Uniform circular motion?
914198
first we must determine circular acceleration             \n\na=(v^2/r), v=velocity and r=radius\n\nwe know r=0.11 m and   rev/min = 3\n\nlet's find velocity:\n\nat r=0.11m the path of the coin is 2xPixr = 0.691 m\n\nso the coin travells   0.691 meters / revolution\n\nvelocity = 0.691 (meters/rev) x  (3 rev / min) \n             = 2.07 meters/min\n             = 0.0345 m/s\n\nNow we can solve acceleration:\n\na=(v^2/r)= (0.0345)^2 / 0.11 = 0.01082 m/s2\n\nF=mxa,  since we don't know the mass of the coin\n               lets just use mass=M\n\nF= 0.01082 x M,\n\nthis is the force that the coin feels as it rotates\nwhich is equal to the friction because the coin\nstarted sliding at this point.\n\n\ncoeff friction = (Force of friction) / (Force acting down )\n\nForce acting down is simply the coins weight\n\nForce acting down = W = mxg\n\nremember that mass = M    and   g=9.81\n\ncoeff friction =(0.01082x M)/(9.81xM), the M's cancel out\n                     = 0.0011
251783
depends on the circle Ü
791900
Ok.  From the hint start by figuring out the ratio of wheel speeds\nCircumference of A =2*pi*10\nCircumference of C = 2*pi*25\nso C/A=2.5  ie A must go around 2.5 times to make C go around 1 time.\nThere are 2*pi radians to 1 revolution so 100 rev/min =200*pi rad/min.  A must go 2.5 times as fast as this or 2.5*200*pi=500 rad/min\nAccelleration is the derivative of velocity which means that we must integrate the angular acceleration to obtain the speed at any given point of time.\nIntegrating 1.6rad/s^2 with respect to time and setting velocity equal to zero at time equals zero we obtain angular velocity= 1.6*time elapsed (rad/sec).  Note that this speed is in "per seconds" and we need it to be in "per minutes" so we must multiply by 60 for the units conversion.\nThis gives us the angular speed in rad/min for any chosen passage of time.  We shall set this equal to the goal speed of 500*pi rad/min as mentioned above. So\n1.6*time elapsed*60 =500*pi\nSolving for time elapsed we obtain a time of 1.63 minutes.
262693
This is very similar to your other question (see m responce there).\n\nTo help you figure this out, it would help if you though just exactly how the velocity changes are the plane goes around the circle.  If you cant visualize it, draw a diagram.\nAs the plane travels around the circle, its velocity vector is always pointed straight out, tangent to the edge of the circle.\nBy the time the plane gets 1/4 way around the circle, it no longer has any of its velocity in the original "positive" direction, it is all directed perpendicular to the original direction.
810893
Oscar. You have written too much. This 125m tower stuff does not matter compared to r=6.38*10^6. Why is this talk about 1.00 km and angle theta? See what “zipper” above says. Besides I answer your last question << Why is the trajectory of the cannonball not parabolic?>>. Why should it be if both you and Newton demand it to be circular? Assume that V1 of cannonball is the 1st cosmic speed corresponding to a circular track. Let V2 be > than V1, then you receive an oval track. Take V3 > than V2, then the track may become parabolic, so that cannonball would never return back to the cannon to linger out among planets and stars. If you are intrigued calculate a set of velocities and corresponding orbits of the cannonball. If cannon is not available consider a football! I like your inventive wits – to be honest!
659614
In order to get better responses, you will need to more specific and explain what you mean by that "question".\n\nThe first thing that came to my mind was centripetal force which maintains circular motion.  Without centripetal force, the object would not be in circular motion (thus linear motion).
867211
The basic formula that we know in circular motion is:\n\nF=mv^2/r, \n\nwhere F is the centripetal force, m the mass, v is the linear velocity, and r is the radius.\n\nWe also know that v=wr\n\nwhere v is the linear speed, w (supposed to be omega) is the angular velocity, and r is the radius.\n\nSubstitute the value of v in our first formula:\n\nF=m(wr)^2/r\n F =mw^2r      Formula 1)\n\nIn this problem, we also know that w which is the angular velocity is the same for m1 as for m2 because they are located along the same line which is rotating about one end of the rod.\n\nWe are also given the value of the tension in the inner section as 3x that of the outer section. Let's draw a simple free body diagram:\n                  \nT1=3F2<----0-->T2=F2<---0\n\nThe first 0 is m1 and the second 0 is m2.\n\nThe diagram might get distorted when I send it, but let's just cross our fingers it comes out clear enough.\n\nThe net force acting on m1 is (3F2-T2). But T2=F2. \nTherefore, net force on m1 =3F2-F2=2F2. Let's call\nthis net force as F1. Thus F1=2F2. Now we also know\nthat \n\nF1=m1w^2r    \nbased on formula 1) above.  Now substitute the value of F1=2F2, and we get:\n\n2F2=m1w^2r   Equation 1)\n\nAnd based on the same formula F=mw^2r:\n\nF2=m2w^2(2r)    Equation 2)\n\nRemember that for m2 the radius is twice that of m1, which is a given condition in this problem.\n\nNow divide equation 2 by equation 1:\n\nF2=m2w^2(2r)\n------------------\n2F2=m1w^2r\n\nF2,w^2, and r will cancel out leaving:\n\n1/2=2m2/m1\n\nDivide both sides by 2:\n\n1/4=m2/m1\n\nm2/m1=1/4\n          =0.25
784822
According to Newton's first law of motion, a moving body travels along a straight path with constant speed (i.e., has constant velocity) unless it is acted on by an outside force.\n\nFor circular motion to occur there must be a constant force acting on a body, pushing it toward the centre of the circular path. This force is the centripetal (“centre-seeking”) force. The centrifugal (centre-fleeing) force will, therefore, be the opposing force which pushes away from the centre\n\nThe two forces are equal in magnitude and opposite in direction.
