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How Does Newton's Second Law Explain Force and Motion?
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[QUOTE="Greg Bernhardt, post: 4804907, member: 1"] [SIZE="4"][U][B]Definition/Summary[/B][/U][/SIZE] Net force = rate of change of momentum (= mass times acceleration of the centre of mass, if the mass is constant) Net impulse = change of momentum These are vector equations, so they apply to each direction individually The rotational versions are: net torque = rate of change of angular momentum net impulsive torque = change of angular momentum [SIZE="4"][U][B]Equations[/B][/U][/SIZE] Newton's second law: [tex]\sum_{\text{all}}\mathbf{F}\,=\,\frac{d(m\mathbf{v}_{c.o.m.})}{dt}[/tex] If m is constant: [tex]\sum_{\text{all}}\mathbf{F}\,=\,m\frac{d(\mathbf{v}_{c.o.m.})}{dt}\,=\,m \mathbf{a}_{c.o.m.}[/tex] Torque version: [tex]\sum_{\text{all}}\mathbf{\tau}\,=\,\frac{d\mathbf{\mathcal{L}}}{dt}[/tex] Torque version for rigid body: [tex]\sum_{\text{all}}\mathbf{\tau}\,=\,I\mathbf{\alpha}[/tex] [SIZE="4"][U][B]Extended explanation[/B][/U][/SIZE] [B]Net force = mass times acceleration:[/B] This [I]only[/I] applies when the mass is constant. [SIZE="1"]For example, it does [I]not[/I] apply to a rocket, whose mass is continually reduced by the burning of fuel.[/SIZE] "Acceleration" means [I]the acceleration [B]of the centre of mass[/B][/I]: this [I]even[/I] applies when the body is [I]rotating[/I]: the sum of the (linear) forces equals the mass times the (linear) acceleration of the c.o.m.: [CENTER][tex]\sum\mathbf{F}\ =\ m\,\mathbf{a}_{c.o.m.}[/tex][/CENTER] [B]Standard (non-impulsive) version:[/B] [INDENT][COLOR="Red"]Newton's second law[/COLOR] states that the total vector sum of the forces on a body is equal to the rate of change of its momentum. This formulation is valid when mass changes, and is even valid at relativistic speeds.[/INDENT] [CENTER][tex]\sum\mathbf{F}\ =\ \frac{d(m\mathbf{v})}{dt}[/tex][/CENTER] [B]Component version:[/B] Force and momentum are vectors, so Newton's second law is a vector equation. That means that it applies to the components in any direction, and so the law can also be written: [INDENT]In any direction, the total sum of the [I]components[/I] of the forces on a body in that direction is equal to the rate of change of the [I]component[/I] of its momentum in that direction.[/INDENT] [CENTER][tex]\sum\mathbf{F}_k\ =\ \frac{d(m\mathbf{v}_k)}{dt}[/tex][/CENTER] [B]Torque:[/B] The vector nature of the equation also means that its cross-product with any fixed vector gives another vector equation: [INDENT]Net [COLOR="Red"]torque[/COLOR] on a point particle = rate of change of angular momentum:[/INDENT] [CENTER][tex]\mathbf{\tau}_{total}\equiv\mathbf{r} \times \sum\mathbf{F}=\frac{d}{dt}(\mathbf{r} \times \mathbf{p})[/tex][/CENTER] Combining this with the strong version of [COLOR="Red"]Newton's third law[/COLOR] (action is equal and opposite to reaction, [I]and[/I] in the same line) gives: [INDENT]Net torque on a rigid body = rate of change of angular momentum:[/INDENT] [CENTER][tex]\mathbf{\tau}_{total}\equiv\sum\mathbf{r} \times \mathbf{F}=\frac{d}{dt}\left(\int\mathbf{r} \times d\mathbf{p}\right)= \frac{d \mathbf{\mathcal{L}}}{dt}[/tex][/CENTER] This is also a vector equation, of course, and so we may take one component at a time, to give the scalar equation: [INDENT]Net torque about any axis = rate of change of angular momentum about that axis.[/INDENT] [B]Impulsive version:[/B] [COLOR="red"]Impulse[/COLOR] is the integral of force times time. [SIZE="1"]By comparison, [COLOR="Red"]work done[/COLOR] is the integral of force times distance.[/SIZE] Sometimes (for example, when a bat hits a ball), the force changes quickly, and it is difficult to measure it, but it is easy to measure the overall effect of the force. The impulse is that overall effect. [INDENT][COLOR="Red"]Newton's second law[/COLOR] states that the total vector sum of the impulses on a body is equal to the change of its momentum.[/INDENT] [B]Collisions, and conservation of momentum:[/B] A collision or explosion is impulsive, and there are no [I]external[/I] impulsive forces in a collision or explosion (in particular, the sum of the external forces is zero :wink:), so the impulsive version of the law states: [INDENT]Total momentum immediately before = total momentum immediately after:[/INDENT] [CENTER][tex]\sum m_i\,u_i\,=\,\sum m_i\,v_i[/tex][/CENTER] In other words: [COLOR="Red"]conservation of momentum[/COLOR] applies to all collisions or explosions. [SIZE="1"]By comparison, conservation of energy applies only to [COLOR="red"]elastic[/COLOR] collisions. Most collisions are [I]not[/I] elastic.[/SIZE] [B]Static problems:[/B] When nothing moves, the momentum is constant, and so Newton's second law states: [CENTER]The total sum of the forces on a body in equilibrium is zero.[/CENTER] [SIZE="1"]This applies [I]separately[/I] to each body in a system, or to the [I]external[/I] forces on any combination of bodies.[/SIZE] When there are only three forces, a [COLOR="Red"]vector triangle[/COLOR] may be used. [B]Sliding problems (normal components):[/B] When one body slides against another along a common [I]flat[/I] surface, the acceleration [I]normal[/I] (perpendicular) to that surface is zero. If the surface is [I]curved[/I], the acceleration normal to the surface is [I]not[/I] zero, but is [I]centripetal[/I] towards the instantaneous centre of curvature of the surface (note: this is a matter of [I]geometry[/I], not physics :wink:). Accordingly, Newton's second law for normal components states: [INDENT]The total sum of the components of the [I]normal[/I] forces on a [I]sliding[/I] body on a [I]flat[/I] surface is zero, and on a [I]curved[/I] surface equals the mass times the [COLOR="Red"]centripetal acceleration[/COLOR].[/INDENT] [B]String problems:[/B] When a number of bodies are connected by a [COLOR="Red"]string[/COLOR] (an inextensible "massless" cable), in a straight line, all have the same (vector) acceleration, and so the tension in the string can be ignored (as an [I]internal[/I] force), and [B]the bodies can be treated together as a single body[/B] with the same total mass and acceleration and subject only to the [I]external[/I] forces. When the string goes over a "massless" pulley, so that the string is in two parts in different directions, and the tensions either side of the pulley are the same, the bodies still all have the same [I]scalar[/I] acceleration, and the bodies can be treated together as a single body with the same total mass and acceleration and subject only to the [I]components of the external forces[/I] along the string. When the string goes over a [I]massive pulley[/I], so that the string is in two parts in different directions, and the tensions either side of the pulley are [I]different[/I], the previous paragraph applies except that the pulley must be treated as a body with the same acceleration and with an [COLOR="Red"]effective mass[/COLOR] equal to the moment of inertia divided by the radius squared: [itex]m_{eff}\ =\ I/r^2[/itex] (for a [I]uniform[/i] cylindrical pulley, that is [I]half[/I] the actual mass: [itex]m_{eff}\ =\ m/2[/itex]). * This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks! [/QUOTE]
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