course PHY201 i?????H????????assignment #029????????????X?|?Physics I
......!!!!!!!!...................................
21:50:38 Query class notes #28. Explain how we can calculate the average angular velocity and the angular acceleration of an object which rotates from rest through a given angle in a given time interval, assuming constant angular acceleration.
......!!!!!!!!...................................
RESPONSE --> Angular velocity is calculated by the change in angle divided by the change in time. v = d(theta) / dt Angular acceleration is the change in angular velocity divided by the change in time. a = dv / dt
.................................................
......!!!!!!!!...................................
21:50:48 **This situation is strictly analogous to the one you encountered early in the course. As before acceleration is change in velocity / change in clock time. However now it's angular acceleration. We have angular acceleration = change in angular velocity / change in clock time. The average angular velocity is change in angular position / change in clock time. This question assumes you know the angle through which the object rotates, which is its change in angular position, as well as the change in clock time. So you can calculate the average angular velocity. If angular accel is uniform and initial angular velocity is zero then the final angular velocity is double the average angular velocity. In this case the change in angular velocity is equal to the final angular velocity, which is double the average angular velocity. From this information you can calculate angular acceleration. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
22:01:49 Principles of Physics and General College Physics Problem 7.46: Center of mass of system 1.00 kg at .50 m to left of 1.50 kg, which is in turn .25 m to left of 1.10 kg.
......!!!!!!!!...................................
RESPONSE --> To find the center of mass, we must use the formula: Xcm = [(m1 * x1) + (m2 * x2) + (m3 * x3)] / [m1+ m2 + m3] Xcm = [(1kg * 0m) + (1.5kg * .5m) + (1.1kg * .25m)] / [1kg+ 1.5kg + 1.1kg] = [0 + .75kg * m + .275kg * m] / 3.6kg = [1.025kg * m] / 3.6kg = .285m Xcm = .285m
.................................................
......!!!!!!!!...................................
22:03:16 Using the position of the 1.00 kg mass as the x = 0 position, the other two objects are respectively at x = .50 m and x = .75 m. The total moment of the three masses about the x = 0 position is 1.00 kg * (0 m) + 1.50 kg * (.50 m) + 1.10 kg * (.75 m) = 1.58 kg m. The total mass is 1.00 kg + 1.50 kg + 1.10 kg = 3.60 kg, so the center of mass is at position x_cm = 1.58 kg m / (3.60 kg) = .44 meters, placing it a bit to the left of the 1.50 kg mass.
......!!!!!!!!...................................
RESPONSE --> I entered in .25 for x3 instead of .75.
.................................................
......!!!!!!!!...................................
22:19:09 Query problem 7.50 3 cubes sides L0, 2L0 and 3L0; center of mass.
......!!!!!!!!...................................
RESPONSE --> According tot eh diagram in the book... 1lo^3 = 1 2lo^3 = 8 3lo^3 = 27 2lo is 8 times greater than lo 3lo is 27 times greater than lo For each block the center of mass relative to the x-line. x1 = 1/2 x2 = 2 x3 = 4.5 The equation: Xcm = [(m1 * x1) + (m2 * x2) + (m3 * x3)] / [m1 + m2 + m3] Xcm = [(m * .5) + (8m * 2) + (27m * 4.5)] / [m + 8m + 27m] = [ .5m + 16m + 121.5m] / 36m =138m / 36m = 3.83 Xcm = 3.83 lo
.................................................
......!!!!!!!!...................................
22:19:13
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
22:19:58 ** The mass of the second will be 2^3 = 8 times as great as the first. It takes 8 1-unit cubes to make a 2-unit cube. The mass of the third will be 3^3 = 27 times as great as the first. It takes 27 1-unit cubes to make a 3-unit cube. In the x direction the distance from left edge to center of first cube is 1/2 L0 (the center of the first cube). In the y direction the distance is from lower edge to center of the first cube is 1/2 L0 (the center of the first cube). In the x direction the distance from left edge to center of the second cube is L0 + L0 (the L0 across the first cube, another L0 to the center of the second), or 2 L0. In the y direction the distance from lower edge to center of the second cube is L0 (the center of the second cube). In the x direction the distance from left edge to center of the third cube is L0 + 2 L0 + 3/2 L0 (the L0 across the first cube, another 2 L0 across the second and half of 3L0 to the center of the third), or 9/2 L0. In the x direction the distance from lower edge to center of the first cube is 3/2 L0 (the center of the third cube). Moments about left edge and lower edge of first cube: If m1 is the mass of the first cube then in the x direction you have total moment m1 * L0/2 + 8 m1 * (2 L0) + 27 M1 * (9/2 L0) = 276 m1 L0 / 2 = 138 m1 L0. The total mass is m1 + 8 m1 + 27 m1 = 36 m1 so the center of mass is at center of mass in x direction: 138 m1 L0 / (36 m1) = 3.83 L0. In the y direction the moment is m1 * L0/2 + 8 m1 * (L0) + 27 m1 * ( 3/2 L0) = 45 m1 L0 so the center of mass is at center of mass in y direction: 45 m1 L0 / (36 m1) = 1.25 L0. **
......!!!!!!!!...................................
RESPONSE --> I did not do ""y.""
.................................................
......!!!!!!!!...................................
22:22:14 What is the mass of the second cube as a multiple of the mass of the first?
......!!!!!!!!...................................
RESPONSE --> If the first cube is lo and the second one 2lo, then that side is twice as long. 1 lo cubed is 1 2 lo cubed is 8 cube 2 is 8 times larger than cube 1.
