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If a body moves only because of the influence of a torque, then it will rotate about the center of gravity.

There is no location for torques, only directions. You you take the equations of motion as seen here (https://physics.stackexchange.com/a/80449/392) you will see that the location of the torque does not enter into the equations. Only the location of the forces.

As a result the acceleration of the center of mass is zero, and only angular velocity will exist. The body will rotate about its center of mass.

Note that these two statements are equivalent:

1. A pure force thorugh the center of gravity (with no net torque) will purely translate a rigid body (any point on the body).
2. A pure torque any point on the body (with no net force) will purely rotate a rigid body about its center of gravity

Consider a motionless rigid body with a pure instantenous torque $\vec{\tau}$ applied on it. The motion of any point A not on the center of gravity is

$$ 0 = m \vec{a}_A - m \vec{c}\times \vec{\alpha} \\ \vec{\tau} = I_c \vec{\alpha} - m \vec{c} \times \vec{c} \times \vec{\alpha} + m \vec{c} \times \vec{a}_A $$

where $\vec{c}$ the position vector of the center of gravity relative to point A. The soltution to the above is

$$ \vec{a}_A = \vec{c} \times \vec{\alpha} \\ \vec{\tau} = I_c \vec{\alpha}- m \vec{c} \times \vec{c} \times \vec{\alpha}+ m \vec{c} \times \vec{c} \times \vec{\alpha} = I_c \vec{\alpha} $$

$$ \vec{\alpha} = I_c^{-1} \vec{\tau} \\ \vec{a}_A = \vec{c} \times I_c^{-1} \vec{\tau} $$

From the above is it obvious that the only point A not moving is at $\vec{c}=0$ and all points parallel to $ \vec{\alpha}$ through the center of mass.

If a body moves only because of the influence of a torque, then it will rotate about the center of mass.

There is no location for torques, only directions. You you take the equations of motion as seen here (https://physics.stackexchange.com/a/80449/392) you will see that the location of the torque does not enter into the equations. Only the location of the forces.

As a result the acceleration of the center of mass is zero, and only angular velocity will exist. The body will rotate about its center of mass.

Note that these two statements are equivalent:

1. A pure force thorugh the center of mass (with no net torque about the center of mass) will purely translate a rigid body (any point on the body).
2. A pure torque any point on the body (with no net force) will purely rotate a rigid body about its center of mass.

Consider a motionless rigid body with a pure instantenous torque $\vec{\tau}$ applied on it. The motion of any point A not on the center of mass is

$$ 0 = m \vec{a}_A - m \vec{c}\times \vec{\alpha} \\ \vec{\tau} = I_c \vec{\alpha} - m \vec{c} \times \vec{c} \times \vec{\alpha} + m \vec{c} \times \vec{a}_A $$

where $\vec{c}$ the position vector of the center of gravity relative to point A. The soltution to the above is

$$ \vec{a}_A = \vec{c} \times \vec{\alpha} \\ \vec{\tau} = I_c \vec{\alpha}- m \vec{c} \times \vec{c} \times \vec{\alpha}+ m \vec{c} \times \vec{c} \times \vec{\alpha} = I_c \vec{\alpha} $$

$$ \vec{\alpha} = I_c^{-1} \vec{\tau} \\ \vec{a}_A = \vec{c} \times I_c^{-1} \vec{\tau} $$

From the above is it obvious that the only point A not moving is at $\vec{c}=0$ and all points parallel to $ \vec{\alpha}$ through the center of mass.

If a body moves only because of the influence of a torque, then it will rotate about the center of gravity.

There is no location for torques, only directions. You you take the equations of motion as seen here (http://physics.stackexchange.com/a/80449/392) you will see that the location of the torque does not enter into the equations. Only the location of the forces.

As a result the acceleration of the center of mass is zero, and only angular velocity will exist. The body will rotate about its center of mass.

Note that these two statements are equivalent:

1. A pure force thorugh the center of gravity (with no net torque) will purely translate a rigid body (any point on the body).
2. A pure torque any point on the body (with no net force) will purely rotate a rigid body about its center of gravity

Consider a motionless rigid body with a pure instantenous torque $\vec{\tau}$ applied on it. The motion of any point A not on the center of gravity is

$$ 0 = m \vec{a}_A - m \vec{c}\times \vec{\alpha} \\ \vec{\tau} = I_c \vec{\alpha} - m \vec{c} \times \vec{c} \times \vec{\alpha} + m \vec{c} \times \vec{a}_A $$

where $\vec{c}$ the position vector of the center of gravity relative to point A. The soltution to the above is

$$ \vec{a}_A = \vec{c} \times \vec{\alpha} \\ \vec{\tau} = I_c \vec{\alpha}- m \vec{c} \times \vec{c} \times \vec{\alpha}+ m \vec{c} \times \vec{c} \times \vec{\alpha} = I_c \vec{\alpha} $$

$$ \vec{\alpha} = I_c^{-1} \vec{\tau} \\ \vec{a}_A = \vec{c} \times I_c^{-1} \vec{\tau} $$

From the above is it obvious that the only point A not moving is at $\vec{c}=0$ and all points parallel to $ \vec{\alpha}$ through the center of mass.

