#$&* course Phy 241 November 20 around 10:45am. 032. `query 32*********************************************
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Given Solution: `a** The bolts toward the outside of the wheel are moving at a greater velocity relative to some fixed point, so their kinetic energy is greater since k = 1/2 m v^2 ** STUDENT COMMENT: Oh... I need to think of it in terms of angular velocity INSTRUCTOR RESPONSE: Think in terms of angular velocity as well as velocity. At any instant all masses on the wheel have the same angular velocity, but the masses further from the center have greater velocity (and therefore greater KE) than those closer to the center. STUDENT COMMENT: i had the right idea here, but had it backward, i thought the closer to the fixed point the greater the velocity INSTRUCTOR RESPONSE: That would be the case for a satellite orbiting a planet. However in this case the entire wheel is rotating at a single angular velocity, so closer points don't move as fast as distinct points. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: `qWhat is the moment of inertia of the Styrofoam wheel and its bolts? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Bolt: mr^2 Styrofoam: 1/2MR^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The moment of inertia for the center of its mass=its radias times angular velocity. Moment of inertia of a bolt is m r^2, where m is the mass and r is the distance from the center of mass. The moment of inertia of the styrofoam wheel is .5 M R^2, where M is its mass and R its radius. The wheel with its bolts has a moment of inertia which is equal to the sum of all these components. ** STUDENT COMMENT: I = .5mr^2 for the disk I = .5mr^2 for each of the bolts INSTRUCTOR RESPONSE: The moment of inertia of a particle of mass m at distance r from the axis of rotation is m r^2. A particle has all its mass concentrated at one specific location. A hoop consists of a collection of particles, all at the same distance from the axis of rotation. If we add up the m r^2 contributions from all the particles in the hoop, we get M R^2, where M is the mass and R the radius of the hoop. Thus the moment of inertia of the hoop is M R^2. The disk consists of a collection of particle spread out at many different distances from the axis. If we 'cut up' the disk into individual particles, we find that the sum of the m r^2 contributions of the particles is 1/2 M R^2, where M is the mass of the disk and R its radius. Thus the moment of inertia of the disk is 1/2 M R^2. The mass of a bolt isn't all concentrated at a single distance from the axis, but all the particles that make up the bolt are pretty close to the center of the bolt, so it doesn't differ from a particle by much. Its moment of inertia is pretty close to m r^2, where m is the mass of the bolt and r its distance from the axis. You add the moment of inertia of the disk to the moments of inertia of the bolts, and you end up with the moment of inertia of the system. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK, I just needed to specify them as a whole, which would be it’s center of mass: radius*angular velocity ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: `qHow do we determine the angular kinetic energy of of wheel by measuring the motion of the falling mass? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If you know moment of inertia, then you can find angular KE of the wheel: =1/2*(moment of inertia)*(angular velocity^2) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** STUDENT ANSWER AND INSTRUCTOR CRITIQUE: The mass falls at a constant acceleration, so the wheel also turns this fast. INSTRUCTOR CRITIQUE: Acceleration isn't the rate at which something moves, or turns. It is the rate at which the velocity (which is itself the rate at which the object moves, or turns) changes. We don't use the acceleration to find the angular KE, we use the velocity. The acceleration, if known, can be used to find the velocity. The question was how we use measurements of the motion of the descending mass to find the angular KE: By observing position vs. clock time we can estimate velocities, and determine the velocity of the descending mass at any point. The string is wound around the rim of the wheel. So the rim of the wheel moves at the same speed as the string, which is descending at the same speed as the mass. So if our measurements give us the speed of the descending mass, we know the speed of the wheel. If we divide the velocity of the rim of the wheel by its radius we get the angular velocity of the wheel. Recall that angular velocity is designated by the symbol omega. • Assuming we know the moment of inertia of the wheel, we find its KE, which is equal to 1/2 I omega^2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I got part of the answer right, I should’ve shown HOW to get what I got. ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: `qQuery problem 8.60 uniform disk at 2.4 rev/sec; nonrotating rod of equal mass, length equal diameter, dropped concentric with disk. Resulting angular velocity? