******************************************************************************* Name: steven burgess Submit: Remote User: Date: 02/17/2005 Time: 10:51 AM Period: Radius: Acceleration: bbangle: ******************************************************************************* Name: Jeff Submit: Submit Query Remote User: Date: 02/17/2005 Time: 11:09 AM Period: 74 frames for a half rotation, which is 2.47 seconds. Making the period of rotation 4.93 seconds in a period. Radius: one feet is .3048m. So 6 feet is 1.8288 meters. Acceleration: centripital acceleration = v^2/r v=d/t distance is 2*pi*r=11.4907m v=11.4907m/4.93s=2.33m/s v^/r=2.97 m/s^2 and since g= 9.81 m/s^2 centripital acc = .303 bbangle: we know that tan(theta) equals to the horizontal acceleration. so tan^-1(0.303) = 16.8568 degrees. ******************************************************************************* Name: Yuan Yang Submit: Submit Query Remote User: Date: 02/17/2005 Time: 11:10 AM Period: It took 75 frames for half a period of rotation and with the movie clip running at 30 frames of second, the period was 5.0 seconds because (75 frams * 2 / 30 frames per second). I measured the half rotation by selection of a fixed point and seeing when one person passes there and then the next time a person passes that same point. Radius: d = 12 ft r = 6 ft = 72 in = 182.88 cm = 1.83 m Acceleration: v = 2 * pi * radius of MGR / period v = 2 * 3.14 * 1.83m / 5.0 s v = 2.30 m/s a = v^2 / radius a = (2.30 m/s)^2/1.83m a = 2.89 m/s^2 a = 0.29 g bbangle: a = g tan(angle) a / g = tan(angle) angle = arctan(a/g) angle = arctan(.29) angle = 16.2° ******************************************************************************* Name: Matthew Submit: Submit Query Remote User: Date: 02/17/2005 Time: 11:11 AM Period: I chose a point on the path and counted the number of frames untill the person got back to that exact same point. I counted 150 frames, and we know that there are 30 frames in a second. Therefore, the period for one rotation is 5 seconds. Radius: The distance from one rider to the center is approximately 1.82 meters. I found this by converting feet to meters. We are given the diameter, so we know that the radius is half that. We simply mulitiply by the conversion factor to get the radius, ie distance to the center of the MGR, in meters. Acceleration: a/g= 0.29 I found this by substituting v=(2pi*r)/period into the equation a=v^2/r bbangle: The bb's would be at the angle 16.4 degrees. I found this by knowing that g*tan(theta)=a. I know the ratio of a/g, (as discovered above), so I took the inverse tangent of this. (I was also careful to be in degree mode as a horizontal accelerometer typically measures in degrees, not radians). ******************************************************************************* Name: Lucie Submit: Submit Query Remote User: Date: 02/17/2005 Time: 11:12 AM Period: The MGR makes one complete rotation, and it takes 152 frames to do so. If the movie clip runs at 30 frames per second, then the period would be 5.1 seconds. Radius: 6 feet is 1.8 meters Acceleration: v^2/r is the centripetal acceleration, which is (2pi*r/p)^2/r, and that is 2.8 m/s^2, which is about 0.28g. bbangle: I know that tan(theta)=a from my force diagrams and net force equations. We would expect an angle of 15.6 degrees, after taking the inverse tangent of the acceleration in g's. ******************************************************************************* Name: Amena Submit: Submit Query Remote User: Date: 02/17/2005 Time: 11:12 AM Period: The period of rotation of the MGR is 5 cycles per second. I found the answer by timing how counting 150 frames, and it was given that the movie clip runs 30 frames per second. Radius: The distance of one of the riders to the center of the MGR is 6 feet, which is about 1.83 meters. I found the answer by finding the radius of the circle the riders were turning in and converting that value into meters. Acceleration: v=2*pi*r/P v=2*(22/7)*1.83/5 v=2.3 m/s a=v2/r a=(2.3)^2/1.83 a=2.89 I found the answer by using the equations for velocity and acceleration and plugging in the values. bbangle: I think the angle that the bb's in the accelerometer would indicate is 90 degrees because the MGR is perpendicular to the ground and the rider would ideally be holding the accelerometer level. ******************************************************************************* Name: Eric Freeman Submit: Submit Query Remote User: Date: 02/17/2005 Time: 11:13 AM Period: period =1/f 150frames/(30frames/seconds) p=5.0sec Radius: 1ft=.3048m 12*.3048=3.7m=diameter 1.8m from a rider to the center. Acceleration: v=2pir/p v=2*3.14*1.8/5.0 v=2.3m/s a=v^2/r a=2.3^2/1.8 a=2.9m/s^2 bbangle: 90degrees ******************************************************************************* Name: Christian Submit: Submit Query Remote User: Date: 02/17/2005 Time: 11:14 AM Period: 5 s, i counted the number of frames it took for a full rotation to occur and i got 150 which i then divided by 30 to find the number of seconds it took. Radius: the distance from the rider to the center would be the radius which would be half the diameter, making it 6 feet. to find this in meters we use the conversion rate and find that it is 1.8288 meters Acceleration: the acceleration would be V^2/r and to find V I used distance/time which would be 2*pi*r/t where r is 1.8288 m, and t is 5 s so V is 2.298 m/s so the acceleration is 2.888 m/s^2 which i then divided by g or 9.81 m/s^2 to find the acceleration in g's which came out as .294 g's bbangle: 16.38 because using a horizantal acclelerometer acceleration=g*tan(x) where x is the angle and because x=tan^-1(a/g) so x equals 16.38 ******************************************************************************* Name: Submit: Submit Query Remote User: Date: 02/17/2005 Time: 11:16 AM