750239
In half the rotation, the direction of velocity will become perpendicular to original direction and will be equal in magnitude to the original as no force acts in  direction of motion(if the air-resistance is negligible).\nsuppose initial direction is along positive z-axis\n=>/&#92; V=V2-V1\n=>/&#92; V=100(-k^) -100(k^)=200(-k^)\n/&#92;V=delta V(change in velocity)\nk^=unit vector along z-axis(axis perpendicular to the x-y plane)
140474
Use the formula Ac = v^2/r = r w^2, where w = angular speed.\n\nSince you already have the angular speed, you can use the second form of the equation above.\n\nAc= r w^2 =(1.0 m) (2.0 rad/s)^2
534094
It is because the object is "trying" to move in a straight line. \n\nImagine an object on a disk and then rotate the disk through a very small angle. The object moves *almost* in a straight line. Draw a straight arrow along that path and it will point to where the momentum of the object will push it - towards the edge of the disk.\n\nIn order to make the object move in a circular manner a force is required to act at 90 degrees to that straight line (i.e. towards the centre of the disk).  That is your centripetal force. \n\nBecause every action has an equal and opposite reaction there is an *apparent* force that is called the centrifugal force. But, as your teacher correctly states, there isn't actually any force - just the appearance of one.
806325
360 degrees because only zero degree or 360 degrees places the angle in the same quadrant. if u are asking multiples of then the answer would be a finite number i.e. 360 degrees.\n\nexplaination:\ntake a circle and hold its radius at the circumference, now move it round the circle with its center point fixed. it will come at the same place only after 360 degrees, or it will substend the same angle as it was when u started the movement.
412281
circular segment\n\nhttp://en.wikipedia.org/wiki/Circular_segment
96858
erm the volume of the circle would be the same, the volume of the cylinder would be (pi)r*r*h (r=radius, h=height)
1082685
Simply equate gravity function with centripetal acceleration.\n\nGMe/(r+h)^2 = V^2/(r+h)....where r is earth radius and h is height\n\nsolve for v
753766
First we determine the total moment of inertia of the system (with everyone on the rim).\n\nPeople: I = m r^2 = (65*150) (96^2) = 8.986 * 10^7\nSpace Station: I = 5.08 * 10^8 (it says 5.08 108 in your question, so I'm gonna assume that this is what you meant)\nSum = 5.679 * 10^8 kg*m^2\n\nIf the apparent acceleration = 1g\nThen, using a = v^2 / r , we can get the linear velocity of the edge\n\n9.8 = v^2 / 96\nv = 30.672 m/s\n\nTo get the angular velocity, use w = v / r \n(w is angular vel.)\n\nw = 30.672 / 96 = 0.3195 radians/sec\n\nSince angular momentum is conserved, we can equate the following:\nI * w (before meeting) = I * w (during meeting)\n\nObtain I (during meeting)\n\nPeople: I = m * r^2 = (65*50) * (96^2) = 2.995 * 10^7\nSpace Station: I = remains the same\nSum = 5.380 * 10^8 kg*m^2\n\nConservation of angular momentum:\nBefore = During\n(5.679*10^8) * (0.3195) = (5.380*10^8 ) * w\nw = 0.337 radians/sec\n\nFind the linear velocity:\nv = w * r = 0.337 * 96 = 32.377 m/s\n\nAcceleration to maintain circular motion:\na = v^2 / r\na = 32.377^2 / 96 = 10.92 m/s^2\n\n(that's 1.11g, if you're wondering)\n\nTada!
908729
(a) ω = ωo + αt\n0 = 150 + α * 132\nα = -150/132\n≈ -1.136 rev/s²\n\n(b) θ = ωot + ½αt²\n= 150 *132 - ½*1.136*132²\n≈ 9903 revolutions (nearest rev)\n\n(c) See (b)\n\n(d) 75 rpm = 75*2π rad/min\nα = rω²\n= 50 *(75*2π)²\n≈ 11103300 cm/min²\n= 111033 m/min²\n= 111033/3600 m/s²\n≈ 30.84 m/s²\n\n(e) ???? I do not understand what it is you are seeking.