Can someone answer a physics problem concerning circular motion?
914856
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.
892916
I think the key here is that the objects will be rolling down the plane; thus the gravitational potential energy they have is not only converted into translational kinetic energy, but also into rotational kinetic energy.\n\nAnyway, I would approach this from energy considerations.  Start by calculating the amount of potential energy the objects will lose by falling a distance of 3 m.  (For both, this will simply be PE = m g h).  Equate this to the total amount of kinetic energy the objects have at the bottom of the ramp (translational plus rotational, or 0.5 m v^2 + 0.5 I w^2 where I is the moment of inertia, and w is the angular velocity.)\n\nSo, you need to know the moment of inertia for a disk and a hoop.  The hoop I know off the top of my head, I(hoop) = m R^2.  The disk, I looked up to find I(disk)=0.25 m R^2.\n\nFinally, recognize that as they do not slip, you can relate the angular velocity w, to the translational velocity v:  v=2 Pi R w.\nOnce you do this, you find that the kinetic energy depends only on the velocity v, so we have one equation with one unknown.  Just solve for v with the two different values of I, and I think you ought to find that the disk will be moving faster than the hoop.\n\nSorry for being long winded.... but I hope it helps.
757387
(a) Assuming the base is a half-circle and that there are no other forces acting on the mass (friction, drag, etc), then the kinetic energy at the base is equal to the potential energy at the top. Note that the change in potential energy is not affected by the semi-circular path the mass follows, but rather the net change in height. This net change is simply the radius of the half-circle container:\n\nKE = mgR = (.71 g * 1 kg/1000 g)(9.8 m/s²)(25 cm * 1 m/100 cm)\n\nKE = .00174 J\n\n(b) The kinetic energy associated with rotation is given as:\n\nKER = (1/2) I w², where w is the rotational velocity. Note that since the sphere rolls without slipping, the translational velocity of its cener of mass is simply:\n\nv = w * r\n\nSubstituting this into the expression for the rotational kinetic energy:\n\nKER = (1/2) (I / r²) v²\n\nNote that the expression for the translational kinetic energy is:\n\nKET = (1/2) m v²\n\nThus, the total kinetic energy of the ball at the base of the hemisphere is;\n\nKE = KER + KET = (v²/2)( I/r² + m)\n\nThe fraction of the rotational kinetic energy over the total is:\n\nf = KER / KE = (v²/2)( I/r²) / (v²/2)( I/r² + m)\n\nf = ( I/r²)/( I/r² + m)\n\nFinally, the moment of inertia, I, of a sphere is:\n\nI = (2/5)mr²\n\nSubstituting this into the expression for f:\n\nf = (2/5)m/( (2/5)m + m)\n\nf = (2/5)/( (2/5) + 1)\n\nf = 2/7\n\n(c) At the base of the hemisphere, there are two forces acting on the ball: the force of gravity and the centripetal force. At the base, these forces are coincident (act in the same direction). Thus the normal force, which opposes these two, is simply:\n\nN = Fgrav + Fcent\n\nN = mg + mv²/R\n\nTo find v, recall that KET / KE is 2/7. Using this, we can manipulate the terms to get v:\n\n(1/2) (I / r) v² = (2/7)(mgR)\n\n(1/2) ((2/5)m) v² = (2/7)(mgR)\n\n(1/5) v² = (2/7)(gR)\n\nv = sqrt((10/7)(gR))\n\nv = sqrt((10/7)(9.8 * 25 cm * 1m/100 cm))\n\nv = sqrt((10/7)(9.8 * 25 cm * 1m/100 cm))\n\nv = 1.871 m/s\n\nFinally, the normal force is computed as:\n\nN = mg + mv²/R\n\nN = (.71 g)(9.81 m/s²) + (.71 g)(1.871 m/s)²/(25 cm)\n\nN = (.71 g)(9.81 m/s²) + (.71 g)(1.871 m/s)²/(25 cm)\n\nN = .0169 Newtons\n\n\nGood Luck! :)
733738
Look up your rotational mechanics equations.\n\n1 revolution also equals 2*pi radians.\n\nI can't write the equations out because I ca't enter pictures in yahoo answers but check it out on this website.\n\nhope this helps
1067758
In physics, circular motion is rotation along a circle: a circular path or a circular orbit. The rotation around a fixed axis of a three-dimensional body involves circular motion of its parts. We can talk about circular motion of an object if we ignore its size, so that we have the motion of a point mass in a plane.