.................................................
......!!!!!!!!...................................
22:22:20 ** 3 dimensions: the mass will be 2^3 = 8 times as great. It takes 8 1-unit cubes to make a 2-unit cube. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
22:23:47 What is the mass of the third cube as a multiple of the mass of the first?
......!!!!!!!!...................................
RESPONSE --> Same concept as the last problem. 1lo cubed = 1^3 = 1 3lo cubed = 3^3 = 27 Cube 3 is 27 times the mass of cube 1.
.................................................
......!!!!!!!!...................................
22:23:50 ** The mass of the third cube is 3^3 = 27 times the mass of the first. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
22:25:55 How far from the outside edge of the first cube is its center of mass?
......!!!!!!!!...................................
RESPONSE --> If CM is the center of a cube, then for cube 1 with a length of ""lo"" will be half of ""lo"". 1/2 lo or .5 lo
.................................................
......!!!!!!!!...................................
22:26:02 ** In the x direction the distance is 1/2 L0 (the center of the first cube). In the y direction the distance is also 1/2 L0 (the center of the first cube). **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
22:29:14 How far from the outside edge of the first cube is the center of mass of the second cube?
......!!!!!!!!...................................
RESPONSE --> Same as before... x-direction: half of 2 lo is 1 lo. 1 lo + 1 lo = 2 lo
.................................................
......!!!!!!!!...................................
22:29:17 ** In the x direction the distance is L0 + L0 (the L0 across the first cube, another L0 to the center of the second), or 2 L0. In the x direction the distance is L0 (the center of the second cube). **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
22:32:02 How far from the outside edge of the first cube is the center of mass of the third cube?
......!!!!!!!!...................................
RESPONSE --> Similiar to before.... 3 / 2 = 1.5 1 lo + 2 lo + 1.5 lo = 4.5 lo
.................................................
......!!!!!!!!...................................
22:32:08 ** In the x direction the distance is L0 + 2 L0 + 3/2 L0 (the L0 across the first cube, another 2 L0 across the second and half of 3L0 to the center of the third), or 9/2 L0. In the x direction the distance is 3/2 L0 (the center of the third cube). **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
22:33:16 How do you use these positions and the masses of the cubes to determine the position of the center of mass of the system?
......!!!!!!!!...................................
RESPONSE --> Plug them into the formula: Xcm = [(m1 * x1) + (m2 * x2) + (m3 * x3)] / [m1 + m2 + m3]
.................................................
......!!!!!!!!...................................
22:33:24 ** In the x direction you have moment m1 * L0/2 + 8 m1 * (2 L0) + 27 M1 * (9/2 L0) = 276 m1 L0 / 2 = 138 m1 L0. The total mass is m1 + 8 m1 + 27 m1 = 36 m1 so the center of mass is at 138 m1 L0 / (36 m1) = 3.83 L0. In the y direction the moment is m1 * L0/2 + 8 m1 * (L0) + 27 m1 * ( 3/2 L0) = 45 m1 L0 so the center of mass is at 45 m1 L0 / (36 m1) = 1.25 L0. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
22:49:25 Univ. 8.94 (8.82 10th edition). 45 kg woman 60 kg canoe walk starting 1 m from left end to 1 m from right end, moving 3 meters closer to the right end. How far does the canoe move? Water resistance negligible.
......!!!!!!!!...................................
RESPONSE --> woman = 45kg canoe = 60kg canoe = 5m 1 meter from the left: Xcm = [(m1 * x1) + (m2 * x2)] / [m1 + m2] = [(45kg * 1m) + (60kg * 2.5m)] / [45kg + 60kg] = 195kg * m / 105kg = 1.857m 1 meter from the right: Xcm = [(m1 * x1) + (m2 * x2)] / [m1 + m2] = [(45kg * 4m) + (60kg * 2.5m)] / [45kg + 60kg] = 330kg * m / 105kg = 3.14m 3.14m - 1.857m = 1.283m Since CM does not move, so the canoe must move 1.283m.
.................................................
......!!!!!!!!...................................
22:50:05 ** Since water resistance is negligible the net force acting on the system is zero. Since the system is initially stationary the center of mass of the system is at rest; since zero net force acts on the system this will continue to be the case. Assuming that the center of mass of the canoe is at the center of the canoe, then when the woman is 1 m from the left end the center of mass of the system lies at distance c.m.1 = (1 m * 45 kg + 2.5 m * 60 kg) / (45 kg + 60 kg) = 195 kg m / (105 kg) = 1.85 m from the left end of the canoe. A similar analysis shows that when the woman is 1 m from the right end of the canoe, then since she is 4 m from the left end the center of mass lies at c.m.2 = (4 m * 45 kg + 2.5 m * 60 kg) / (45 kg + 60 kg) = 310 kg m / (105 kg) = 2.97 m. The center of mass therefore changes its position with respect to the left end of the canoe by about 1.1 meters toward the right end of the canoe. Since the center of mass itself doesn't move the canoe must move 1.1 meters toward the left end, i.e., backwards. Note that since the woman moves 3 m forward with respect to the canoe and the canoe moves 1.3 m backwards the woman actually moves 1.7 m forward. The sum -1.3 m * 60 kg + 1.7 m * 45 kg is zero, to within roundoff error. This is as it should be since this sum represents the sum of the changes in the centers of mass of the canoe and the woman, which is the net change in the position of center of mass. **
......!!!!!!!!...................................
RESPONSE -->
.................................................