Period: Radius: Acceleration: bbangle: ******************************************************************************* Name: Athena Submit: Submit Query Remote User: Date: 02/17/2005 Time: 11:17 AM Period: I counted how many frames there were in one full period, I found it to be 160 frames in one second. Because I know that the movie clip runs at 30 frames per second, I divided 160/30, and found the period to be approximately 5.3s. Radius: The diameter of the plank is 12 feet. The distance from one of the riders to the center is then 6 feet. Converting 6 feet to meters, I yielded 1.83m. Acceleration: The overall forces on the rider are the normal force and weight, which cancel out, and the centripedal force, which is mv^2/r. Then, using F=ma=mv^2/r, I found acceleration to be v^2/r. Velocity is change in distance over change in time. Velocity is 2.17m/s. Acceleration, therefore is 2.57m/s^2. In g's, it is .262g's. bbangle: a=gtan(theta). Therefore, theta=tan^-1(a/g). Theta is equal to 14.7. ******************************************************************************* Name: Steven Burgess Submit: Submit Query Remote User: Date: 02/17/2005 Time: 11:17 AM Period: One period on the MGR is 5.5 seconds. I did this by picking a point on the screen were the person in red sat and timed how long it took for the person to come back to the same position. This time would be one period. Radius: Each rider is approximately 1.8 meters from the center of the MGR. I first converted found out how many inches were in 12 feet. I then used the fact that 2.54 centimeters were in an inch. By multiplying the number inches by how many centimeters were in an inch, you could arrive at the diameter in centimeters. Acceleration: The centripetal acceleration is given by the expression a=V^2/r. I used the formula for the circumference of a circle and then divided the distance traveled for one period by the time to complete one period to get a value for the velocity. This method produced a value of 2.1 m/s as the velocity of the rider. You then plug this velocity into the above equation and it pruduces an acceleration of 2.3m/s^2. This value in g's would be .23g's. bbangle: We used the following equation to find theta: a=gtan(theta). When the values are plugged in, theta comes out to be 13 degrees. ******************************************************************************* Name: Donovan Patterson Submit: Submit Query Remote User: Date: 02/17/2005 Time: 11:17 AM Period: I started counting frames when the boy in the red shirt was closest to the camera. I then advanced the video one frame at a time to determine the number of frames for the MGR to rotate once. I counted 160 frames. I then divided this by the 30 frames per second. This gives me a period of 5.33 seconds. Radius: If the diameter of the plank is 12 feet, then the radius is half of that, or 6 feet. I know that there is 3.29 feet in every meter. To convert the radius into feet, I divided the 6 feet by 3.29. This gives me a radius of 1.83 meters. Acceleration: The centripetal acceleration of the rider is equal to velocity squared over radius. The veloicty is equal to the circumfrence times the period. So, the acceleration is (2*pi*r/P)^2/r. Plugging in the known values, we know that the acceleration is 2.54 meters per second sqaured. If you divide this value by 9.81, you will find the acceleration in g's. This value is .259 g's. bbangle: We know that the acceleration is equal to g*tan of the angle of the bb's. Solving for that angle, I found it equal to tan^-1 (a/g). This gives us an angle of 14.8 degrees. ******************************************************************************* Name: Amena Ross Submit: Submit Query Remote User: Date: 02/17/2005 Time: 11:23 AM Period: The period of rotation of the MGR is 5 cycles per second. I found the answer by timing how counting 150 frames, and it was given that the movie clip runs 30 frames per second. Radius: The distance of one of the riders to the center of the MGR is 6 feet, which is about 1.83 meters. I found the answer by finding the radius of the circle the riders were turning in and converting that value into meters. Acceleration: v=2*pi*r/P v=2*(22/7)*1.83/5 v=2.3 m/s a=v2/r a=(2.3)^2/1.83 a=2.89 I found the answer by using the equations for velocity and acceleration and plugging in the values. bbangle: The angle that I would expect the bb's in the accelerometer would indicate about 16.4 degrees. I found this answer by taking the inverse tangent of the acceleration/gravity. ******************************************************************************* Name: Isaac Owolabi Submit: Submit Query Remote User: Date: 02/17/2005 Time: 11:23 AM Period: The period is five seconds. I counted the number of frames it took for the student in the red shirt to go all the way around and divided this by thirty frames per second to find number of seconds. 160 frames divided by 30 frames per second equals 5 seconds. 160/30=5 Radius: They are about six feet away from the center of the plank and one foot is approximately .3048 meters so r=6*.3048=1.8288 Acceleration: a=v^2/r and v=2*pi*r/P and we know that P=5 and r=1.8288 so: v=2.30 and a=2.89 bbangle: tan^-1(a/g) where a=2.89 and g=9.81 so theta equals 16 degrees. I would expect sixteen degrees. ******************************************************************************* Name: joe Submit: Remote User: Date: 03/02/2006 Time: 09:37 AM Period: Radius: Acceleration: bbangle: ******************************************************************************* Name: tina Submit: Remote User: Date: 09/28/2006 Time: 06:03 PM Period: Radius: Acceleration: bbangle: ******************************************************************************* Name: rajah Submit: Remote User: Date: 03/22/2008 Time: 09:32 AM Period: Radius: Acceleration: bbangle: