A kid on a merry go round is 5 meters from the center. He goes around once. How many degress has he gone though? How many radians has he spun through? Thank you :)
210045
davidosterberg1 is correct\n\nbelow is a link to the explanation\n\nhttp://id.mind.net/~zona/mmts/trigonometryRealms/radianDemo1/RadianDemo1.html
427651
One revolution, 360 degrees, pays out an amount of cable equal to the drum's circumference.  C = 2 pi r = 180 pi = 565.2 cm = 5.652 m\n\n.50 is .50/5.652 of a revolution or .50/5.652 of 360 which is 31.847 degrees per second  (angular speed)\n\n6/5.652 is how many revolutions
349544
Say that a boy runs 1000 km in 5 minutes.  A unit rate must have the second quantity at 1 unit, so divide both by 5 to get 200 km per min.\n\nThe height decreases so it shows a negative rate of change, perhaps -2 mm  per minute
547619
5,25rad/s = 2πf\nf= 0,836Hz\nVt= (2πr*f)\nVt= 2π*0,0600m*0,836Hz= 0,315m/s\nAc= (V^2)/r\nAc= (0,315m/s)^2/0,0600m= 1,66m/s^2\n\nBoth points have the same frecuency, since they have the same angular velocity. So, for point 2 you already have a velocity of 5,25rad/s.\nThen, Ac= (V^2)/r →  (4π^2*r^2*f^2)/r→ (2πf)^2*r\nTherefore, Ac α r.\nIf r is half, then Ac is half (1,66m/s^2)/2= 0,83m/s^2\nthe same for the velocity.\n Vt= (2πr*f), because the radius is half and Vt α r\nVt= 2π*(r/2)*f = π*r*f= π*0,0600m*0,836Hz= 0,158 m/s\nOr just (0,315m/s)/2= 0,158m/s
525542
Every object in circular motion, in fact, tries to go along a tangential path. But the centripetal force which acts perpendicular to this motion makes it move in circular path. \n\nTangential path means away from the center of the circle.\n\nIn a merry-go- round, every object tries to go away from the center (along a tangential path)\n\nBy Newton’s third law for every action, there is an equal but opposite reaction.\n\nHence, when every object tries to go away from the center, reaction forces are developed the resultant of which is the centripetal force.\n\nTherefore, in merry-go-round we feel as if we are made to move away from the center.\n\nIf there were no such reaction force, then there will be no centripetal force and we will be moving along a tangent at the point at which the centripetal force vanishes.\n\nIt can be easily understood in the case of a car taking a curve. When a car is moving very fast, some times the frictional force may not be enough to provide the necessary centripetal force and the car is skidding away from the center of the curved road.
403987
We'll use a similar approach as to your other question.\nAs before, the big hand moves at a rate of 6 degrees per minute, and the little hand at 0.5 degrees per minute.\nSo t minutes after midnight, the big hand has moved 6t degrees, and the little hand has moved 0.5t degrees. Now, we want the two angles to add up to a multiple of 360 degrees to make them in the required positions.\nSo we want 6.5t = multiple of 360 degrees. So the required positions occurs every 360/6.5 minutes from midnight (lots of answers).\n\nNote that that was basically the same question as your last one - if the hands were aligned when one was moving the wrong way; if it was actually moving the right way, they will be perfectly symmetrical as here. So thats why the answers are the same.\n\nFor the second question, its very similar. This time we want the two angles to add to 180 more than a multiple of 360. \nSo 6.5t + 180 is a multiple of 360.\nNow, since the little hand is near the 8, we know that t (which is in minutes since midnight) should be approximately 8 hours * 60 = 480 minutes. 6.5 * 480 + 180 is 3300, and the nearest multiple of 360 is 3240. So t = (3240-180)/6.5 = 470.769 minutes since midnight.\nThat works out to be about 7:50:46, to the nearest second.
484875
So I'm assuming there is no static or kinetic friction in this problem.\n\nWell, since the cart experiences a force in the horizontal direction, then: \nF = ma\na = F/m\na = 11N/26kg = 11/26 m/s^2\n\nHow far will it move in 2 seconds starting from rest?  What do we know?\n\na = 11/26 m/s^2\nt = 2 secs\nVelocity initial = 0 m/s (since it starts from rest)\nSo what is d (for distance although I know some schools use X)?\n\nd = V(initial)t + 1/2at^2\nd = 0t + (1/2)(11/26)(2)^2\nd = 11/52*4\nd = 11/13 m\n\n30 N child in the cart before he begins to push it?\n\nNewtons is just another way of saying weight but in order to find the acceleration we need mass and not weight.\nWeight (in N) = mg\nm = weight/g\nm = 30/10\nm = 3kg  <-------- this is the mass of the child\n\nRepeat as above with little modification.\na = F/m\na = 11N/(26kg + 3kg) = 11/29 m/s^2\n\nd = V(initial)t + 1/2at^2\nd = 0t + (1/2)(11/29)(2)^2\nd = 11/58*4\nd = 44/58 m\n\nGood luck!
468550
This is one of those look hard but not really math riddles. The answer is actually 1800 degrees. Also known as a five full spin turn. Where will it be? You just leave it there!\n\n1440 is 4 full spins. But that under 1441 so go to the next full spin, 1800 degrees. 1800 is between 1441 and 1892 so that's the answer.\n\nI can't actually think any other easier spin than a full spin. You don't have to move or rotate it at all. You just leave it in it's original spot.