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Sphere: Inertia=2/5MR^2 Rod: Inertia=1/12ML^2 L=R (axis rotation) 24/60+5/60=29/60 (29/60)/(1/2)=29/30 Where do I go from here? confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The moment of inertia of the disk is I = 1/2 M R^2; the moment of inertia of the rod about its center is 1/12 M L^2. The axis of rotation of each is the center of the disk so L = R. The masses are equal, so we find that the moments of inertia can be expressed as 1/2 M R^2 and 1/12 M R^2. The combined moment of inertia is therefore 1/2 M R^2 + 1/12 M R^2 = 7/12 M R^2, and the ratio of the combined moment of inertia to the moment of the disk is ratio = (7/12 M R^2) / (1/2 M R^2) = 7/12 / (1/2) =7/12 * 2 = 7/6. Since angular momentum I * omega is conserved an increase in moment of inertia I results in a proportional decrease in angular velocity omega so we end up with final angular velocity = 6/7 * initial angular velocity = 6/7 * 2.4 rev / sec = 2.1 rev/sec, approximately. STUDENT COMMENT: I had no idea to do the ratio. I probably wouldn’t have ever thought of that either. INSTRUCTOR RESPONSE: You don't need to use the idea of ratio, it's simply convenient to do so. You could equivalently obtain the expression for the angular momentum in terms of initial angular momentum omega_0: • initial angular momentum of disk = I_disk * omega_0 = 1/2 M R^2 * omega_0 If omega_f is the angular momentum of the system after the rod is dropped then we have • final angular momentum of system = I_system * omega_f = 7/12 M R^2 * omega_f. No external torque acts on the system so its angular momentum remains constant. Thus initial angular momentum of disk = final angular momentum of system 1/2 M R^2 * omega_0 = 6/12 M R^2 * omega_f We solve this to get • omega_f = (1/2 M R^2) / (7/6 M R^2) * omega_0 = 6/7 * 2.4 rev/s = 2.1 rev / s, approx.. CORRECTION BY INSTRUCTOR The given solution incorrectly solve the problem for a sphere and a rod, not a disk and a rod. It used the moment of inertia 2/5 M R^2 for the sphere where it should have use 1/2 M R^2, the moment of inertia of a disk. The above solution can be easily revised to correct for this error. STUDENT QUESTION I am not sure how we know to use the values 1/2 or 1/12 in this situation??? I do see how we add the two values together though. I am still having a little trouble understanding this. Is there another example I can look at.??? INSTRUCTOR RESPONSE A uniform disk rotating about an axis through its center and perpendicular to its plane has moment of inertia 1/2 M R^2. A uniform sphere rotating about an axis through its center has moment of inertia 2/5 M R^2. A rod rotating about its center has moment of inertia 1/12 M L^2; rotating about its end the moment of inertia is four times as great, 1/3 M L^2. All students should know these formulas. Physics 231 students are expected to be able to derive these formulas, and other, using calculus. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK I just needed to multiply the final and initial angular velocities together and then times that by the given 2.4rev/sec. GOT IT! ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: `qUniv. 10.64 (10.56 10th edition). disks 2.5 cm and .8 kg, 5.0 cm and 1.6 kg, welded, common central axis. String around smaller, 1.5 kg block suspended. Accel of block? Then same bu wrapped around larger. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Moment of Inertia: (0.5)*[0.8*0.025^2]+(0.5)*[1.6*0.05^2]+[1.5*0.025] =0.0051 Torque= F*’dx =0.025*(1.5)*(9.807) =0.368mN Angular acceleration: (0.368)/(0.005)=73 Acceleration: 73*0.025=1.825m/s^2 (SMALL DISK) Moment of Inertia: (0.5)*[0.8*0.025^2]+(0.5)*[1.6*0.05^2]+[1.5*0.05^2] =0.006 Torque=F*’dx =0.05*(1.5)*(9.807) =0.7mN Angular acceleration: (0.7)/(0.006)=120 Acceleration: 120*0.05=6m/s^2 (LARGE DISK) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The moment of inertia of each disk is .5 M R^2; the block lies at perpendicular distance from the axis which is equal to the radius of the disk to which it is attached. So the moment of inertia of the system, with block suspended from the smaller disk, is I = .5 (.8 kg) * ( .025 m)^2 + .5 * 1.6 kg * (.05 m)^2 + 1.5 kg * (.025 m)^2= .0032 kg m^2 approx. The 1.5 kg block suspended from the first disk results in torque tau = F * x = .025 m * 1.5 kg * 9.8 m/s^2 = .37 m N approx. The resulting angular acceleration is alpha = tau / I = .37 m N / (.0032 kg m^2) = 115 rad/s^2 approx. The acceleration of the block is the same as the acceleration of a point on the rim of the wheel, which is a = alpha * r = 115 rad/s^2 * .025 m = 2.9 m/s^2 approx. The moment of inertia of the system, with block suspended from the larger disk, is I = .5 (.8 kg) * ( .025 m)^2 + .5 * 1.6 kg * (.05 m)^2 + (1.5 kg * .05 m)^2= .006 kg m^2 approx. The 1.5 kg block suspended from the first disk results in torque tau = F * x = .05 m * 1.5 kg * 9.8 m/s^2 = .74 m N approx. The resulting angular acceleration is alpha = tau / I = .74 m N / (.006 kg m^2) = 120 rad/s^2 approx. The acceleration of the block is the same as the acceleration of a point on the rim of the wheel, which is a = alpha * r = 120 rad/s^2 * .05 m = 6 m/s^2 approx. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ self-critique rating #$&*: