosen, $\vec r$ is the relative position with respect to this point, and $\vec \omega$ is the angular velocity. To simplify things a bit, we can break down the position vector into two components: $\vec r = \vec R + \vec r_\parallel$, where $\vec R$ is perpendicular to $\vec \omega$ (and hence the axis of rotation), and $\vec r_\parallel$ is parallel to it. Then, we can write $$\vec v = \vec v_0 +\vec \omega\times\vec R.$$

Taking the derivative with respect to time, $$\vec a = \vec a_0 + \vec \alpha \times \vec R + \vec \omega\times (\vec v - \vec v_0)$$ $$=\vec a_0 + \vec \alpha\times \vec R + \vec \omega\times (\vec \omega \times \vec R).$$ The second term on the right hand side is the tangential acceleration, while the third term is the centripetal acceleration. Using the vector identity $\vec A \times (\vec B \times \vec C) = (\vec A \cdot \vec C)\vec B-(\vec A \cdot \vec B)\vec C$, we may simplify the centripetal acceleration term to obtain $$\vec a=\vec a_0 + \vec \alpha\times \vec R-\omega ^2\vec R.$$

Now to get the out-of-plane component of the torque, $$\require{cancel}\vec \tau =\int\rho(\vec r)\vec R\times \vec a \ dV$$ $$=\left(\int\rho(\vec r)\vec R\ dV\right)\times \vec a_0 + \int\rho(\vec r)\vec R\times (\vec \alpha \times \vec R) \ dV - \int\rho(\vec r)\omega^2\cancelto{0}{(\vec R \times \vec R)}\ dV$$ $$=\left(\int\rho(\vec r)\vec R\ dV\right)\times \vec a_0 + \left(\int\rho(\vec r)R^2\ dV\right)\vec\alpha$$ $$=M\vec R_\text{CM}\times\vec a_0 + I\vec\alpha.$$ We see that $\vec\tau = I\vec \alpha$ holds if only if $M\vec R_\text{CM}\times\vec a_0 = 0$, where $\vec R_\text{CM}$ is the position of the center of mass relative to the reference point you pick. If you pick $A$, this point has no downward acceleration, so $M\vec R_\text{CM}\times\vec a_0 = 0.$ If you pick $B$ (the center of mass), $\vec R_\text{CM} = 0$, so $M\vec R_\text{CM}\times\vec a_0 = 0.$

Finally, for point $C$, the downward (tangential) component of $\vec a_0$ is $2\alpha R$, so $M\vec R_\text{CM}\times\vec a_0$ is $-2MR^2\alpha$, out-of-plane (counter-clockwise). We can write in the counter-clockwise direction $$\tau_C = -2MR^2\alpha + I_C \alpha.$$ Plugging in $\tau_C = \frac13MRg$ and $I_C=\frac32MR^2$, we find the correct result $\alpha = -\frac{2g}{3R}$.

Let's derive $\tau = I\alpha$ to see the conditions of its validity.

At any instant, we may describe the motion of a rigid body as a rotation about an arbitrary point, on top of a translation. Specifically, the velocity of any point on the rigid body may be expressed as $$\vec v = \vec v_0 +\vec \omega\times\vec r,$$ where $\vec v_0$ is the velocity of the center of rotation you have chosen, $\vec r$ is the relative position with respect to this point, and $\vec \omega$ is the angular velocity. To simplify things a bit, we can break down the position vector into two components: $\vec r = \vec R + \vec r_\parallel$, where $\vec R$ is perpendicular to $\vec \omega$ (and hence the axis of rotation), and $\vec r_\parallel$ is parallel to it. Then, we can write $$\vec v = \vec v_0 +\vec \omega\times\vec R.$$

Taking the derivative with respect to time, $$\vec a = \vec a_0 + \vec \alpha \times \vec R + \vec \omega\times (\vec v - \vec v_0)$$ $$=\vec a_0 + \vec \alpha\times \vec R + \vec \omega\times (\vec \omega \times \vec R).$$ The second term on the right hand side is the tangential acceleration, while the third term is the centripetal acceleration. Using the vector identity $\vec A \times (\vec B \times \vec C) = (\vec A \cdot \vec C)\vec B-(\vec A \cdot \vec B)\vec C$, we may simplify the centripetal acceleration term to obtain $$\vec a=\vec a_0 + \vec \alpha\times \vec R-\omega ^2\vec R.$$

Now to get the out-of-plane component of the torque, $$\require{cancel}\vec \tau =\int\rho(\vec r)\vec R\times \vec a \ dV$$ $$=\left(\int\rho\vec R\ dV\right)\times \vec a_0 + \int\rho\vec R\times (\vec \alpha \times \vec R) \ dV - \int\rho\omega^2\cancelto{0}{(\vec R \times \vec R)}\ dV$$ $$=\left(\int\rho\vec R\ dV\right)\times \vec a_0 + \left(\int\rho R^2\ dV\right)\vec\alpha$$ $$=M\vec R_\text{CM}\times\vec a_0 + I\vec\alpha$$ or $$\tau - M\vec R_\text{CM}\times\vec a_0 = I\vec\alpha.$$

We see that $\vec\tau = I\vec \alpha$ holds if only if $- M\vec R_\text{CM}\times\vec a_0$, where $\vec R_\text{CM}$ is the position of the center of mass relative to the reference point you pick. $- M\vec R_\text{CM}\times\vec a_0$ is the fictitious torque term that you need to account for to be able to use $\vec\tau = I\vec\alpha$ around an arbitrary point.

If you pick $A$, this point has no downward acceleration, so $-M\vec R_\text{CM}\times\vec a_0 = 0.$ If you pick $B$ (the center of mass), $\vec R_\text{CM} = 0$, so $-M\vec R_\text{CM}\times\vec a_0 = 0.$ Finally, for point $C$, the downward (tangential) component of $\vec a_0$ is $2\alpha R$, so $-M\vec R_\text{CM}\times\vec a_0$ is $2MR^2\alpha$, out-of-plane (counter-clockwise). We can write in the counter-clockwise direction $$\tau_C + 2MR^2\alpha = I_C \alpha.$$ Plugging in $\tau_C = \frac13MRg$ and $I_C=\frac32MR^2$, we find the correct result $\alpha = -\frac{2g}{3R}$.