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183_notes:examples:predicting_the_motion_of_system_subject_to_a_spring_interaction [2014/07/22 02:49] – caballero | 183_notes:examples:predicting_the_motion_of_system_subject_to_a_spring_interaction [2014/07/22 05:23] – pwirving | ||
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- | * The force of the spring is given by $F_{spring} = -k_s (|\vec{L}| - L_0) \hat{L}$ | + | * The force of the spring is given by $\vec{F}_{spring} = -k_s (|\vec{L}| - L_0) \hat{L}$ |
- | * The gravitational force is given by $F_{Earth}=-mg$ | + | * The gravitational force is given by $\vec{F}_{Earth}=-mg$ |
* The momentum of the system is given by $\vec{p_f}=\vec{p_i}+\vec{F_{net}} \Delta t$ | * The momentum of the system is given by $\vec{p_f}=\vec{p_i}+\vec{F_{net}} \Delta t$ | ||
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<WRAP todo> | <WRAP todo> | ||
- | Set-up force equations for both spring and force due to gravity | + | To solve this problem we must first set-up force equations for both spring and force due to gravity. To begin this process we must first determine the position vector ($\vec{L}$) of the mass and the length of the position vector ($|\vec{L}|$). |
$\vec{L}=\langle 0,0.1,0 \rangle - \langle 0,0,0 \rangle = \langle 0,0.1,0 \rangle m$ | $\vec{L}=\langle 0,0.1,0 \rangle - \langle 0,0,0 \rangle = \langle 0,0.1,0 \rangle m$ | ||
$\vec{|L|}=0.1$ | $\vec{|L|}=0.1$ | ||
+ | |||
+ | These can be used to compute the unit (direction) vector for the stretch ($\hat{s}$) (the difference between $|\vec{L}|$ and the relaxed distance \langle 0,.2,0 \rangle, which is in the same direction as the position vector. | ||
$\hat{L}=\langle 0,1,0 \rangle$ | $\hat{L}=\langle 0,1,0 \rangle$ | ||
+ | |||
+ | As expected it is acting only in the y direction. | ||
+ | |||
+ | You can now input the unit vector and rewrite the representation for the spring force equation so that it is acting solely in the y direction as indicated by the unit vector. | ||
$F_{spring} = -k_s(|\vec{L}|-L_0)\langle 0,1,0 \rangle = \langle 0, | $F_{spring} = -k_s(|\vec{L}|-L_0)\langle 0,1,0 \rangle = \langle 0, | ||
+ | |||
+ | We know that the force due to gravity acts solely in the y direction also so we can write the representation for the force equation to represent this: | ||
$F_{Earth} = \langle 0,-mg,0 \rangle$ | $F_{Earth} = \langle 0,-mg,0 \rangle$ | ||
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$F_{Earth}=-mg$ | $F_{Earth}=-mg$ | ||
+ | |||
+ | The addition of these two forces is the net force acting on the system: | ||
$F_{net,y} = F_{spring, | $F_{net,y} = F_{spring, | ||
- | For 0.1 seconds | + | We needed the net force in order to be able to calculate the momentum at different time intervals as the momentum of a system is dependent on the net force on that system: |
+ | |||
+ | $\vec{p_f}=\vec{p_i}+\vec{F_{net}} \Delta t$ | ||
+ | |||
+ | So we are now going to calculate the y position of the system for 0.1 seconds. | ||
+ | |||
+ | As found earlier: | ||
$\vec{|L|} = 0.1m$ | $\vec{|L|} = 0.1m$ | ||
+ | |||
+ | The stretch is equal to \vec{|L|} - the relaxed distance (0.2m). | ||
$s = 0.1m - 0.2m = -0.1m$ | $s = 0.1m - 0.2m = -0.1m$ | ||
+ | |||
+ | Compute the value of the force of the spring. | ||
$F_{spring_y} = -8N/ | $F_{spring_y} = -8N/ | ||
+ | |||
+ | Compute the value of the force due to gravity. | ||
$F_{Earth, | $F_{Earth, | ||
+ | |||
+ | Add the force of spring to force due to gravity to obtain the net force. | ||
$F_{net,y} = .212N$ | $F_{net,y} = .212N$ | ||
+ | |||
+ | Input this net force into the equation for momentum using 0.1s as the time to find the momentum at this instance. | ||
$p_{fy} = 0 + (.212N)(0.1s)\quad(Momentum\ Principle)\\$ | $p_{fy} = 0 + (.212N)(0.1s)\quad(Momentum\ Principle)\\$ | ||
+ | |||
+ | Compute momentum: | ||
$p_{fy} = 0.0212 kg * m/s$ | $p_{fy} = 0.0212 kg * m/s$ | ||
+ | |||
+ | In order to find the change in position of the system we must find the velocity of the system for the time interval of 0.1s using the momentum just computed. | ||
$v_{avg,y} \approx v_{fy}$ | $v_{avg,y} \approx v_{fy}$ |