**Unformatted text preview: **Physics for Scientists & Engineers 1
Spring Semester 2011 Lecture 16 Conservation of Mechanical Energy February 16, 2011 Physics for Scientists&Engineers 1 1 1 Mechanical Energy
Define the mechanical energy (or also called “total mechanical energy”) as the sum of kinetic and potential energy E = K +U
Define an isolated system as a system of objects exerting forces forces on each other, but for which no external force causes energy changes • No energy is transferred into or out of the system February 16, 2011 Physics for Scientists&Engineers 1 2 2 Conservation of Energy
For any mechanical process inside an isolated system that involves only conservative forces, the total mechanical energy is conserved, i.e. remains constant in time ∆E = ∆K + ∆U = 0
Alternative way of writing the same result: K + U = K 0 + U0
Law of total energy conservation is a cornerstone of all of physics More conservation laws to come as we go along • Momentum, angular momentum, charge, baryon number, …
February 16, 2011 Physics for Scientists&Engineers 1 3 3 Conservation of Energy - Derivation
For conservative forces, we have found: ∆U = −W
Last week, we had seen that this work is also equal to the change in kinetic energy ∆K = W
Combining these two results, we then find ∆K = − ∆U ⇔ ∆K + ∆U = 0 Now we use ∆U = U − U 0 , ∆K = K − K 0 and get:
0 = ∆U + ∆K = U − U 0 + K − K 0 = U + K − (U 0 + K 0 ) U + K = U0 + K0
February 16, 2011 Physics for Scientists&Engineers 1 4 4 Energy Conservation - Remarks
We did not make reference to any particular path • Since the force has to be conservative for energy conservation, this result also needs to be path independent We do not need to know any details of the conservative force to be able to use energy conservation An object in free fall near the Earth’s surface is not really an isolated system. But we can still employ energy conservation for this case, because • That object + Earth form an isolated system • Changes in the kinetic energy and motion of the Earth due to rearrangement of this object can be neglected • All changes in kinetic and potential energy are for this object
February 16, 2011 Physics for Scientists&Engineers 1 5 5 Defending the Castle
Defend the castle from attackers! Shoot rocks from catapult with launch speed of 14.2 m/s, from the courtyard over the castle walls onto the attackers in front of the castle, at elevation 7.2 m below courtyard. Question:
What is speed with which your rocks hit the attackers? Answer:
Hard way: find appropriate launch angle to clear wall; decompose initial velocity vector into components; solve for y-component of velocity as a function of time or altitude, following trajectory to impact; take square root of sum of the squares of the velocity components February 16, 2011 Physics for Scientists&Engineers 1 6 6 Defending the Castle
THINK • Much easier way is to use energy concepts • Once the rock is launched, only the conservative force of gravity is acting on the rock • Total mechanical energy is conserved SKETCH • Define initial and final conditions RESEARCH • Conservation of mechanical energy gives us E = K + U = K0 + U0
February 16, 2011 Physics for Scientists&Engineers 1 7 7 Defending the Castle
• The kinetic energy of the projectile is 1 K = mv2 2 • The potential energy of the projectile is U = mgy SIMPLIFY • Substitute K and U in E 1 12 E = mv2 + mgy = mv0 + mgy0 2 2 • The mass of rock cancels out