394310
Do you already know where the CM of the hemisphere is?  Because I don't and it would take me a while to work it out.  Anyway, it would be on the vertical radius, so the horizontal position of the CM of the composite body is on the radius drawn to the particle, distant 0.2*r/4.2 from the centre.  If the height of the CM of the hemisphere above the centre is h, then the height of the CM of the composite body is\n4*h/4.2
501903
OK as the ball is swung around in a horizontal circle the scale inside the tube is measuring the tension on the string. By knowing the time to complete one revolution  linear velocity can be found from v=r*omega(angular velocity). in the second part the v is given and needs to be converted into radians/second by using 3.7=.5*omega where omega=3.7/5=.74rad/s then using the mass radius and angular velocity you can find the tension in the string. With a= acceleration (into center) omega^2*r=.74^2*.5=.2738m/s^2and using f=ma the tension in the string will equal f which equals .2*.2738=.0548newtons Unless i have a silly math mistake somewhere.
361143
Tell the other person to stop and you run at an angular speed that's the sum: (1.5 + 3.2)*10^-3 rad/s. How long before you cover the whole 2*pi*radians of the lake?
530084
Conservation of angular momentum must be considered as PhysicsDude mentioned. The thing to keep in mind here is the speed of the train given is "with respect to the track". The train is moving in one direction and the wheel/track is the other. If the wheel were not moving the speed of the train would be the 0.15 m/s, however since the wheel is moving counter to the train, and observer standing by the track (that is moving on the wheel along with the track) would see the speed of the train as faster than the actual train speed by an amount equal to the speed of the track. \n\nVr = Vtt + Vtr\n\nwhere Vr (=0.15 m/s) is the relative speed, Vtt is the speed of the toy train, and Vtr is the speed of the track.\n\nNow, considering conservation of angular momentum\n\nLtt = Itt*wtt = Itr*wtr = Ltr\n\nwhere Itt is the moment of inertia of the train, wtt is the angular speed of the train, Itr is the moment of inertia of the track, and wtr is the speed of the track. The two equate since the intial angular momentum is zero so that the two must cancel each other out after the train has started moving. The moments of inertia are:\n\nItt = m*r^2\nItr = 1.1*m*r^2\n\nwhere m is the mass of the train (1.1*m is the mass of the wheel/track), and r is the radius of the wheel. Recall that the angular speed is the linear speed divided by the radius. This means that the above conservation equation is\n\nm*r^2*wtt = m*r^2*(Vr/r - wtr) = 1.1*m*r^2*wtr\n\nSolving for the angular speed of the wheel/track (wtr)\n\nwtr = Vr/(2.1*r)\n\nPlugging in the values given\n\nwtr = (0.15 m/s)/(2.1*0.43 m) = (0.15 m/s)/(0.903 m)\n\nwtr = 0.166 rad/s
162679
Acceleration= 10 rad/s2\n\nAngular vel w= 10 x t rad/s\n\nAngular displacement d= 10 x t^2 / 2  :(t squared)\n\nThey say that angular displacement is 150 rad in 5 seconds, this means in an interval between t and (t+5).\nd= 10 t^2 /2 \nd2-d1= 10/2 (t+5)^2 - 10/2 t^2 =150 rad\nt^2 + 10t + 25 - t^2 = 30\nt= 0.5 sec\n\nSo the wheel totates for 0.5 seconds before they start to count.  \n\nHope its clear enough.
146352
OK, you are right, there are 2(pi) radians in a circle and spinning 10 times is 20 radians.\n\nYour angular velocity is the rate the mass (the clothespin?) spins within a given time. You give 10 revolutions in 5 seconds. or you can say 2 revolutions per second or 120 rpm (revs per minute). Multiply that by 2 to get radians per minute. \n\nAngular acceleration doesn't seem to be considered here since you say the mass has spun 10 times at a constant rate (or speed). So, in what you are asking, the length of the string (the radius of the circle of which the mass spins - your half a meter) Does not come into play here.
344300
The Average Linear Speed is {total distance / total time taken} = 65/ 5.7 = 11.4 m/s. \n\nDivide this value by the radius of the wheel to get the ‘The Average Angular Speed’. \n\nThe Average Angular Speed is = 11.4 / 0.02 = 570 radian per second.\n\nSince the motion is uniform,\n\n The Average Angular Speed = {Initial Angular Speed + Final Angular Speed} / 2   \n\nBut the Initial Angular Speed is zero.\n\nTherefore, the Final Angular Speed   is twice the Average Angular Speed\n= 2 x 570 \n=1140 radian/ second. (The answer for the second question)\n\nSince the Initial Angular Speed is zero,\n\nThe Angular acceleration is, “Final Angular Speed/ total time”.\n= 1140/5.7 = 200 [radian / second] / second. The answer for the first question)\n\n\nThe number of revolutions is found by “total distance / (pi x diameter of the wheel)” = 65 / pi 0.04 = 517.25 revolutions.
385603
It's 8 meters per second.  He's gone 5000 meters in 625 seconds; 5000 meters/625 seconds = 8 meters per second.\n\n v=d/t\nv=5000 m/625 seconds\nv=8m/sec
361922
What are the co-ordinates of the other point? just use the midppint formula for a segment. All the best!\n\nLife's Lovely! Love & Live Life!