Presentation on theme: "Springs A spring can exert a force, store energy, and do work. Think about a dart gun – when a dart gun is loaded it compresses a spring inside the."— Presentation transcript:
Springs A spring can exert a force, store energy, and do work. Think about a dart gun – when a dart gun is loaded it compresses a spring inside the barrel. What happens when the trigger is pulled? The spring exerts a force on the dart and fires it out of the gun!
Springs The exerted force is an example of Hooke’s Law: Hooke’s Law: F s = - kx F s = force exerted by a compressed or stretched spring (in N). k = spring constant, a characteristic of each spring. k is small for a flexible spring, big for a stiff spring (in N/m). x = distance the spring is compressed (negative) or stretched (positive) (in m).
Spring constant from a graph Weights are hung from a spring and the elongation of the spring is measured. Using a graph of force (y-axis) vs distance (x-axis), the spring constant can be calculated as the slope of the line. For this graph, k = 60 N/m
Spring constant from data A 3.0 kg mass is hung from a spring. As a result, the spring stretches 2.0 cm. What is the spring constant for this spring? The spring stretches due to the weight of the 3.0 kg mass: F s =- kx but F s is equal to mg mg = -kx (3.0 kg) (9.8 m/s 2 ) = - k (.020 m) k = 1500 N/m (it’s not a vector so the sign doesn’t matter) A bigger k corresponds to a stiffer spring
Springs Hooke’s Law: F s = - kx The negative sign in the equation is due to the fact that a spring force is a restorative force – it tends to restore things back to equilibrium. If I stretch a spring and release it, it will go back to its original unstretched state. If I compress a spring and release it, it will also go back to its original unstretched state.
Spring force points in opposite direction In this picture, the spring is stretched in the direction of f But the force felt by the spring is in the opposite direction Hence the negative sign in the equation: F = -kx
Hooke’s Law example What is the force exerted by a spring in a dart gun (k =15 N/m) when it is compressed 2.5 cm? F s = -kx F s = -(15 N/m) (-.025 m) F s = +0.38 N The positive sign tells you that the spring force points in the direction of stretch, not the direction of compression.
Energy of a spring The reason the dart gun fired at all is because the spring stored energy. PE s = ½ kx 2 PE s = potential energy stored in the spring (in J) k = spring constant (in N/m) x = distance spring is compressed or stretched (in m).
Energy stored in spring example What is the energy stored in a spring with a spring constant of 15 N/m when it is compressed 2.5 cm? PE s = 1/2 kx 2 PE s = ½ (15 N/m) (-.025 m) 2 PE s = 0.0047 J We added this energy to the spring when we did work on it by compressing the spring (applying a force) through a distance.
Work done by a spring Similarly, when the compression or stretching is released, the spring does work equal to the potential energy stored in the spring. PE s = Work
Conservation of energy If a spring is used, it must be included in total energy, and conservation of energy.
Conservation of energy - example A 2.0 kg ball starts from rest at the top of a 3.0 meter hill. At the bottom of the hill, it hits a horizontal spring with a spring constant of 900 N/m. How far does the ball compress the spring before it comes to a rest? Remember that energy must be conserved – what kind(s) of energy are present at the top of the hill? What kind(s) of energy are present when the spring is done compressing?
Conservation of energy - example At the top of the hill, all the energy is potential: PE = mgh = (2.0 kg)(9.8 m/s 2 )(3.0 m) = 60 J By the law of conservation of energy, this must also be the total energy stored in the spring when it is done compressing (since the ball isn’t moving). PE s = 1/2 kx 2 60 J = ½ (900 N/m) x 2 x = -.36 m This is the distance the spring compresses.
Centripetal Force Acts on an object in circular motion. Centripetal means “center-seeking”. For an object in circular motion, centripetal force is the force that points towards the center of the circle. Like all forces, it is measured in Newtons.
Centripetal Force Picture Without centripetal force, these amusement park cars would move in the direction of the yellow arrows (with a velocity tangent to the curve). Centripetal force (always pointing towards the center) helps to keep the cars on the track.
Centripetal Force equations F c = mv 2 / r Where F c = centripetal force (N) m = mass (kg) v = linear velocity (m/s) r = radius (m)
Centripetal Force and N 2 Law Remember Newton’s 2 nd Law: F = ma. If the force is centripetal, then the acceleration must be centripetal too. F c = ma c Where F c = centripetal force (N) m = mass (kg) a c = centripetal acceleration (m/s 2 ) Examples
Centripetal acceleration equations a c = v 2 / r Where a c = centripetal acceleration (m/s 2 ) v = linear velocity (m/s) r = radius (m) examples
Centripetal acceleration aka… Sometimes centripetal acceleration (a c ) is called radial acceleration, while linear acceleration (a) is called tangential acceleration.
Period Period ( Τ ) – the time it takes for one revolution. Period is measured in seconds. Remember that velocity = distance/time For an object moving in a circle, the distance traveled is the circumference (2πr) and the time required to complete 1 revolution is Τ.
Another equation for velocity Therefore, for an object moving in a circle: v = 2πr / Τ Where: v = linear velocity (m/s) r = radius (m) Τ = period (s)
Centripetal Force and N 2 Law - FBDs When using Newton’s 2 nd Law to sum up forces on steadily rotating object: ΣF = F c If that weren’t true, the object would not be steadily rotating. Note that since F c is the summation of the forces, it is never drawn on a FBD of a steadily rotating object.
Example 1 What are the forces acting on the rider in the roller coaster on the left? Weight and Normal force. What would the FBD look like?
FBD for example 1 FBD: Sum up the forces: Σ F = N – W Remember that the sum of the forces equals F c. Therefore: N – W = Fc Normal Weight
Example 2 What are the forces acting on the rider in the top roller coaster on the right? Weight and Normal force. What would the FBD look like?
FBD for example 2 FBD: Sum up the forces: Σ F = - N – W Remember that the sum of the forces equals F c. Therefore: - N – W = Fc Normal Weight
Centripetal Force? You and your huge older brother are going to ride the sleigh ride at Santa’s Village (the one that goes in a circle). Where do you want your huge brother to be – on the inside or the outside? Does that seem like a center-seeking force to you?
Centrifugal “Force” That force is centrifugal (center-fleeing) and it is a false force. It is due to Newton’s 3 rd Law – see drawing on the next slide. What you felt on Santa’s sleigh ride is the car turning from under you and causing you to push up against the side of the car. Most people have felt centrifugal “force”, but not centripetal force.
Centrifugal “Force” A body lies in the back of a car. Here is an overhead view of the car driving straight. Now the car turns to the right and the unrestrained body continues in a straight line. The result is the body gets shoved against the side of the car. That’s what you think of as centrifugal “force”!
References for images http://www.cellmigration.org/resourc e/modeling/res_resource_images/fig6.gif http://www.cellmigration.org/resourc e/modeling/res_resource_images/fig6.gif http://www.onlinephys.com/hooke6.g if http://www.onlinephys.com/hooke6.g if