701875
We're given r = 10 m, t = 5.0 s, and that \n\nθ = αt^2, where α = 0.3 rad/s^2\n\n(a) Rotational velocity, ω = θ/t = αt^2/t = αt = (0.3)(5.0) = 1.5 rad/sec\n\n(b) Tranlational velocity, v = ωr = (1.5)(10) = 15 m/s\n\n(c) Tangential acceleration, a = ω^2r = (1.5)^2(10) = 22.5 m/s^2\n\n(d) Radial acceleration, α = ω/t = 0.3 rad/s^2
818860
I'm not going to give you the complete answer because you'll learn more by working through it yourself, but I'll give a few pointers.\n\nAny object travelling in a circle at a constant speed has a centripital acceleration that can be calculated from the velocity and the radius of the circle.\n\nNewton tells us that F[net] = ma, so the object must be experiencing a net centripetal force equal to its mass multiplied by its centripetal acceleration.\n\nWhere does this force come from?  In your problem, it sounds as though the only force that could be pulling the box toward the center of the circle (the centripetal force) is the friction between the box and the bed of the truck.  So, when you add up all the (horizontal) forces to get the net force, the olny one is going to be friction.  F[net] = F[friction] \n\nNow all you need to do set F[friction]  equal to the F[net] that you calculated above (remember that F[net] was equal to your centripetal force), and solve for the coefficient of friction (your textbook shold have the formula for static friction, which will include the coefficient of static friction).\n\nGood luck!
235638
Remember the formula for centripetal acceleration,\na_c = v^2/r\n\nso if r is doubled but v remains constant, than the centripetal acceleration is halfed.\n\nTo prove this, lets put in some numbers.\nSay v = 2 m/s, r = 1 meter, thus, a_c = 4 m/s^2.\nNow double r to be 2 meters, the new a_c = 2 m/s^2, or 1/2 the original value.
850378
Centripetal acceleration a=r*w^2, r=0.35 m, w=2500*2pi/60 rad/s is angular speed, v=w*r is linear speed;
113237
I'll answer the second part first. RIme taken to get to the other side is distance/speed = 2km/20kph = 6 mins. The downstream motion of water does not affect te perpendicular motion.\nHow far will it reach = distance covered downstream at the speed of the river = speed x time = 6 kph x 1/10 hr = 0.6 km.
811425
This is more of a geometry/arithmetic problem than physics.\n\nLet's reason through it.\n\nThe reel has a radius of 3 cm, so its circumference is 2 x pi x 3 cm.\n\nThe line is reeled in at a rate of (2.5 / 9.5) m/s, \nor 100 x (2.5 / 9.5) cm/s.\n\nBy doing these calculations, you will have the centimeters per second at which the line is being reeled in, and the centimeters per revolution at which it is being reeled in. If you divide centimeters per second by centimeters per revolution, the result will be the revolutions per second at which the reel is turning. This is the angular speed that you need to calculate.
886116
You need to look at the normal acceleration versus the acceleration due to gravity at that angle.\n\nThe normal acceleration = v^2/R (where R is the radius and v is the velocity) and if this is greater than the component of gravity pointed towards the center of circle then airborne. So as it slides down the circle the velocity, v, is increasing, at the same time the component of gravity gets smaller.\n\nAt 0 degrees  (on top) v is zero but all the gravity is pointed towards the center of the circle. At 90 degrees the velocity is equal to square root of (2 * g * R) but the gravity component passing through the center of circle is 0. So somewhere between 0 and 90 degrees it takes off.\n\nI will let you do the math, but I would check 48 degrees as your instructor implied. (wink wink), use conservation of energy to find v.
612424
I haven't done any physics problems for ages, but I'll give it a try.\n\nThe centripetal acceleration = V^2/R \n\nBut, because the acceleration due to gravity is about 9.8m/sec^2 and this is important to the answer, you first want to express V in m/sec;\n\nV = 360km/hr = 360,000m/3,600sec = 100m/sec.\n\nSo the centripetal acc. = (100m/sec)^2 /500m = 10000/500 = 20m/sec^2, which is about 20/9.8 = 2.04 g-forces. This force is always directed towards the center of rotation. \n(Note: if you were on a spinning wheel space station where gravitational forces were negligible, and it had a radius of 500m and had a tangential V of 100m/sec, and you were walking around the inside of the station at the 100m extremity, you would have the apparent weight of 2.04 gs)   \n\nAs far as his "weight" is concerned: if the seat of the plane measured his weight at different points around the loop then it would vary. \n\nAt the top of the loop: his recorded weight would be 70kg x (20-9.8)/9.8 = 72.86kg. This is because gravity is reducing the  weight. \n\nAt the bottom: it would be 70x(20+9.8)/9.8 = 212.86 kg\nThis is because gravity is adding to the weight.\n\nAt the mid-points gravity and the centripetal acc. would be perpendicular to each other and the recorded weight would simply be 70x20/9.8 = 142.86kg\n\nI hope that helps.
