Sunday 8 September 2019

classical mechanics - Deriving $T = F r = Ialpha$ for a rigid body


For a single point mass : $\tau=F_{t}r=ma_tr=(m r^2)\alpha = I\alpha$


For multiple point masses bound together : $\sum \tau_i = (m_ir_i^2)\alpha = I\alpha$


But how do we go from that to $I\alpha = F_tr_{app} $ (for the multiple masses case), where $r_{app}$ is the point where force is applied? That is saying that the entire sum of torques of individual masses $\sum \tau_i$ is physically equivalent to (i.e. produces the same angular acceleration of the bound rigid object as) a single torque $F_tr_{app}$ applied to the point $r_{app}$. (Or putting it another way, that applying a single force $F_t r_{app}$ will result in many $F_{t\_i}$ forces on each mass element such that $\sum F_{t\_i}r_{i}$ will be equal to $I\alpha$.)


Every place I've looked for a derivation just jumps from $\sum \tau_i =I\alpha$, to $\tau=I\alpha$, as if they are the same thing.




No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...