* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Lecture 5 Forces that are a function of Position
Jerk (physics) wikipedia , lookup
Lagrangian mechanics wikipedia , lookup
Internal energy wikipedia , lookup
Brownian motion wikipedia , lookup
Specific impulse wikipedia , lookup
Wave packet wikipedia , lookup
Relativistic quantum mechanics wikipedia , lookup
Theoretical and experimental justification for the Schrödinger equation wikipedia , lookup
Renormalization group wikipedia , lookup
Electromagnetism wikipedia , lookup
Mass versus weight wikipedia , lookup
Centrifugal force wikipedia , lookup
Heat transfer physics wikipedia , lookup
Fundamental interaction wikipedia , lookup
Fictitious force wikipedia , lookup
Newton's theorem of revolving orbits wikipedia , lookup
Matter wave wikipedia , lookup
Equations of motion wikipedia , lookup
Relativistic mechanics wikipedia , lookup
Classical mechanics wikipedia , lookup
Hunting oscillation wikipedia , lookup
Work (thermodynamics) wikipedia , lookup
Rigid body dynamics wikipedia , lookup
Centripetal force wikipedia , lookup
Lecture 5 Forces that are a function of Position Beginning with the expression that we derived from last time, F (x) = dT dx (5.1) where T = 1/2mv 2 is the kinetic energy, we may “solve” for T given a particular force F (x). That is, Z Z x T − To = x dT = F (x)dx (5.2) where To is the kinetic energy evaluated at V (xo ). Note that, Z x F (x)dx W = (5.3) xo xo xo is just the work done on the particle by the applied force F (x). • This is an important result which states that the work is just the change in kinetic energy. • Let us define a function V (x) such that, − then it is clear that, Z x Z F (x)dx = − xo dV (x) = F (x) dx (5.4) x dV = −V (x) + V (xo ) = T − To xo T + V (x) = To + V (xo ) = constant = E Thus, 1 2 mv + V (x) = E 2 (5.5) which is just the Energy Equation where we identify V (x) as the potential energy of the particle. LECTURE 5. FORCES THAT ARE A FUNCTION OF POSITION 18 Important • ∴ if the applied force is only a function of position, then the total energy remains constant throughout the motion. • Such a force is called a conservative force and these forces play an important role in many areas of physics. • Note that from the energy equation we can solve for the motion of the particle as follows: r dx 2 v= =± [E − V (x)] dt m • in integral form, this becomes Z x t − to = xo dx q ± m2 [E − V (x)] which gives us the time as a function of x! So what are the allowed values of x? • we are restricted to values of x for which E ≥ V (x). • also note that the velocity,v, vanishes when E = V (x). Here, the particle comes to rest and reverses direction and these points are called turning points of the motion: V(x) Allowed Region E Turning Points x Example 5.1 Variation of Gravity with Height a. LECTURE 5. FORCES THAT ARE A FUNCTION OF POSITION 19 • Newton’s Law of Gravity states that, Fr = − GM1 M2 r2 (5.6) where the force is directed along the line connecting the two bodies and only depends on the separation between the two bodies. • Consider now a mass m. The force due to gravity at the surface of the earth is: −mg • Take the mass of the earth to be M , then GM m re2 GM g = re2 mg = • If the mass is a distance z above the earth, then relative to the center of the earth r = re + z and, F (z) = − GM m mgre2 GM m = − = − = mz̈ r2 (re + z)2 (re + z)2 where we have used GM m = mgre2 . • Now write, z̈ = dv dz dv dv = = v dt dz dt dz and therefore, Z −mgre2 z zo dz = (re + z)2 Z Z v v mvdv = vo ¡ ¢ d ø12mv 2 vo Solving the above integrals we find, µ ¶ 1 1 1 1 2 mgre − = mv 2 − mvo2 r e + z r e + zo 2 2 • This is again just the energy equation, where we identify the potential energy as, · ¸ re2 V (z) = −mg (re + z) which is in a different form than encountered before for the potential energy (V (z) = mgz) b. Escape Speed LECTURE 5. FORCES THAT ARE A FUNCTION OF POSITION 20 • if the mass, m, is projected upward with velocity vo at the surface of the earth, zo = 0, then the energy equation becomes: µ ¶ −z 2 2 2 v = vo + 2gre r (r + z) Ãe e ! 