- #1
Soren4
- 128
- 2
For the rotation of a rigid body about a fixed axis [itex]z[/itex] the following holds.
$$\vec{\tau_z}=\frac{d\vec{L_z}}{dt}= I_z \vec{\alpha} \tag{1}$$
Where [itex]\vec{\tau_z}[/itex] is the component parallel to the axis [itex]z[/itex] of a torque [itex]\vec{\tau}[/itex] exerted in the body; [itex]\vec{L_z}[/itex] is the component parallel to the rotation axis [itex]z[/itex] of the angular momentum and [itex] \vec{\alpha}[/itex] is the angular acceleration.
Can I interpret (1) as follows?
If there is an angular acceleration [itex] \vec{\alpha}[/itex] there must be an exerted torque [itex]\vec{\tau}[/itex] with non zero component [itex]\vec{\tau_z}[/itex] along the axis of rotation [itex]z[/itex]: this last mentioned component [itex]\vec{\tau_z}[/itex] is the only one responsible for the present angular acceleration [itex] \vec{\alpha}[/itex]. If the applied torque [itex]\vec{\tau}[/itex] has no component along the axis of rotation [itex]z[/itex] (i.e. it is completely perpendicular to it) there is no way that an angular acceleration [itex] \vec{\alpha}[/itex] appears.
$$\vec{\tau_z}=\frac{d\vec{L_z}}{dt}= I_z \vec{\alpha} \tag{1}$$
Where [itex]\vec{\tau_z}[/itex] is the component parallel to the axis [itex]z[/itex] of a torque [itex]\vec{\tau}[/itex] exerted in the body; [itex]\vec{L_z}[/itex] is the component parallel to the rotation axis [itex]z[/itex] of the angular momentum and [itex] \vec{\alpha}[/itex] is the angular acceleration.
Can I interpret (1) as follows?
If there is an angular acceleration [itex] \vec{\alpha}[/itex] there must be an exerted torque [itex]\vec{\tau}[/itex] with non zero component [itex]\vec{\tau_z}[/itex] along the axis of rotation [itex]z[/itex]: this last mentioned component [itex]\vec{\tau_z}[/itex] is the only one responsible for the present angular acceleration [itex] \vec{\alpha}[/itex]. If the applied torque [itex]\vec{\tau}[/itex] has no component along the axis of rotation [itex]z[/itex] (i.e. it is completely perpendicular to it) there is no way that an angular acceleration [itex] \vec{\alpha}[/itex] appears.