1cat2ferrets Posted February 23, 2010 Share Posted February 23, 2010 I think we got the first question so I'll type it out because it's part of other questions that we're having trouble with. 1. A child riding a horse on the outer edge of a merry-go-round has a rotational speed of 5RPM and a tangential speed of 8m/s. A child riding a horse halfway between the outside and center of the merry-go-round would have a. the same rotational speed and half the tangential speed, b. the same rotational speed but twice the tangential speed, c. half the rotational speed and half the tangential speed,d. half the rotational speed but the same tangential speed. My son chose answer B. Here are the two questions we're having trouble with: 2. What factor would INCREASE the angular MOMENTUM experienced by the child riding on a horse at the outer edge of the merry-go-round? a. The child changes horses and moves to one nearer the center of the merry-go-round, b. The child wears a 3kg backpack, c. The merry-go-round slows down, d. none of these. My son and I think it's A. Are we right? 3. If the child riding on the outer edge of the merry-go-round in the first question above(my question#1) has a mass of 25kg and is 15m from the center of the ride, what would be his or her angular momentum? the choices are: a. 3kg*m^2/s b. 30kg*m^2/s c. 300kg*m^2/s d. 3000kg*m^2/s the angular momentum formula in the textbook is: angular moment= rotational inertia * rotational velocity PLEASE help us figure these two questions out! TIA! Quote Link to comment Share on other sites More sharing options...
mpcTutor Posted February 23, 2010 Share Posted February 23, 2010 I think we got the first question so I'll type it out because it's part of other questions that we're having trouble with. 1. A child riding a horse on the outer edge of a merry-go-round has a rotational speed of 5RPM and a tangential speed of 8m/s. A child riding a horse halfway between the outside and center of the merry-go-round would have a. the same rotational speed and half the tangential speed, b. the same rotational speed but twice the tangential speed, c. half the rotational speed and half the tangential speed, d. half the rotational speed but the same tangential speed. My son chose answer B. Answer (a) Tangential speed = radius x rotational speed. (Recall v = r*w) Since w (omega) is constant the tangential speed will vary directly with radius alone. Half way from axis of rotation means half tangential speed. Here are the two questions we're having trouble with: 2. What factor would INCREASE the angular MOMENTUM experienced by the child riding on a horse at the outer edge of the merry-go-round? a. The child changes horses and moves to one nearer the center of the merry-go-round, b. The child wears a 3kg backpack, c. The merry-go-round slows down, d. none of these. My son and I think it's A. Are we right? Answer (b) Angular momentum = Moment of inertia x angular velocity. With 3 kg bag Moment of inertia increases and hence angular momentum increases proportionately. The problem must assume that angular velocity is constant.) 3. If the child riding on the outer edge of the merry-go-round in the first question above(my question#1) has a mass of 25kg and is 15m from the center of the ride, what would be his or her angular momentum? the choices are: a. 3kg*m^2/s b. 30kg*m^2/s c. 300kg*m^2/s d. 3000kg*m^2/s the angular momentum formula in the textbook is: angular moment= rotational inertia * rotational velocity PLEASE help us figure these two questions out! TIA! Answer (d)Angular momentum = Moment of inertia x angular velocity. = (25 kg x 225 m^2) x (10 Pi/60) = 2945.24 kg*m^2/s = approx. 3000 kg*m^2/s (Recall I =m*r^2) Best regards. MPCTutor --------------------------------------------------------------------------- AP Calculus, AP Physics, Singapore Math Grades 7-12 --------------------------------------------------------------------------- US Central Time: 2:25 PM 2/23/2010 Quote Link to comment Share on other sites More sharing options...
1cat2ferrets Posted February 23, 2010 Author Share Posted February 23, 2010 Thanks so very much for the help, we truly appreciate it, mpcTutor! You're a godsend. Quote Link to comment Share on other sites More sharing options...
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