Example: Predicting the motion of a system that is subject to a spring interaction/Predicting the final location of an object moving under a non-constant force
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
- Spring has relaxed length of (0.2m) $L_{0} = 0.2m$
- Spring has compressed length of (0.1m) $|\overset{\rightarrow}{L}| = 0.1m$
- Spring has spring constant of $8N/m$
- Block of mass (.06)kg attached to top of the spring
- The block has no initial momentum, $\overset{\rightarrow}{p_{i}} = \langle 0,0,0\rangle$
- There are two forces acting on the spring, the gravitational force and the spring force
Lacking
- The $y$ position of the block after 0.1 and 0.2 seconds.
Approximations & Assumptions
- Assume block is at rest when released from compressed position therefore the initial momentum of block is zero
- Assume that the origin is at the base of the spring
- Over the time interval investigated drag forces are negligible.
Representations
- The force of the spring is given by ${\overset{\rightarrow}{F}}_{spring} = - k_{s}(|\overset{\rightarrow}{L}| - L_{0})\widehat{L}$
- The gravitational force is given by ${\overset{\rightarrow}{F}}_{Earth} = - mg$
- The momentum of the system is given by $\overset{\rightarrow}{p_{f}} = \overset{\rightarrow}{p_{i}} + \overset{\rightarrow}{F_{net}}\Delta t$
- Momentum equation = $m(\overset{\rightarrow}{v}) = \overset{\rightarrow}{p}$
- Postion update = $\overset{\rightarrow}{r_{f}} = \overset{\rightarrow}{r_{i}} + {\overset{\rightarrow}{v}}_{avg}\Delta t$
Solution
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 ($\overset{\rightarrow}{L}$) of the mass and the length of the position vector ($|\overset{\rightarrow}{L}|$).
$$ \overset{\rightarrow}{L} = \langle 0,0.1,0\rangle - \langle 0,0,0\rangle = \langle 0,0.1,0\rangle m $$$$ \overset{\rightarrow}{|L|} = 0.1 $$These can be used to compute the unit (direction) vector for the stretch ($\widehat{s}$) (the difference between $|\overset{\rightarrow}{L}|$ and the relaxed distance \langle 0,.2,0 \rangle, which is in the same direction as the position vector.
$$ \widehat{L} = \langle 0,1,0\rangle $$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.
$$ F_{spring} = - k_{s}(|\overset{\rightarrow}{L}| - L_{0})\langle 0,1,0\rangle = \langle 0, - k_{s}(|\overset{\rightarrow}{L}| - L_{0}),0\rangle $$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
$$ \overset{\rightarrow}{p_{i}} = \langle 0,0,0\rangle $$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.
$$ F_{spring} = - k_{s}(|\overset{\rightarrow}{L}| - L_{0})\widehat{L} $$$$ F_{Earth} = - mg $$The addition of these two forces is the net force acting on the system:
$$ F_{net,y} = F_{spring,y} + F_{Earth,y} $$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:
$$ \overset{\rightarrow}{p_{f}} = \overset{\rightarrow}{p_{i}} + \overset{\rightarrow}{F_{net}}\Delta t $$So we are now going to calculate the y position of the system for 0.1 seconds.
As found earlier:
$$ \overset{\rightarrow}{|L|} = 0.1m $$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.
$$ F_{spring_{y}} = - 8N/m( - 0.1m) = + 0.8N $$Compute the value of the force due to gravity.
$$ F_{Earth,y} = - 0.06kg*9.8N/kg = - 0.588N $$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:
$$ p_{fy} = 0.0212kg*m/s $$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. Assume v_{avg,y} is approximate to v_{fy} for the time period of 0.1s.
$$ v_{avg,y} \approx v_{fy} $$Using the equation $m(\overset{\rightarrow}{v}) = \overset{\rightarrow}{p}$ we can calculate $v_{fy}$ using momentum just found and the mass of the system.
$$ v_{fy} = \frac{p_{fy}}{m} = \frac{0.0212kg*m/s}{0.06kg} = + 0.353m/s $$Using the position update equation $\overset{\rightarrow}{r_{f}} = \overset{\rightarrow}{r_{i}} + {\overset{\rightarrow}{v}}_{avg}\Delta t$ we can compute the new position of the system:
$$ y_{f} = 0.1m + (0.353m/s)(0.1s)\quad(position\ update) $$$$ y_{f} = 0.135m\quad(position\ in\ y\ direction\ after\ .1s) $$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}.
Use the y position of the system for 0.1 seconds as the initial position for 0.2s.
$$ \overset{\rightarrow}{|L|} = 0.135m $$Compute the new s based on this new \vec{|L|}
$$ s = 0.135m - 0.2m = - 0.0647m $$Calculate the new force of the spring based on the new s distance. The force due gravity remains the same.
$$ F_{spring,y} = + 0.520N $$Add the force due to the spring to the force due to gravity.
$$ F_{net,y} = - 0.0707N $$Calculate the momentum using this new net force.
$$ p_{fy} = (0.0212kg*m/s) + ( - 0.0707N)(0.1s) $$$$ p_{fy} = 0.0141kg*m/s $$Calculate the new $v_{fy}$ based on this new momentum.
$$ v_{fy} = 0.236m/s $$Update position based on this new velocity.
$$ y_{f} = 0.159m\quad(position\ in\ y\ direction\ after\ .2s) $$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.