course phy201 assignment #036?€?o????????Physics I
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18:27:27 ** Position at clock time is x = Acos(`omega* t) Velocity = -`omega *A*sin(`omega* t) Accel = -`omega * A * cos(`omega* t) University Physics students should note that velocity and acceleration are the first and second derivatives of the position function. **
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RESPONSE --> ** Position at clock time is x = Acos(`omega* t) Velocity = -`omega *A*sin(`omega* t) Accel = -`omega * A * cos(`omega* t) University Physics students should note that velocity and acceleration are the first and second derivatives of the position function. **
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18:27:54 STUDENT ANSWER: a = -`omega A sin(`omega *t) and aCent = v^2/r for the circle modeling SHM INSTRUCTOR AMPLIFICATION: ** The centripetal acceleration of the point on the reference circle, which acts toward the center of the circle, has two components, one in the x direction and one in the y direction. The component of the centripetal acceleration in the direction of the motion of the oscillator is equal to the acceleration of the oscillator. If the oscillator is at position theta then the centripetal acceleration has direction -theta (back toward the center of the circle, opposite to the position vector). The centripetal acceleration is aCent = v^2 / r; so the x and y components are respectively ax = aCent * cos(-theta) = v^2 / r * cos(theta) and ay = aCent * sin(-theta) = -v^2 / r * sin(theta). **
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RESPONSE --> STUDENT ANSWER: a = -`omega A sin(`omega *t) and aCent = v^2/r for the circle modeling SHM INSTRUCTOR AMPLIFICATION: ** The centripetal acceleration of the point on the reference circle, which acts toward the center of the circle, has two components, one in the x direction and one in the y direction. The component of the centripetal acceleration in the direction of the motion of the oscillator is equal to the acceleration of the oscillator. If the oscillator is at position theta then the centripetal acceleration has direction -theta (back toward the center of the circle, opposite to the position vector). The centripetal acceleration is aCent = v^2 / r; so the x and y components are respectively ax = aCent * cos(-theta) = v^2 / r * cos(theta) and ay = aCent * sin(-theta) = -v^2 / r * sin(theta). **
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18:28:10 ** The PE of the pendulum at displacement x is .5 k x^2. By conservation of energy, if nonconservative forces are negligible, we find that the KE of the pendulum at position x is.5 k A^2 - .5 k x^2. This result is obtained from the fact that at max displacement A the KE is zero, and the KE change from displacement A to displacement x is the negative of the PE change between these points. Thus .5 m v^2 = .5 k A^2 - .5 k x^2. Solving for v we have v = +- sqrt( .5 k / m * (A^2 - x^2) ) . **
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RESPONSE --> ** The PE of the pendulum at displacement x is .5 k x^2. By conservation of energy, if nonconservative forces are negligible, we find that the KE of the pendulum at position x is.5 k A^2 - .5 k x^2. This result is obtained from the fact that at max displacement A the KE is zero, and the KE change from displacement A to displacement x is the negative of the PE change between these points. Thus .5 m v^2 = .5 k A^2 - .5 k x^2. Solving for v we have v = +- sqrt( .5 k / m * (A^2 - x^2) ) . **
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18:28:42 GOOD STUDENT ANSWER: If we pullback a pendulum of length L a distance x (much smaller than L), and stop the motion at the equilibrium point (vertical limit of motion) a washer on the pendulum will become a projectile and project off the pendulum, to land at a distance from which we can determine the horizontal velocity of the washer. That velocity is the same as the max velocity of the pendulum, since the max velocity is that which is at the lowest point in its path.
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RESPONSE --> GOOD STUDENT ANSWER: If we pullback a pendulum of length L a distance x (much smaller than L), and stop the motion at the equilibrium point (vertical limit of motion) a washer on the pendulum will become a projectile and project off the pendulum, to land at a distance from which we can determine the horizontal velocity of the washer. That velocity is the same as the max velocity of the pendulum, since the max velocity is that which is at the lowest point in its path.
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18:29:07 From the weight of the driver and the compression of the spring, we determine the spring constant (the 'stiffness' of the spring in N / m): driver weight of 68 kg * 9.8 m/s^2 = 670 N compresses the spring .05 meters, so since | F | = k | x | we have k = | F | / | x | = 670 N / (.05 m) = 13,400 N / m. Now from the force constant and the mass of the system we have omega = sqrt(k / m) = sqrt( (13,400 N/m) / (1570 kg) ) = 3 sqrt( (N/m) / kg) ) = 3 sqrt( (kg / s^2) / kg) = 3 s^-1, or 3 cycles / second.
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RESPONSE --> From the weight of the driver and the compression of the spring, we determine the spring constant (the 'stiffness' of the spring in N / m): driver weight of 68 kg * 9.8 m/s^2 = 670 N compresses the spring .05 meters, so since | F | = k | x | we have k = | F | / | x | = 670 N / (.05 m) = 13,400 N / m. Now from the force constant and the mass of the system we have omega = sqrt(k / m) = sqrt( (13,400 N/m) / (1570 kg) ) = 3 sqrt( (N/m) / kg) ) = 3 sqrt( (kg / s^2) / kg) = 3 s^-1, or 3 cycles / second.