2gre −z = vo2 + re 1 + rze µ ¶−1 z 2 2 v = vo − 2gz 1 + re (Note:if z << re , then we obtain the standard result for a body moving in a uniform gravitational field) What is the turning point of the motion? • this is just the point for which v = 0 which occurs at the maximum height zmax = h. vo2 zmax = 2gzmax re vo2 vo2 zmax + = zmax 2g 2gre µ ¶ vo2 v2 zmax 1 − = o 2gre 2g µ ¶−1 vo2 vo2 zmax = h = 1− 2g 2gre ∴ vo2 + To obtain the escape speed, we need to obtain the value of h that is infinite. We can get this by setting the expression above in brackets to zero, 2 vO = 2gre → vescape = (2gre )1/2 for g = 9.8m/s2 and re = 6.4 × 106 m we find, vescape ≈ 11km/sec (5.7) Questions: i. Why does the earth retain its atmosphere where as the moon does not have an atmosphere? ii. What would happen if the earth’s atmosphere were composed of hydrogen? 5.1 Force as a function of time We consider now the case where F = F (t) such that we have: F (t) = m dv dt (5.8) LECTURE 5. FORCES THAT ARE A FUNCTION OF POSITION which may be integrated to give, Z t F (t0 )dt0 = mv(t) − mv(to ) = ∆p 21 (5.9) o where ∆p is the change of momentum. • This is just the impulse: change of momentum of a body due to a force F (t) acting over a given time interval. Z t dx F (t0 ) 0 = v(t) = v(to ) + dt (5.10) dt m o ¸ Z t Z t ·Z t F (t0 ) 0 0 0 dt dt (5.11) x − xo = v(t )dt = v(to )t + m o o o Example 5.2 Consider a force of the following form: F (t) = ct (think about driving a car and pressing down increasingly on the accelerator). Hence, dv = ct dt Z t ct ct2 ∴v = dt = m 2m Zo t 2 ct ct3 x = dt = 6m o 2m m • v increases quadratically with t • x increases as the cube of t • note that the “jerk”, defined as, d3 x c da = 3 = = constant dt dt m 5.2 Velocity-Dependent Forces • There are many cases where the force on a body depends on the velocity F = F (v). • If we also include a constant force that does not depend on either v or t, then Fo + F (v) = m dv dv = mv dt dx (5.12) • Such forces arise, for example, in the case of viscous resistance exerted on a body moving through a fluid. LECTURE 5. FORCES THAT ARE A FUNCTION OF POSITION 22 • This is not necessarily some simple function, but we can make a first approximation as, F (v) = −c1 v − c2 v|v| (5.13) where the negative signs are because the force is opposite to v. The constants c1 and c2 depend on the size and shape of the body. a. Example 5.3 • Consider a block of mass m which is initially projected with velocity vo along some smooth surface (i.e. ignore friction) and there is air resistance (linear). Then, taking Fo = 0 in 5.12 and using the leading term in 5.13, we have, µ ¶ Z v dv mdv −mdv −m v −c1 v = m , dt = − →t= = ln dt cv cv c1 vo vo we can now invert this to find v: v = vo e−c1 t/m and then integrate to find: Z t ¢ mvo ¡ 1 − e−c1 t/m x= vo e−c1 t/m dt = c1 o • Note that the velocity of the block decreases exponentially with time and reaches a limiting position of: · ¸ ¢ mvo ¡ mvo −c1 t/m xlim = lim x = lim 1−e = t→∞ t→∞ c1 c1 b. Terminal Velocity • We consider now vertical fall through a fluid so that our constant force, Fo , is just the force due to gravity, Fo = −mg, and thus, −mg − c1 v = m then, Z v t= vo hence, or dv dt mdv m = − ln (−mg − c1 v) c1 µ (5.14) mg + c1 v mg + c1 vo ¶ mg + c1 v −tc1 ln = mg + c1 vo m µ ¶ −mg mg v= + + vo e−c1 t/m c1 c1 ¶ µ (5.15) LECTURE 5. FORCES THAT ARE A FUNCTION OF POSITION 23 for t À c1 /m, the velocity of the body reaches the limiting value, vterm = −mg c1 (5.16) referred to as the terminal velocity • this is the condition where the total force on the body is zero → so there is no acceleration. • the above equation can be slightly modified as, v = −vt + (vt + vo ) et/τ (5.17) where vt is the terminal velocity and τ = m/c1 is the characteristic time. • for a body falling from rest (i.e. at time t = 0, vo = 0) we have, ¡ ¢ v = −vt 1 − e−t/τ (5.18) c. Some Comments • For spherical objects, the constants c1 and c2 are: c1 = 1.55 × 10−4 D, C2 = 0.22D2 where D is the diameter of the sphere in meters. • if we take the ratio of the two terms in 5.13: 0.22v|v|D2 = 1.4 × 103 |v|D 1.55 × 10−4 vD • hence, for objects the size of a basketball (D ∼ 0.07m), the quadratic term will dominate when, 1.4 × 103 |v|D ≥ 1, v ∼ 0.01m/s (5.19) • for speeds less than this, the linear term will dominate. ¨