Rotational mechanics

by Urmi Roy
Tags: mechanics, rotational
 P: 52 it would be better if you start a new thread for that
P: 288
 Quote by BobbyBear Thank you sganesh for clearing my doubt about 'torque of a solid'. But still, if you were to have a solid suspended in space, and apply only one force say at one end of the solid (not through its COM), would it rotate at all? (I'm assuming there are no physical constraints upon the solid at all - no physical pivot or anything). And about which point would it rotate? I can't seem to imagine why, if it should rotate, it should do so about any particular point.
I too had that doubt a year before. I thought rotation is possible only in the presence of a pivot constraint or two opposite forces whose lines of actions are separated by a distance. But the body has to rotate if the line of action of the force doesn't pass through the COM (even if its just a single force). Why? Check whether conservation of angular momentum is conserved if the body doesn't rotate but just translates as a whole when a force acts on its tip. (Though i get a doubt. Why should angular momentum be conserved? Going to read about it now.)

Regarding the axis of rotation, draw a line from the COM to the point of application of the force; the perpendicular to this line, passing through the COM will be the axis.
P: 162
 Quote by sganesh88 Regarding the axis of rotation, draw a line from the COM to the point of application of the force; the perpendicular to this line, passing through the COM will be the axis.
Hmm, if that is the axis of rotation, then there is a net torque with respect to that axis, and as you said, the 'torque' is always with respect to the COM . . .
D H said that in an inertial frame, we have: $d\boldsymbol L/dt = \boldsymbol {\tau}_{\text{ext}}$, so the solid must acquire an angular momentum with respect to an inertial frame? And also as D H said, this does not mean that it must acquire an angular velocity, as the inertia tensor is not constant in an inertial frame - though again, if the body does not rotate, then the inertia tensor would be constant, so it must rotate . . . . Hence if the body does not rotate at all, it would mean that angular momentum is conserved, no? So that can't be true if there's an external force acting on the solid with a certain eccentricity with respect to the centre of mass. Anyhow, I'm a bit dodgy with all of this, I think I need to follow some text book that develops the general equations and explains all this with certain depth xD
 P: 288 Torque needn't always be w.r.t COM but in this case of a solid suspended in space, this is preferable. And i know nothing about this inertia tensor thing. Yet to study about it.
P: 162
 Quote by sganesh88 Torque needn't always be w.r.t COM but in this case of a solid suspended in space, this is preferable. And i know nothing about this inertia tensor thing. Yet to study about it.
*flails* that's just the point, you can consider the torque produced by a force with respect to any point, but the motion of the solid cannot depend on which point you choose to consider to apply Newton's law of rotation . . . I think I need to study about it too *wink* if you come across any good books let me know their names xD
 Mentor P: 15,065 BobbyBear: To answer your first question regarding the decoupling of translational and rotational equations of motion, consider a system comprising a fixed number of constant mass particles. Such a system will have a constant mass and the individual components won't change (e.g., there are no chemical interactions that change the number of particles). Denote the mass and position with respect to some inertial frame of the ith particle as mi and xi. The mass, center of mass of the system, and system center of mass velocity are \aligned m_{\text{tot}} &\equiv \sum_i m_i \\ m_{\text{tot}} \boldsymbol x_{\text{CM}} &\equiv \sum_i m_i \boldsymbol x_i \\ m_{\text{tot}} \boldsymbol v_{\text{CM}} &\equiv \sum_i m_i \dot{\boldsymbol x}_i \endaligned The total linear momentum is the sum of the linear momentum for each particle: $$\boldsymbol p_{\text{tot}} = \sum_i m_i \dot{\boldsymbol x}_i$$ Note that this is the total mass times the system center of mass velocity. Differentiating with respect to time, $$\dot{\boldsymbol p}_{\text{tot}} = \sum_i m_i \ddot{\boldsymbol x}_i$$ Assume each particle obeys Newton's laws of motion: $$m_i\ddot{\boldsymbol r}_i = \boldsymbol F_{\text{tot},i}$$ External and internal forces will act on each particle. The external forces come from outside the system; the internal forces are interactions between pairs of particles. The total force on the ith particle is the sum of the external force on that particle plus the interactions with the other particles in the system: $$\boldsymbol F_{\text{tot},i} = \boldsymbol F_{\text{ext},i} + \sum_{j\ne i} \boldsymbol F_{\text{int},ij}$$ Thus \aligned \dot{\boldsymbol p}_{\text{tot}} &= \sum \boldsymbol F_{\text{tot},i} \\ &= \sum_i \left(\boldsymbol F_{\text{ext},i} + \sum_{j\ne i} \boldsymbol F_{\text{int},ij} \right) \\ &= \boldsymbol F_{\text{ext},\text{tot}} + \sum_{\text{pairs\,}i,j} (\boldsymbol F_{\text{int},ij} + \boldsymbol F_{\text{int},ji} By the weak form of Newton's third law, interactions between particles are equal but opposite: $\boldsymbol F_{\text{int},ji} = - \boldsymbol F_{\text{int},ij}$ and thus the change in linear momentum of the system as a whole is equal to the sum of the external forces acting on elements of the system: $$\dot{\boldsymbol p}_{\text{tot}} = \boldsymbol F_{\text{ext},\text{tot}}$$ Using the definition of the system center of mass, $$m_{\text{tot}} \ddot{\boldsymbol x}_{\text{CM}} = \boldsymbol F_{\text{ext},\text{tot}}$$ The system as a whole obeys Newton's second law so long the particles obey the Newton's second and third laws. What about the angular momentum? The total angular momentum of the system is the sum of that of the individual particles: $$\boldsymbol L_{\text{tot}} = \sum_i \boldsymbol L_i = \sum_i m_i \boldsymbol x_i \times \dot{\boldsymbol x}_i$$ Differentiating with respect to time, \aligned \dot{\boldsymbol L}_{\text{tot}} &= \sum_i m_i \boldsymbol x_i \times \ddot{\boldsymbol x}_i \\ &= \sum_i \boldsymbol x_i \boldsymbol F_{\text{ext},\text{tot}} \\ &= \sum_i \boldsymbol x_i\times \left(\boldsymbol F_{\text{ext},i} + \sum_{j\ne i} \boldsymbol F_{\text{int},ij} \right) \\ &= \left(\sum_i \boldsymbol x_i\times \left(\boldsymbol F_{\text{ext},i}\right) + \left(\sum_{\text{pairs\,}i,j} (\boldsymbol x_i\times \boldsymbol F_{\text{int},ij} + \boldsymbol x_j\times \boldsymbol F_{\text{int},ji}\right) \\ &= \left(\sum_i \boldsymbol x_i\times \left(\boldsymbol F_{\text{ext},i}\right) + \left(\sum_{\text{pairs\,}i,j} \boldsymbol x_i \times (\boldsymbol F_{\text{int},ij} +\boldsymbol F_{\text{int},ji}) + (\boldsymbol x_j-\boldsymbol x_i)\times \boldsymbol F_{\text{int},ji}\right) \endaligned The terms $\boldsymbol F_{\text{int},ij} +\boldsymbol F_{\text{int},ji}$ vanish per the weak form of Newton's third law. The final set of terms, $(\boldsymbol x_j-\boldsymbol x_i)\times \boldsymbol F_{\text{int},ji}$, vanish per the strong form of Newton's third law. Thus the total angular momentum depends only on the external forces acting on individual particles: $$\dot{\boldsymbol L}_{\text{tot}} = \sum_i \boldsymbol x_i\times \left(\boldsymbol F_{\text{ext},i}$$ The linear and angular momenta for a rigid body can be expressed as \aligned \boldsymbol p &= m \dot{\boldsymbol x}_{\text{CM}} \\ \boldsymbol L &= m \boldsymbol x_{\text{CM}}\times \dot{\boldsymbol x}_{\text{CM}} + \mathbf I \boldsymbol {\omega} \endaligned When expressed in this form, the translational and rotational equations of motion decouple. The translational equations of motion take on the form $$m \ddot{\boldsymbol x}_{\text{CM}} = \boldsymbol F_{\text{ext,tot}}$$ The rotational equations of motion were already described in post #102.
 P: 162 THANK YOU D H, that was very helpful xD

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