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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 [2015/09/29 16:44] (current) – obsniukm | ||
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| ===== Example: Predicting the motion of a system that is subject to a spring interaction/ | ===== Example: Predicting the motion of a system that is subject to a spring interaction/ | ||
| - | A spring has a relaxed length of (0.2m) and it has a spring constant of 8 N/m. Attached to the top of the spring is a block of mass (.06)kg. A force is exerted on the block to compress the spring to a total length of (0.2m). Predict the y position for the block after 0.1 second and 0.2 seconds. | + | A spring has a relaxed length of (0.2m) and it has a spring constant of 8 N/m. Attached to the top of the spring is a block of mass (.06)kg. A force is exerted on the block to compress the spring to a total length of (0.1m). Predict the y position for the block after 0.1 second and 0.2 seconds. |
| === Facts ==== | === Facts ==== | ||
| Line 26: | Line 26: | ||
| {{183_notes: | {{183_notes: | ||
| - | * 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 $\vec{p_f}=\vec{p_i}+\vec{F_{net}} \Delta t$ | ||
| + | * Momentum equation = $m(\vec{v}) = \vec{p}$ | ||
| + | * Postion update = $\vec{r_f} = \vec{r_i} + \vec{v}_{avg} \Delta t$ | ||
| | | ||
| ==== Solution ==== | ==== Solution ==== | ||
| - | <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 ($|\vec{L}|$). |
| - | $\vec{L}=\langle 0,0.1,0 \rangle - \langle 0,0,0 \rangle = \langle 0,0.1,0 \rangle m$ | + | $$\vec{L}=\langle 0,0.1,0 \rangle - \langle 0,0,0 \rangle = \langle 0,0.1,0 \rangle m$$ |
| - | $\vec{|L|}=0.1$ | + | $$\vec{|L|}=0.1$$ |
| - | $\hat{L}=\langle 0,1,0 \rangle$ | + | These can be used to compute the unit (direction) vector for the stretch ($\hat{s}$) (the difference between $|\vec{L}|$ and the relaxed distance |
| - | $F_{spring} = -k_s(|\vec{L}|-L_0)\langle 0,1,0 \rangle | + | $$\hat{L}=\langle 0,1,0 \rangle$$ |
| - | $F_{Earth} = \langle 0,-mg,0 \rangle$ | + | 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. | ||
| + | |||
| + | $$F_{spring} = -k_s(|\vec{L}|-L_0)\langle 0,1,0 \rangle = \langle 0, | ||
| + | |||
| + | We know that the force due to gravity acts solely in the y direction also so we can write the representation for the force equation to represent this: | ||
| + | |||
| + | $$F_{Earth} = \langle 0,-mg,0 \rangle$$ | ||
| The initial momentum of the block is zero, since you approximate that it is at rest when you release it | The initial momentum of the block is zero, since you approximate that it is at rest when you release it | ||
| - | $\vec{p_i}= \langle 0,0,0 \rangle$ | + | $$\vec{p_i}= \langle 0,0,0 \rangle$$ |
| All of the forces acting on the system by the spring and gravitational force are in the y direction, and the initial x and z components of the blocks momentum are zero, so we only need to consider the y components of forces. | All of the forces acting on the system by the spring and gravitational force are in the y direction, and the initial x and z components of the blocks momentum are zero, so we only need to consider the y components of forces. | ||
| - | $F_{spring} = -k_s(|\vec{L}|-L_0)\hat{L}$ | + | $$F_{spring} = -k_s(|\vec{L}|-L_0)\hat{L}$$ |