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18:29:59 ** The PE of the system will be .5 k A^2, where A = .2 m and k = F / x = 80 N / (.2 m) = 400 N / m. The KE of the released ball will in the ideal case, which is assumed here, be .5 m v^2 = .5 k A^2. Solving for v we obtain v = +- sqrt( k A^2 / m ) = +- sqrt( 400 N/m * (.2 m)^2 / (.15 kg) ) = +- sqrt( 106 m^2 / s^2) = +-10.3 m/s, approx. The speed of the ball is the magnitude 10.3 m/s of the velocity. **
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RESPONSE --> ** The PE of the system will be .5 k A^2, where A = .2 m and k = F / x = 80 N / (.2 m) = 400 N / m. The KE of the released ball will in the ideal case, which is assumed here, be .5 m v^2 = .5 k A^2. Solving for v we obtain v = +- sqrt( k A^2 / m ) = +- sqrt( 400 N/m * (.2 m)^2 / (.15 kg) ) = +- sqrt( 106 m^2 / s^2) = +-10.3 m/s, approx. The speed of the ball is the magnitude 10.3 m/s of the velocity. **
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18:30:19 **The solution given here is for restoring force constant 210 N/m and mass .250 kg. You should be able to adapt your solution accordingly, and you should understand why the angular frequency will be sqrt(305 * .250 / (210 * .260)) times as great as that given here.The angular frequency of the oscillation (the angular velocity of the point on the reference circle) is omega = sqrt(k / m), with k = 210 N/m and m = .250 kg. The equation of motion could be y = A sin(omega * t). We obtain omega = sqrt( 210 N/m / (.250 kg) ) = sqrt( 840 s^-2) = 29 rad/s, approx.. A is the amplitude 28 cm of motion. So the equation could be y = 28 cm sin(29 rad/s * t). The motion could also be modeled by the function 28 sin (29 rad/s * t + theta0) for any theta0. The same expression with cosine instead of sine would be equally valid, though for any given situation theta0 will be different for the cosine model than for the sine model. **
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RESPONSE --> **The solution given here is for restoring force constant 210 N/m and mass .250 kg. You should be able to adapt your solution accordingly, and you should understand why the angular frequency will be sqrt(305 * .250 / (210 * .260)) times as great as that given here.The angular frequency of the oscillation (the angular velocity of the point on the reference circle) is
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18:30:35 Univ. 13.74 (13.62 10th edition). 40 N force stretches spring .25 m. What is mass if period of oscillation 1.00 sec? Amplitude .05 m, position and vel .35 sec after passing equil going downward?
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RESPONSE --> problem not required
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18:30:39 GOOD PARTIAL STUDENT SOLUTION WITH INSTRUCTOR COMMENT I am sorry to say I did not get that one--but mostly because I am hurrying through these, and I could not locate in my notes, altough I remember doing extensive work through the T=period problems--let me look at Set 9 for a moment. I think I found something now. If `omega = 2`pi/T and t = 1 sec, `omega = 2pi rad/s If I convert to accel, thenI can find the mass by way of F = ma. a = `omega ^2 * A. I do not know A yet so that is no good. }If A = x then my pullback of x = .25 m would qualify as A, so a = (2`pi rad/s) ^2 * .25 m = 9.87 m/s^2 So m = F/a = 40.0N/9.87 m/s^2 = 4.05 kg THAT IS PART A. INSTRUCTOR COMMENT: ** Good. But note also that you could have found m = k / omega^2 from omega = sqrt(k/m). F = -k x so 40 N = k * .25 m and k = 160 N/m. Thus m = 160 N/m / (2 pi rad/s)^2 = 4 kg approx.. STUDENT SOLUTION TO PART B:For part B If A = .050m and T = 1 sec, then the position can be found using the equation, x = A cos(`omega *t) INSTRUCTOR COMMENT: ** You could model this situation with negative omega, using x = .05 m * sin(-omega * t). This would have the mass passing thru equilibrium at t = 0 and moving downward at that instant. Then at t = .35 s you would have x = .05 m * sin( - 2 pi rad/s * .35 s ) = .05 m * sin( -.22 rad) = -.040 m, approx.. Velocity would be dx/dt = - 2 pi rad/s * .05 m * cos(-2 pi rad/s * .35 s) = -.18 m/s, approx.. Alternatively you might use the cosine function with an initial angle theta0 chosen to fulfill the given initial conditions: x = .05 m * cos(2 pi rad/s * t + theta0), with theta0 chosen so that at t = 0 velocity dx/dt is negative and position is x = 0. Since cos(pi/2) and cos(3 pi/2) are both zero, theta0 will be either pi/2 or 3 pi/2. The velocity function will be v = dx/dt = -2 pi rad/s * .05 m sin(2 pi rad/s * t + theta0). At t = 0, theta0 = pi/2 will result in negative v and theta0 = 3 pi/2 in positive v so we conclude that theta0 must be pi/2. Our function is therefore x(t) = .05 m * cos(2 pi rad/s * t + pi/2). This could also be written x(t) = .05 m * cos( 2 pi rad/s * ( t + 1/4 sec) ), indicating a 'time shift' of -1/4 sec with respect to the function x(t) = .05 m cos(2 pi rad/s * t). **
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