PHYS 218 Chapter 7

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Potential Energy

Possibility of work to be done (represented by U). Measured in Joules (J)

Gravitational potential energy: greater height → greater potential energy

Elastic potential energy: stretching/compressing a spring more → greater potential energy


Work-Energy Theorem

(See PHYS 218 Chapter 6#Work-Energy Theorem→)

Work of all forces is equivalent to the difference in Kinetic Energy.

W=K2K1


Gravitational Energy

Ug=mgy=ΔU

Elastic Energy

Ue=12k(xl0)2


Net Forces

W1+W2+W3=K2K1

Any W's that can be replaced by ΔU's are considered "conservative forces". All others are considred "non-conservative forces".

Result:

Ki+U1i+U2i+mechanical energy+Wnc=Kf+U1f+U2f+


Example

An object is stored in front of a compressed spring. The object will be pushed by the spring and move up a ramp. What is the max height?

kΔx22+mvi22+mgyi+Wnonconservative=mgh


Forces and Potential Energy

Fx=dUdxFy=dUdyFz=dUdzF=U

Example

Potential Energy = U=mgy

F=U=(dUdx,dUdy,dUdz)=(0,mg,0)


Energy Diagrams

Graph the energy equations (results in parabola)

If a system is given to have ye J of energy, then the total area of the graph is bounded by y=ye

Distance from x-axis to graph is potential energy, distance from graph to ye


HARD EXAMPLE

An object is at height h on an inclined plane at angle φ and an x component of d with a friction coefficient of μk. Once it gets to the bottom, it moves across a level, frictionless plane, then goes over a circular bump of radius R and flies off the track.

  1. Find the acceleration a on the inclined plane.
  2. Find the time t1 at which the object hits the frictionless plane.
  3. Find the angle θ at which the object flies off

1. Acceleration of inclined plane

Draw a free body diagram.

F={x^mgsinφ=maxy^Nmgcosφ=may=0

Using Fy, we can find that N=mgcosφ.

ax=g(sinφμkcosφ)


2. Velocity at t1

Use kinematics or conservation of energy

Kinematics

v2=v02+2aΔxv=2g(sinφμkcosφ)h2+d2


Conservation of Energy

Ki+Ui++Wnc=Kf+Ufmghμkmgcosφh2+d2=12mv22g(hμkgcosφh2+d2=v2same as other method


3. Angle of liftoff

Draw a free-body diagram (use r and φ for coordinate system)

The object will leave when the normal N=0

F={r^Nmgcosθ=mvθ2Rφ^mgsinθ=maθ


Using conservation of energy to find vθ,

12mvi2=12mvf2+mg(R+cosθ)



Monday, March 7, 2011

Another Problem

A ramp with friction coefficient of μk makes an angle of φ with the ground and ends at a height h. A rocket at the base of the ramp will fire its engine until it leaves the ramp.

What is the force of the engine if the velocity at height L is vf.

Note: Friction and the force of the engine are non-conservative forces

Ki+Ui+Wnonconservative=Kf+Uf0+0+μkNHsinφ+Fengined=12mvf2+mgLμkmgcosφHsinφ+FeHsinφ=12mvf2+mgL


Yet Another Problem

An 80 kg box sits on a ledge with a rope attached to a pulley and ultimately to a 65 kg bucket. The ground has a friction coefficient μk=0.4 with the box. What is the velocity of the bucket after it descends 2 m?

Ki+Ui+Wnc=Kf+UfμkN(2)=12m2v2+12m1v2m2g(2)


Loop-the-loop

"How high should the ball be to make a full circle without falling off the rail?"

"make a full circle"
Circular motion: a=v2R
"without falling off the rail"
Normal has to be greater than zero (the point at which an object breaks free of circular motion is when the normal is zero).

Nmg=mv2R

If N=0, g=v2R


MOST IMPORTANT!

When is the mechanical energy of a system conserved?
As long as the work of the non-conservative forces is equal to 0