course PHY201 ϖ`óeassignment #003
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22:32:01 Practice Test 2 Problem Number 1. Explain how we used a rubber band and a rail to demonstrate the conservation of the F `ds quantity.
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RESPONSE --> I do not remember using a rubberband and a rail in a lab. I do know that F * ds = dKE. I also know that as the rubberband is stretched out is builds PE. When the rubberband is released, the system begins to lose PE and gain KE. Energy is conserved as no energy is lost in the system. It is made up of PE, KE, and a small portion that is coverted into thermal energy.
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22:32:06 GOOD STUDENT ANSWER: We calculate the potential energy of the system when the rubber band is fully stretched, and compare with the F `ds total as the rail slides across the floor to see if all the potential energy was dissipated against friction.
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22:39:32 Problem Number 2 An Atwood machine consists of 34 paper clips, each of mass .4 grams, suspended from each side of a light pulley. If 4 clips are transferred from one side to the other, what will be the total gravitational force on the system?
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RESPONSE --> If an atwood machine has 34 paper clips on each side; and is then subject to 4 clips being transferred to the other side, that would make one side consist of 30 paper clips, and the other 38 paper clips. If each clips has a mass of .4g we discover... 38 paper clips * .4g = 0.0152kg 30 paper clips * .4g = 0.012kg F = mg F1 = (0.0152kg)(9.8m/s^2) = 0.14896N F2 = (0.012kg)(9.8m/s^2) = 0.1176N Since the masses are attached to each other by means of a cable, the force of gravity of the two masses work against each other. F1 - F2 = Fnet 0.14896N - 0.1176N = 0.03136N
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22:39:40 GOOD STUDENT SOLUTION: Well Im assuming that there are 34 paper clips on each side. So; 34 * .0004 kg= .0136 kg on one side when even 4 * .0004 kg = .0016 kg less on the side that has 4 removed so that gives you .0152 kg on one side and .012 kg on the other now to get the gravitational force on the system you multiply both by 9.8 m/sec^2 and that will give you the force acting on each one, which are: .14896 N and .1176 N respectively. Now to get the net gravitational force just subtract and that will give you a net gravitational force of GravFnet = .03136 N
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22:45:22 If the frictional force exerted by the pulley is .05 times the total weight of the system, then what is the net accelerating force?
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RESPONSE --> To find the frictional force we need to determine the sums of the forces and multiply the sum by the coefficient of friction. F1 + F2 = Ft 0.14896N + 0.1176N = 0.26656 F * fr = Fr 0.26656 * .05 = 0.013328N Fg - Fr = Fnet 0.03136N - 0.013328N = 0.018032N
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22:46:30 well take the weights given above and combine them, then multiply by the .05, that will give us: [(.0136 kg + .0016 kg) * 9.8 m/sec^2] * .05= .007448 N for the frictional force now subtract GravFnet Ff = Fnet, or 0.03136 N - .007448 N = .023912 N net force.
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RESPONSE --> Isnt one side 30 paper clips and the other 38 paper clips?
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22:49:17 What therefore is the acceleration of the system?
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RESPONSE --> To solve this answer we must find the total mass. Paper clips 30 + 38 = 68 68 * .4g = 27.2g = .0272kg Fnet = ma Fnet/m = a 0.018032N / .0272kg = .6629 m/s^2 a = .6629m/s^2
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22:49:23 Divide net force by the total mass of the system to get the acceleration of the system .023912 N / .0152 kg = 1.573 m/sec^2
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22:59:40 Problem Number 3 What will be the tension in the string holding a ball which is being swung in a circle of radius 1 meters, if the ball is making a complete revolution every .3 seconds? Assume that the system is in free fall (e.g., in a freely falling elevator, in orbit, etc.)?
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RESPONSE --> If the ball makes one revolution in .3s, then we should use the period to determine the frequency. f = 1/T f = 1/.3s = 3.33 Hz = 3.33 rps 3.33rps * 2pi = 20.94 rad/s This value is known as angular velocity (w). v = rw v = (1m)(20.94 rad/s) = 20.94m/s a = v^2/r a = (20.94m/s)^2 / 1m = 438.48m/s^2 alpha = a/r alpha = 438.48m/s^2 / 1m = 438.48 rad/s^2 If we assume that the system is in freefall than the mass will experience weightlessness. To find the tension, we need to find F = ma; but we have no value for m. m(438.48m/s^2) = F
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22:59:47 STUDENT ATTEMPT: Alright Im not very good at this YET, so I could use some help!!!!!!!! I know what is given in the problem. From this I can figure: angular velocity = 2pi*(f) where f = 3.33333 rev/sec so angular velocity = 20. 94359102 sec^-1 and because 1m* omega = linear velocity is 20. 944 m/sec AND THE REST I GET LOST ON, IM HAVING PROBLEMS WITH ALL THESE RELATIONSHIPS!!!!!!!!!!!!!!!! INSTRUCTOR NOTE: ** Centripetal acceleration is v^2 / r. That's the key thing you're forgetting here. This gives you aCent = (20.94 m/s)^2 / (1 m)^2 = 440 m/s^2. The centripetal force would therefore be mass_ball * 440 m/s^2. **
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23:06:20 What would be the tension in the string if the system was on and stationary with respect to the surface of the Earth, with the ball being swung in a vertical circle, when the ball is at the top of its arc?
