Massachusetts Institute of Technology - Physics Department

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1 Massachusetts Institute of Technology - Physics Department Physics Assignment #4 October 6, It is strongly recommended that you read about a subject before it is covered in lectures. Lecture Date Material Covered Reading #13 Fri 10/8 Potential Energy - watch PIVoT Page Energy Considerations to derive SHM - watch PIVoT Page #14 Wed 10/13 Escape Velocities - Bound and Unbound Orbits - PIVoT Page 11 8 Circular Orbits (elliptical orbits will be discussed later Page in the course). Various Forms of Energy - Power #15 Fri 10/15 Momentum - Conservation of Momentum - watch PIVoT Page Center of Mass - watch PIVoT Page Due Friday, Oct 15, before 4 PM in 4-339B. Solutions will be posted on the Web Sat, Oct Air drag on very small drops. Watch PIVoT under resistive force. We release an oil drop of radius r in air. The density of the oil is 700 kg/m 3. C 1 and C for 1 atmosphere air at 0 C are (kg/m)/sec and 0.85kg/m 3, respectively. How small should the oil drop be so that the drag force is dominated by the linear term in the speed (in lectures we called this Regime I). In this regime, the terminal velocity is mg/c 1 r. [m is the mass of the drop]. 4. Drag force at very low speeds. Watch PIVoT under resistive force. At low speeds (especially in liquids rather than gases), the drag force is proportional to the speed rather than its square, i.e., F = C 1 rv, where C 1 is a constant. At time t = 0, a small ball of mass m is projected into a liquid so that it initially has a horizontal velocity of u in the +x direction. The initial speed in the vertical direction (y) is zero. The gravitational acceleration is g. a) Write down the differential equations of motion in the x and y direction. b) What is the horizontal component of the ball s velocity at time t? c) What is the vertical component of the ball s velocity at time t? d) After how many seconds is the vertical speed 99% of its maximum value? What would that be for the 1/4 inch steel ball bearing that we dropped in Karo Corn Syrup in lectures? e) Answer the questions under b) and c) for the limiting case that t becomes initely large. 4.3 SHO page 405, problem 1. Watch PIVoT under simple harmonic motion. 4.4 SHO page 405, problem 4. Watch PIVoT under simple harmonic motion. 4.5 Oscillating Spring. Watch PIVoT under simple harmonic motion. A 3 kg mass is attached to a spring. The period of oscillation is 0.4 sec. At t = 0, the mass has a speed of 3 m/sec towards equilibrium, and its displacement from equilibrium is 0.1 m. a) Calculate the position of the mass for all time t > 0. b) When will the mass rst go through equilibrium; what then will be its speed, acceleration, kinetic energy, and potential energy? c) When will the mass rst reach a turning point; what then will be its speed, acceleration, kinetic energy, and potential energy? 4.6 Vertical Spring page 406, problem 11.

2 ß ß t = + = = = ß 1:0 s 6 3 3! 3 :0 rad=s (c) All the times for which x = 0 and x = ±A were already given above. and x = 0 at t = + 1 x = ±A at t = + 6 The part indicates that the particle will move through a turning point or equilibrium exactly at intervals of. But we must b e careful. There are two turning points: half of the times above correspond to x = +A, and the other half correspond to x = A. The time t = corresponds to x = + A. So the possible times for x = + A are 6 The possible times for x = A are Problem 4.5 t = + 6 t = + + = (a) The motion is simple harmonic, so the position as a function of time is x = A cos (!t + f) and the velocity as a function of time is v =!A sin (!t + f) The angular frequency is given by ß ß! = = ß 15:7 rad=s 0:4 If we c hoose the direction for x to point along the same direction as the initial displacement then the initial (t = 0) displacement from origin is x = 0 :1 m, so 0:1 = A cos f (1) 9

3 The initial (t = 0) velocity is v = 3 m=s (it is directed toward the origin from the positive side), so Using equation (1) and equation () we can nd f. 3 =!A sin f () sin f 3 3 tan f = = = cos f 0:1! 0:1 ß Thus 3 f = tan 1 0:1 ß = tan 1 3 0:4 0:1 ß ß 1:088 (Rememb er th a t a n y value f = 1 :088 + nß will do.) Then using equation (1) gives Therefore the motion is given by 0:1 A = ß 0:16 m cos f x = 0 :16 cos (15:7 t + 1 :088) (b) Equilibrium corresponds to x = 0. x = A cos(!t + 1 :088) = 0 =) cos(!t + 1 :088) = 0!t + 1 :088 = ß + nß t = 1 ß + nß 1:088! t = (0:4 8 + nß) ß t = 7: The rst time the mass will pass through equilibrium is t = 7 :7 10 = 3 :1 10 s 10

4 The total energy is conserved. E = K + U with K = 1 mv and U = 1 kx = m! x = ß m x At t =0the energy is given by E = 1 mv 0 + ß m x 0 = 1 m (3) + ß m (0:1) ß 17: J By energy conservation, this will be the total energy at any other time. At equilibrium, x = 0, and the potential energy is U = 1 k(0) =0 Thus the total energy, which is now just the kinetic energy, is s E Thus the kinetic energy is E = K = 1 mv =) jvj = m K = E ß 17: J and the speed is jvj = s E m ß 3:4 m=s The acceleration at x = 0 is given by a = d x dt =! x =) a =0 (c) The turning points corresponds to x = ±A. These points also correspond to v =0. v =!A sin(!t +1:088) = 0 =) 11

5 sin(!t + 1 :088) = 0!t + 1 :088 = nß 1 t = (nß 1:088)! t = (nß 1:088) ß t = 0:18 + The rst time the mass will pass through a turning point i s t = 0:18 + ß 0:13 s By substituting t = 0:13 s into the equation of motion we nd that the corresponding position is x = 0:16 m. (We knew it had to b e x = ±0:16 m by the above.) At a turning point, v = 0, and the kineticenergy is 1 K = m(0) = 0 Thus the total energy, which is now just the potential energy, is U = E ß 17: J The acceleration at x = 0:16 m is given by d x a = =! A =) a = 53 : m=s dt Problem 4.6 (Ohanian, page 406, problem 11) Springs and gravity are discussed in example on page 386. The key p o i n t in that example is that gravity only lowers the equilibrium position of the spring. The motion is still simple harmonicwith the same frequency. The equilibrium length is moved down by an amount mg x = ß 0:7 m k Thus, holding the spring at the unstretched position is actually holding it at a displacement of x up from equilibrium. At this p o i n t, determining the motion will b e very similar to problem 4.5 part (a). Label the vertical direction with x; dene the increasing 1

Oscillations. Phys101 Lectures 28, 29. Key points: Simple Harmonic Motion (SHM) SHM Related to Uniform Circular Motion The Simple Pendulum

Oscillations. Phys101 Lectures 28, 29. Key points: Simple Harmonic Motion (SHM) SHM Related to Uniform Circular Motion The Simple Pendulum Phys101 Lectures 8, 9 Oscillations Key points: Simple Harmonic Motion (SHM) SHM Related to Uniform Circular Motion The Simple Pendulum Ref: 11-1,,3,4. Page 1 Oscillations of a Spring If an object oscillates