book:chap3:3.3_newton_s_axioms_and_equations_of_motion_eom
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| book:chap3:3.3_newton_s_axioms_and_equations_of_motion_eom [2021/11/19 23:22] – created abril | book:chap3:3.3_newton_s_axioms_and_equations_of_motion_eom [2024/02/01 00:13] (current) – fixing typos in mine cart exampe jv | ||
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| Line 103: | Line 103: | ||
| </ | </ | ||
| - | <wrap lo>** Remark 3.4. ** | + | <wrap lo # |
| In general the time dependence of the forces can be decomposed into three contributions\\ | In general the time dependence of the forces can be decomposed into three contributions\\ | ||
| Line 128: | Line 128: | ||
| The resulting relation between the acceleration and the force is called equation of motion of the particle. | The resulting relation between the acceleration and the force is called equation of motion of the particle. | ||
| - | <WRAP box round> | + | <WRAP box round #EOM> |
| Newton' | Newton' | ||
| Line 327: | Line 327: | ||
| stating that the velocity stays constant when the pushing force and the friction force balance. | stating that the velocity stays constant when the pushing force and the friction force balance. | ||
| - | <WRAP box round> | + | <WRAP box round #minecart> |
| The motion of the mine cart is one-dimensional along its track such that the position, $q$, velocity, $x$, and forces are one-dimensional, | The motion of the mine cart is one-dimensional along its track such that the position, $q$, velocity, $x$, and forces are one-dimensional, | ||
| Once the mine cart is moving it experiences a friction force $\mathbf F_f = -\gamma \, \mathbf v$, | Once the mine cart is moving it experiences a friction force $\mathbf F_f = -\gamma \, \mathbf v$, | ||
| Line 336: | Line 336: | ||
| \end{align*} | \end{align*} | ||
| The mine cart travels with constant velocity $\dot v = 0$, when the attacking forces balance, | The mine cart travels with constant velocity $\dot v = 0$, when the attacking forces balance, | ||
| - | i.e., for $v_c = F_M / m\, \gamma$. | + | i.e., for $v_c = F_M / \gamma$. |
| For a different initial velocity, $v(t_0) = v_0$, one finds an exponential approach to the asymptotic velocity, | For a different initial velocity, $v(t_0) = v_0$, one finds an exponential approach to the asymptotic velocity, | ||
| \begin{align*} | \begin{align*} | ||
| - | v (t) = v_c + \bigl( v_0 - v_c \bigr) \; \mathrm{e}^{ - \gamma \, (t-t_0) } | + | v (t) = v_c + \bigl( v_0 - v_c \bigr) \; \mathrm{e}^{ - \gamma \, (t-t_0) |
| \end{align*} | \end{align*} | ||
| After all, | After all, | ||
| Line 345: | Line 345: | ||
| and | and | ||
| \begin{align*} | \begin{align*} | ||
| - | \dot v (t) | + | |
| - | &= \bigl( v_0 - v_c \bigr) \; (-\gamma) \, \mathrm{e}^{ - \gamma \, (t-t_0) } | + | &= \bigl( v_0 - v_c \bigr) \; (-\gamma) \, \mathrm{e}^{ - \gamma \, (t-t_0)/m } |
| \\ | \\ | ||
| - | & | + | &= -\gamma \; \bigl( v(t) - v_c \bigr) = -\gamma \; v(t) + F_M |
| \end{align*} | \end{align*} | ||
| </ | </ | ||
| Line 376: | Line 376: | ||
| and $c_d \simeq 0.5$ the drag coefficient. | and $c_d \simeq 0.5$ the drag coefficient. | ||
| - | - The drag coefficient is a dimensionless number that depends on the shape of the object that experiences drag. For the rest the expression for the drag force follows from dimensional analysis. Verify this claim.\\ | + | - The drag coefficient is a dimensionless number that depends on the shape of the object that experiences drag. For the rest the expression for the drag force follows from dimensional analysis. Verify this claim. |
| - | - A slightly more informed derivation of $\mathbf F_d$ introduces also the diameter $D$ of the golf ball and states that drag arises because the ball has to push air out of its way. When moving it has to push air out of the way at a rate $A\, u$. The air was at rest initially and must move roughly with a velocity $u$ to get out of the way. Subsequently, | + | - A slightly more informed derivation of $\mathbf F_d$ introduces also the diameter $D$ of the golf ball and states that drag arises because the ball has to push air out of its way. When moving it has to push air out of the way at a rate $A\, u$. The air was at rest initially and must move roughly with a velocity $u$ to get out of the way. Subsequently, |
| - What is the terminal velocity of a golf ball that is falling out of the pocket of a careless hang glider? | - What is the terminal velocity of a golf ball that is falling out of the pocket of a careless hang glider? | ||
| - Use dimensional analysis to estimate the distance after which the ball acquires its terminal velocity, and how long it takes to reach the velocity. | - Use dimensional analysis to estimate the distance after which the ball acquires its terminal velocity, and how long it takes to reach the velocity. | ||
| Line 430: | Line 430: | ||
| - Describe the motion of the cow when there is no friction. In the beginning the cow is at rest. | - Describe the motion of the cow when there is no friction. In the beginning the cow is at rest. | ||
| - What changes when there is friction with a friction coefficient of $\gamma = 0.2$, i.e. a horizontal friction force of magnitude $-\gamma m g$ acting on the cow. | - What changes when there is friction with a friction coefficient of $\gamma = 0.2$, i.e. a horizontal friction force of magnitude $-\gamma m g$ acting on the cow. | ||
| + | - A more informed choice for the friction $\gamma N$ takes into account that the normal force $N$ on the surface is reduced to a smaller value than $mg$ when the cow is pulled. How will that change the discussion? | ||
| - Is the assumption realistic that the force remains constant and will always act in the same direction? What might go wrong? | - Is the assumption realistic that the force remains constant and will always act in the same direction? What might go wrong? | ||
book/chap3/3.3_newton_s_axioms_and_equations_of_motion_eom.1637360555.txt.gz · Last modified: 2021/11/19 23:22 by abril