| + | |||
| + | $$F_{Earth}=-mg$$ | ||
| + | |||
| + | The addition of these two forces is the net force acting on the system: | ||
| + | |||
| + | $$F_{net,y} = F_{spring, | ||
| + | |||
| + | 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: | ||
| + | |||
| + | $$\vec{|L|} = 0.1m$$ | ||
| + | |||
| + | The stretch is equal to \vec{|L|} - the relaxed distance (0.2m). | ||
| + | |||
| + | $$s = 0.1m - 0.2m = -0.1m$$ | ||
| + | |||
| + | Compute the value of the force of the spring. | ||
| + | |||
| + | $$F_{spring_y} = -8N/ | ||
| + | |||
| + | Compute the value of the force due to gravity. | ||
| + | |||
| + | $$F_{Earth, | ||
| + | |||
| + | Add the force of spring to force due to gravity to obtain the net force. | ||
| + | |||
| + | $$F_{net,y} = .212N$$ | ||
| + | |||
| + | Input this net force into the equation for momentum using 0.1s as the time to find the momentum at this instance. | ||
| + | |||
| + | $$p_{fy} = 0 + (.212N)(0.1s)\quad(Momentum\ Principle)\\$$ | ||
| + | |||
| + | Compute momentum: | ||
| - | $F_{Earth}=-mg$ | + | $$p_{fy} = 0.0212 kg * m/s$$ |
| - | $F_{net,y} = F_{spring,y} + F_{Earth,y}$ | + | 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. Assume v_{avg,y} is approximate to v_{fy} for the time period of 0.1s. |
| - | For 0.1 seconds | + | $$v_{avg,y} \approx v_{fy}$$ |
| - | $\vec{|L|} = 0.1m$ | + | Using the equation |
| - | $s = 0.1m - 0.2m = -0.1m$ | + | $$v_{fy} |
| - | $F_{spring_y} = -8N/ | + | Using the position update equation |
| - | $F_{Earth, | + | $$y_f = 0.1m + (0.353m/s)(0.1s) \quad(position\ update)\\$$ |
| - | $F_{net, | + | $$y_f = 0.135m |
| - | $p_{fy} = 0 + (.212N)(0.1s)\quad(Momentum\ Principle)\\$ | + | For the time of 0.2s we just have to repeat the process but use the new values for \vec{|L|}, s, F_{spring_y} and F_{net,y}. |
| - | $p_{fy} = 0.0212 kg * m/s$ | + | Use the y position of the system for 0.1 seconds as the initial position for 0.2s. |
| - | $v_{avg, | + | $$\vec{|L|} = 0.135m$$ |
| - | $v_{fy} = \dfrac{p_{fy}}{m} = \dfrac{0.0212 kg * m/ | + | Compute the new s based on this new \vec{|L|} |
| - | $y_f = 0.1m + (0.353m/s)(0.1s) \quad(position\ update)\\$ | + | $$s = 0.135m -0.2m = -0.0647m$$ |
| - | $y_f = 0.135m \quad(position\ in\ y\ direction\ after\ | + | Calculate the new force of the spring based on the new s distance. The force due gravity remains the same. |
| - | For the time of 0.2s | + | $$F_{spring, |
| - | $\vec{|L|} = 0.135m$ | + | Add the force due to the spring to the force due to gravity. |
| - | $s = 0.135m -0.2m = -0.0647m$ | + | $$F_{net, |
| - | $F_{spring, | + | Calculate the momentum using this new net force. |
| - | $F_{net,y} = -0.0707N$ | + | $$p_{fy} = (0.0212 kg * m/s) + (-0.0707N)(0.1s)$$ |
| - | $p_{fy} = (0.0212 kg * m/s) + (-0.0707N)(0.1s)$ | + | $$p_{fy} = 0.0141 kg * m/s$$ |
| - | $p_{fy} = 0.0141 kg * m/s$ | + | Calculate the new $v_{fy}$ based on this new momentum. |
| - | $v_{fy} = 0.236 m/s$ | + | $$v_{fy} = 0.236 m/s$$ |
| - | $y_{f} = 0.159m \quad(position\ in\ y\ direction\ after\ | + | Update |
| + | $$y_{f} = 0.159m \quad(position\ in\ y\ direction\ after\ .2s)\\$$ | ||
| + | If you wished to calculate to position after 0.3s you repeat the same calculations again based on the y position that was computed for .2s. | ||