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RESPONSE --> When the ball is at the top of the circle both the tension force and gravity are pointing in the direction toward earth. Therefore, Ft + mg = ma Ft = ma - mg Ft = m(438.48m/s^2) - m(9.8m/s^2) Ft = m(428.68m/s^2)
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23:06:31 The net force on the ball is always equal to the centripetal force Fcent = mass_ball * 440 m/s^2. Gravity exerts a force equal to mass_ball * 9.8 m/s^2. If the ball is at the top of its arc the net force consists of the tension plus the gravitational force, both of which act in the same direction, which is downward. So using downward as the positive direction we have tension + mass_ball * 9.8 m/s^2 = mass_ball * 440 m/s^2. We find that tension = mass_ball * 440 m/s^2 - mass_ball * 9.8 m/s^2 = mass_ball * 430.2 m/s^2. You could substitute a reasonable mass (e.g., .4 kg) for the ball and obtain all quantities in Newtons. Note that these calculations are not in fact accurate to 4 significant figures. The numbers shown here are intended to demonstrate how the tension differs from the centripetal force.
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23:08:36 What if the ball is at the bottom of its arc?
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RESPONSE --> If the ball is at the bottom of the circle, the tension force is pointing back toward teh center, and gravity is pointing toward the earth. Ft - mg = ma Ft = ma + mg Ft = m(448.28m/s^2)
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23:08:41 If the ball is at the bottom of its arc the net force consists of the tension plus the gravitational force, with tension up and gravity down. Again using downward as the positive direction and noting that the upward centripetal acceleration is with this assumption negative we have tension + mass_ball * 9.8 m/s^2 = -mass_ball * 440 m/s^2. We find that tension = -mass_ball * 440 m/s^2 - mass_ball * 9.8 m/s^2 = -mass_ball * 449.8 m/s^2.
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23:12:58 What if the ball is at its halfway height?
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RESPONSE --> When the ball is at the halfway point the tension force is pointing toward teh center of the circle and gravity is acting downward. The forces here are perpendicular to each other. Therefore, only the tension force is the acting force.
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23:13:03 At the halfway height the centripetal force and gravitational acceleration are perpendicular and hence independent. The only force acting toward the center is the tension in the string, which must therefore supply the entire centripetal force. The tension is mass_ball * 440 m/s^2 toward the center.
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23:15:02 Problem Number 4 If we set the expression G M / r^2 for the gravitational acceleration at distance r from the center of a planet of mass M equal to the centripetal acceleration v^2 / r and solve for v in terms of r, what is the result?
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RESPONSE --> GM / r^2 = v^2 / r (multiply by r) GMr / r^2 = v^2 (simplify) GM/r = v^2 (find square root) v = sqrt(GM / r)
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23:15:36 Setting the two expressions equal we have the equation G M / r^2 = v^2 / r. To solve for v we first multiply both sides by r to get G M r = v^2, Taking the square root of both sides and reversing sides of the equation gives us v = sqrt(G M r).
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23:17:53 Problem Number 5 A disk of negligible mass and radius 30 cm is constrained to rotate on a frictionless axis about its center. On the disk are mounted masses of 7 gram at a distance of 23.1 cm from the center, 24 grams data distance of 16.8 cm from the center and 47 grams at a distance o 9 cm from the center. A uniform force of .06983 Newtons is applied at the rim of the disk in a direction tangent to the disk.
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RESPONSE --> What is the question?
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23:25:13 What will be the angular acceleration of the disk?
