- #1

Leo Liu

- 353

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Question: If we place the frame of reference on an accelerating point, does the total rotational momentum still remain the same?

I attempted to solve this question by manipulating the equations as shown below.

$$\text{Define that }\vec r_i=\vec R+\vec r_i'\text{, where r is the position vector starting from a noninertial frame, R is the position of CM in that frame, r' is the distance between the particle to the CM}$$

$$\text{According to Konig theorem: }\vec\tau=\vec R\times M\ddot{\vec R}+\sum_i \vec r_i'\times m_i\ddot{\vec r}_i'$$

$$\text{Since the position vector does not depend on the accelerating frame, it is a constant, and we obtain:}$$

$$\vec\tau=\vec R\times M\ddot{\vec R}+\vec{Constant}$$

It seems that as long as the equation above equals to 0, the rotational momentum of this system is conserved.

Does this imply that the frame doesn't have to be noninertial to make the rotational momentum conserved?

Thank you and merry early Christmas!

I attempted to solve this question by manipulating the equations as shown below.

$$\text{Define that }\vec r_i=\vec R+\vec r_i'\text{, where r is the position vector starting from a noninertial frame, R is the position of CM in that frame, r' is the distance between the particle to the CM}$$

$$\text{According to Konig theorem: }\vec\tau=\vec R\times M\ddot{\vec R}+\sum_i \vec r_i'\times m_i\ddot{\vec r}_i'$$

$$\text{Since the position vector does not depend on the accelerating frame, it is a constant, and we obtain:}$$

$$\vec\tau=\vec R\times M\ddot{\vec R}+\vec{Constant}$$

It seems that as long as the equation above equals to 0, the rotational momentum of this system is conserved.

Does this imply that the frame doesn't have to be noninertial to make the rotational momentum conserved?

Thank you and merry early Christmas!

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