12 12 v + gy = v0 + gy0 2 2
February 16, 2011 Physics for Scientists&Engineers 1 8 8 Defending the Castle
• Solve for the speed when the rock lands in the courtyard
2 v = v0 + 2 g ( y0 − y ) CALCULATE • y0 - y = 7.20 m and v0 = 14.2 m/s, so
v=
2 (14.2 m/s ) + 2 ( 9.81 m/s2 ) ( 7.20 m ) = 18.517667 m/s ROUND v = 18.5 m/s DOUBLE-CHECK • The speed when the rock hits the ground is 18.5 m/s, which is faster than the initial speed of 14.2 m/s • Note that our answer is independent of θ0
February 16, 2011 Physics for Scientists&Engineers 1 9 9 Defending the Castle
• Let’s do the same problem using kinematics • The components of the initial velocity are vx 0 = v0 cosθ0 vy 0 = v0 sin θ0 • The x-component of the final velocity is the same as the x-component of the initial velocity vx = vx 0 = v0 cosθ0
• The y-component of the final velocity can be obtained using free fall in the y-direction
2 2 vy = vy 0 − 2 g ( y − y0 ) February 16, 2011 Physics for Scientists&Engineers 1 10 10 Defending the Castle
• The final velocity of the rock is
2 2 v = vx + vy v= ( v0 cosθ0 ) 2 2 + ( vy 0 − 2 g ( y − y0 ) ) 2 2 v = v0 cos2 θ0 + v0 sin2 θ0 − 2 g ( y − y0 ) • Remembering that sin2θ + cos2θ = 1
2 v = v0 ( cos2 θ0 + sin2 θ0 ) − 2 g ( y − y0 ) 2 2 v = v0 − 2 g ( y − y0 ) = v0 + 2 g ( y0 − y ) • We get the same answer, but had to work really hard February 16, 2011 Physics for Scientists&Engineers 1 11 11 Clicker Quiz
Assume that the rock is launched with an angle of θ = 45°. With what angle with respect to the horizontal does the rock strike the ground in front of the castle? v0 = 14.2 m/s v f = 18.5 m/s A. 25.4° B. C. D. E. 37.9° 45.0° 57.1° 65.2° February 16, 2011 Physics for Scientists&Engineers 1 12 12 Clicker Quiz Solution
v0 = 14.2 m/s v f = 18.5 m/s cos φ = vx v0 cosθ = vf vf
−1 φ = cos v0 cosθ = cos −1 vf
D. (14.2 m/s )cos 45°
18.5 m/s φ = 57.1° February 16, 2011 Physics for Scientists&Engineers 1 13 13 Race of Two Balls
Two balls are released from the top of two ramps with different shapes Energy Energy conservation tells us that the speed of both balls at the bottom end of the ramps has to be the same
v = 2 g (y0 − y ) This result would be difficult to obtain using kinematics You can see that the lighter colored ball arrives at the bottom first
February 16, 2011 Physics for Scientists&Engineers 1 14 14 Clicker Quiz
The lighter-colored ball arrives first because A. The darker-colored ball cannot accelerate well on the flat track B. The acceleration of gravity works better on a curved track C. The darker-colored ball has more air resistance D. The lighter-colored ball spends more time at higher speeds
February 16, 2011 Physics for Scientists&Engineers 1 15 15 Trapeze Artist
PROBLEM • A circus trapeze artist starts her motion with the trapeze at rest at an angle of 45.0° relative to the vertical • The trapeze ropes have a length of 5.00 m • What is her speed at the lowest point in her trajectory? THINK • Initially, the trapeze artist has only gravitational potential energy • We can choose a coordinate system such that y = 0 is at her trajectory’s lowest point, so the potential energy is zero at that lowest point • When the trapeze artist is at the lowest point, her kinetic energy will be a maximum • We can then equate the initial gravitational potential energy to the final kinetic energy of the trapeze artist February 16, 2011 Physics for Scientists&Engineers 1 16 16 Trapeze Artist
SKETCH RESEARCH • The trapeze is pulled back to an initial angle θ0 • The trapeze artist is then at a height h = ℓ(1-cosθ0) • The trapeze artist has potential energy U = mgℓ(1-cosθ0)
February 16, 2011 Physics for Scientists&Engineers 1 17 17 Trapeze Artist
• The total energy of the trapeze artist at any angle is 1 E = mg ℓ (1 − cos θ ) + mv2 2
SIMPLIFY • Equating the total energy at maximum deflection to the total energy at any angle • Solving for |v| 1 mg ℓ (1 − cos θ0 ) = mg ℓ (1 − cos θ ) + mv2 2 12 mv ⇒ v = 2 g ℓ ( cos θ − cos θ0 ) 2 mg ℓ ( cos θ − cos θ0 ) = • We want the speed for θ = 0° and θ0 = 45° v = 2 g ℓ ( cos 0° − cos 45° )
February 16, 2011 Physics for Scientists&Engineers 1 18 18 Trapeze Artist
CALCULATE v = 2 ( 9.81 m/s2 ) ( 5.00 m )( cos 0° − cos 45° ) = 5.3603008 m/s
ROUND v = 5.36 m/s DOUBLE-CHECK • Look at limiting cases for our result v = 2 g ℓ (1 − cos θ0 )
• θ0 = 0° v = 2 g ℓ (1 − 1) = 0
• θ0 = 90° v = 2 g ℓ (1 − 0 ) = 2 g ℓ
February 16, 2011 Physics for Scientists&Engineers 1 19 19 Sliding Chain
PROBLEM • A uniform chain of total mass m is laid out straight on a frictionless table and held stationary so that one-third of its length, L = 1.00 m, is hanging vertically over the edge of the table • The chain is then released • Determine the speed of the chain at the instant when only one-third of its length remains on the table February 16, 2011 Physics for Scientists&Engineers 1 20 20 Sliding Chain
SKETCH RESEARCH • Initially the mechanical energy is E = K + U = 0 + ( m / 3) ( g ) ( − L / 6) • Finally the mechanical energy is 1 E = K + U = mv2 + (2m / 3) ( g ) ( − L / 3) 2
February 16, 2011 Physics for Scientists&Engineers 1 21 21 Sliding Chain
THINK • Before the chain is let go, 2/3 of the chain is on the table and 1/3 hangs vertically • The chain is let go and the vertical part of the chain accelerates downward • When 1/3 of the chain is on the table and 2/3 is hanging vertically, the amount of chain hanging has increased and the center of mass of the the hanging chain has moved down February 16, 2011 Physics for Scientists&Engineers 1 22 22 Sliding Chain
SIMPLIFY (m / 3) ( g ) (− L / 6) = 12 mv + (2m / 3) ( g ) ( − L / 3) 2 gL 1 2 2 gL 12 3 − = v − gL ⇒ v = gL ⇒ v = 18 2 9 2 18 3 CALCULATE v= ( 9.81 m/s ) (1.00 m ) = 1.808314 m/s
2 3 ROUND v = 1.81 m/s DOUBLE-CHECK • Units are good and speed seems reasonable February 16, 2011 Physics for Scientists&Engineers 1 23 23 Clicker Quiz
A ball of mass m is thrown vertically into the air with an initial speed v. Which of the following equations correctly describes the maximum height, h, of the ball? A) h = v 2g B) h = g 12 v 2
2 mv g C) h = mv2 D) h = g v2 E) h = 2g
February 16, 2011 Physics for Scientists&Engineers 1 24 24 Energy Conservation, Mass on Spring (1)
Potential energy for mass on a spring: U s = 1 kx 2 2
Total mechanical energy: E = K + U s = 1 mv 2 + 1 kx 2 2 2
At maximum displacement (= amplitude, x = A), the mass turns around => v = 0 => K = 0 at this point So we find the important result that the energy in simple harmonic oscillations is E = 1 kA 2 2
February 16, 2011 Physics for Scientists&Engineers 1 25 25 Energy Conservation, Mass on Spring (2)
Total mechanical energy:
1 2 kA 2 = 1 mv 2 + 1 kx 2 2 2 Solve for speed as function of displacement k v = (A − x ) m
2 2 Note: we did not need to solve the equation of motion to come to this result! Energy conservation alone was sufficient. February 16, 2011 Physics for Scientists&Engineers 1 26 26 ...

View
Full Document