If a body moves only because of the influence of a torque, then it will rotate about the center of gravity.

There is no location for torques, only directions. You you take the equations of motion as seen here (https://physics.stackexchange.com/a/80449/392) you will see that the location of the torque does not enter into the equations. Only the location of the forces.

As a result the acceleration of the center of mass is zero, and only angular velocity will exist. The body will rotate about its center of mass.

Note that these two statements are equivalent:

1. A pure force thorugh the center of gravity (with no net torque) will purely translate a rigid body (any point on the body).
2. A pure torque any point on the body (with no net force) will purely rotate a rigid body about its center of gravity

Consider a motionless rigid body with a pure instantenous torque $\vec{\tau}$ applied on it. The motion of any point A not on the center of gravity is

$$ 0 = m \vec{a}_A - m \vec{c}\times \vec{\alpha} \\ \vec{\tau} = I_c \vec{\alpha} - m \vec{c} \times \vec{c} \times \vec{\alpha} + m \vec{c} \times \vec{a}_A $$

where $\vec{c}$ the position vector of the center of gravity relative to point A. The soltution to the above is

$$ \vec{a}_A = \vec{c} \times \vec{\alpha} \\ \vec{\tau} = I_c \vec{\alpha}- m \vec{c} \times \vec{c} \times \vec{\alpha}+ m \vec{c} \times \vec{c} \times \vec{\alpha} = I_c \vec{\alpha} $$

$$ \vec{\alpha} = I_c^{-1} \vec{\tau} \\ \vec{a}_A = \vec{c} \times I_c^{-1} \vec{\tau} $$

From the above is it obvious that the only point A not moving is at $\vec{c}=0$ and all points parallel to $ \vec{\alpha}$ through the center of mass.

If a body moves only because of the influence of a torque, then it will rotate about the center of gravity.

There is no location for torques, only directions. You you take the equations of motion as seen here (http://physics.stackexchange.com/a/80449/392) you will see that the location of the torque does not enter into the equations. Only the location of the forces.

As a result the acceleration of the center of mass is zero, and only angular velocity will exist. The body will rotate about its center of mass.

Note that these two statements are equivalent:

1. A pure force thorugh the center of gravity (with no net torque) will purely translate a rigid body (any point on the body).
2. A pure torque any point on the body (with no net force) will purely rotate a rigid body about its center of gravity

Consider a motionless rigid body with a pure instantenous torque $\vec{\tau}$ applied on it. The motion of any point A not on the center of gravity is

$$ 0 = m \vec{a}_A - m \vec{c}\times \vec{\alpha} \\ \vec{\tau} = I_c \vec{\alpha} - m \vec{c} \times \vec{c} \times \vec{\alpha} + m \vec{c} \times \vec{a}_A $$

where $\vec{c}$ the position vector of the center of gravity relative to point A. The soltution to the above is

$$ \vec{a}_A = \vec{c} \times \vec{\alpha} \\ \vec{\tau} = I_c \vec{\alpha}- m \vec{c} \times \vec{c} \times \vec{\alpha}+ m \vec{c} \times \vec{c} \times \vec{\alpha} = I_c \vec{\alpha} $$

$$ \vec{\alpha} = I_c^{-1} \vec{\tau} \\ \vec{a}_A = \vec{c} \times I_c^{-1} \vec{\tau} $$

From the above is it obvious that the only point A not moving is at $\vec{c}=0$, or at the center of gravity.

If a body moves only because of the influence of a torque, then it will rotate about the center of gravity.

There is no location for torques, only directions. You you take the equations of motion as seen here (http://physics.stackexchange.com/a/80449/392) you will see that the location of the torque does not enter into the equations. Only the location of the forces.

As a result the acceleration of the center of mass is zero, and only angular velocity will exist. The body will rotate about its center of mass.

Note that these two statements are equivalent:

1. A pure force thorugh the center of gravity (with no net torque) will purely translate a rigid body (any point on the body).
2. A pure torque any point on the body (with no net force) will purely rotate a rigid body about its center of gravity

Consider a motionless rigid body with a pure instantenous torque $\vec{\tau}$ applied on it. The motion of any point A not on the center of gravity is

$$ 0 = m \vec{a}_A - m \vec{c}\times \vec{\alpha} \\ \vec{\tau} = I_c \vec{\alpha} - m \vec{c} \times \vec{c} \times \vec{\alpha} + m \vec{c} \times \vec{a}_A $$

where $\vec{c}$ the position vector of the center of gravity relative to point A. The soltution to the above is

$$ \vec{a}_A = \vec{c} \times \vec{\alpha} \\ \vec{\tau} = I_c \vec{\alpha}- m \vec{c} \times \vec{c} \times \vec{\alpha}+ m \vec{c} \times \vec{c} \times \vec{\alpha} = I_c \vec{\alpha} $$

$$ \vec{\alpha} = I_c^{-1} \vec{\tau} \\ \vec{a}_A = \vec{c} \times I_c^{-1} \vec{\tau} $$

From the above is it obvious that the only point A not moving is at $\vec{c}=0$ and all points parallel to $ \vec{\alpha}$ through the center of mass.