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RESPONSE --> Okay. A 30cm disk has a tangent force of .06983N applied to it, and the disk contains 3 mounted masses. First we must determine the total inertia. I1 = mr^2 (7g)(23.1cm)^2 = 3.74 * 10^-4 kg * m^2 I2 = mr^2 (24g)(16.8cm)^2 = 6.77 * 10^-4 kg * m^2 I3 = mr^2 (47g)(9cm)^2 = 3.81 * 10^-4 kg * m^2 Now add them together: It = I1 + I2 + I3 It = 3.74 * 10^-4 kg * m^2 + 6.77 * 10^-4 kg * m^2 + 3.81 * 10^-4 kg * m^2 It = 1.432 * 10^-3 Now we know Tau = I * alpha, so we must find Tau. Tau = r * F * sin(90deg) Tau = (.30m)(.06983N)(1) = .020949 mN alpha = Tau / I alpha = .020949 mN / 1.432 * 10^-3 alpha = 1.463 * 10^-5 rad/s^2
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23:25:24 ** The moment of inertia of the system is the sum of all the m r^2 contributions of the individual particles. The net torque is the product of the net force and the moment arm. Newton's 2d law F = m a, expressed in in angular form, is `tau = I `alpha, where `tau is net torque (analogous to force), I is moment of intertia (sum of all mr^2, analogous to mass) and `alpha is angular acceleration in rad/s^2. **
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23:27:52 If the force is applied for 4 seconds with the disk initially at rest, what angular velocity with the disk attain?
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RESPONSE --> alpha * dt = dw 1.432*10^-5rad/s^2 * 4s = 5.852 * 10^-5 rad/s w = 5.852 * 10^-5 rad/s
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23:28:05 ** Once you know angular acceleration it's easy to find change in angular velocity **
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RESPONSE --> Yep.
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23:29:18 What then will be the speed of each of the masses?
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RESPONSE --> They will all be the same. Angular velocity is the same for all points since they cover the same angle change during a time interval.
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23:29:37 ** You know angular velocity and distance of each mass from the axis of rotation. Angular velocity is the velocity of the mass along the arc divided by the radius. So what is the velocity of each of the masses along its arc? **
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23:31:37 What will be their total kinetic energy?
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RESPONSE --> Rotational KE = .5 * I * w^2 KE = (.5)(1.432 * 10^-3 kg * m^2)(5.852 * 10^-5) KE = 2.452 * 10^-12 J
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23:32:02 *&*& Add up the individual kinetic energies.
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RESPONSE --> I just used the total inertia.
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23:32:35 Compare the total kinetic energy to the change in the quantity .5 I `omega^2.
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RESPONSE --> Should be the same. That is the formula that i used.
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23:32:38 *&*& The two quantities should be the same.
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23:34:39 Problem Number 6 A gun fires a bullet of mass 33 grams out of a barrel 38 cm long, from which it exits at 267 m/s. Assuming that the bullet accelerates uniformly from rest along the length of the barrel
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23:37:37 How long does it take the bullet to exit the barrel after the powder ignites?
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RESPONSE --> If a bullet starts at rest and accelerates to 267m/s in a distance of 38cm. Then we can calculate the time interval from start to finish. (Vo + Vf)/2 = vAve (0m/s + 267m/s) / 2 = 133.5m/s dt = ds / vAve dt = .38m / 133.5m/s = .00285s
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23:37:39 Well I know that v0= 0 and vf= 267 m/sec and `ds= .38 m and from there I can use vf^2 = v0^2 + 2a`ds so I figure a= 93801 m/ sec^2 From there I can find the change in time to be 2.846441948 * 10^- 3 ** OK but it's easier to average init and final vel then divide into displacement (e.g., vAve = (0 + 267) / 2 m/s = 133.5 m/s so to move .38 meters requires .38 m / (133.5m/s) = .028 sec approx. **
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23:40:20 Using the Impulse-Momentum Theorem determine the average force exerted on the bullet as it accelerates along the length of the barrel.
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RESPONSE --> Momentum is calculated.. p = mv p = (.033kg)(267m/s) = 8.811 F = p / dt F = 8.811 / .00285s = 3091.58N
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23:40:25 By the impulse-momentum theorem Fave `dt = m `dv we get Fave = (m`dv)/ `dt so from there I can substitute in the values already determined and find the force to be 3095.4434 N ** Good work. Impulse-momentum is appropriate here. It would also have been possible to find acceleration and multiply by mass, but impulse-momentum is more direct. **
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23:42:56 Using Newton's Second Law determine the average force exerted on the bullet as it accelerates along the length of the barrel.
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RESPONSE --> We have to find the acceleration. Then use F = ma. dv / dt = a (Vf - Vo) / dt = a (267m/s - 0m/s) / .00285s = 93684.21m/s^2 a = 93684.21m/s^2 F = ma F = (.033kg)(93684.21m/s^2) = 3091.58N They are the same.
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23:43:00 i dont know what to do here. ** Sure you do. You found the acceleration above when you didn't need it. You need it now. **
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