835037
Let the two case be identified as having  moment-of-inertia I1 and I2. \nTorque T on to the each of the two bodies produces angular accelerations (alfa1 and alfa2) given  by the relation T= I x alpha \n\nalfa1 = T/ I1 and alfa2 = T/I2\n\nStarting form rest(wi =0), let the time taken to reach final angular velocity wf be t1\nusing:  wf =- wi + alfa x t  \n \nwf/alfa1 = t1 and wf/alfa2 = t2 \nleading to \n\nt2  = alfa1x t1/alfa2 \n\n     = I2 xt1/I1\n     = (I2/I1) x t1\n     =. ......Ans \n(Hoping you will put the values and get the answer you want)
860005
I finished the problem.  The answer is s = 0.48517m.  It was hard, although now that I've worked it, it doesn't seem as difficult anymore.  Had to do some serious thinking though.\n\nThe collar A will tend to slide <down> the bar CD.  The vector CD (from C to D) is [-0.2, -0.3, 0.25] with magnitude 0.43875 and direction cosines [-0.45584, -0.68376, 0.56980].  The weight of collar A, directed downward, is 8g = 78.4N and is represented as [0, -78.4, 0].\n\nWe don't have to worry about the collar being supported by bar CD (the normal force).  The component of the collar directed down the bar is the inner (scalar, dot) product of the weight vector and the direction cosines.  This force is 78.4 x .68376 = 53.60706N.\n\nLet E be the point on the bar CD such that BE is perpendicular to CD.  Then BEA is a right triangle, and by the Pythagorean Theorem, the force along BE is sqrt(150^2 - 53.60706^2) = 140.09384N.  (I'm using numbers stored in the calculator and showing five decimal digits.  All this can be rounded off at the end.)\n\nSince the force triangle BEA is similar to the distance triangle BEA, we just have to get one of the legs to find our distance s = AB.  We will get the distance BE.\n\nTriangle BEA is in the plane defined by the points B, C, and D.  Vector CB = [-0.4, 0.2, 0.15].  The vector cross product CB x CD (see above) is normal to the plane defined by points B, C, and D.  CB x CD = [0.095, 0.07, 0.16].\n\nNow, the cross product CD x (CB x CD) lies in the plane defined by B, C, and D; is perpendicular to CD; and is parallel to the line segment BE.\n\nCD x (CB x CD) = [-0.0655, 0.05575, 0.0145] with magnitude 0.08723 and direction cosines [-0.75091, 0.63914, 0.16623].\n\nNow we can get parametric equations of the lines CEAD and BE, and find the point of intersection E.  Starting with point C and using vector CD, the first line is\n\nx = 0.4 - 0.2t\ny = 0.3 - 0.3t\nz = 0 + 0.25t\n\nStarting from point B and using the direction cosines from vector BE, the second line is\n\nx = 0 + 0.75091u\ny = 0.5 - 0.63914u\nz = 0.15 - 0.16623u\n\nSetting like terms equal to each other, we have three equations in two unknowns.  The three equations are not linearly independent.  Solving the x and y equations for t and u, we get\n\nt = 0.29871 and u = 0.45313\n\n(and these results are confirmed in the z equations).  Plugging these results into the x, y, and z equations, we find that the intersection point E is (0.34026, 0.21039, 0.07468).\n\nWe know the coordinates of point B, so by the Pythagorean Theorem, the distance BE is 0.45313m.\n\nComparing the force and distance vectors in similar triangles BEA, we have\n\ns/150 = 0.45313 / 140.09384\n\nand the length s of the cable AB is 0.48517m.  (Answer)\n\nA difficult, but fun, problem.
970605
For your first problem...\n\nYou know that T = Fc = (mv^2) / r\n\nWhere T is the Tension, Fc is the Centripetal Force, m is the mass, v is the velocity, and r is the radius.\n\nYou're given m = 80 kg, r = 0.90 m, and T = 600N.\n\nSo...\n\n600 N = (80kg * v^2) / (0.90 m)\nSolve for v...\nv = +/- 2.598 m/s\n\nSo since -2.598 m/s doesn't make sense (a negative speed), the solution would be ~ 2.6 m/s.\n\nFor your second problem...\n\nIts been a long time since I've done circular motion.  This site has the equation you need to use (its too messy to type here): http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/carbank.html if you can figure out the angle.  So if you know what the angle should be, that equation will get you the coefficient of friction.\n\nIf you can't figure out the angle, at least attempting to solve it using that equation and some reasonable angle should get you some points I'd imagine.\n\nIf I can think of the correct angle I'll post an edit to this.
757331
The motion is deccelerated by the action of gravitational and frictional force. Hence, \nm*a=Fg+Ff\nm*a=w*sinx+f*w*cosx\na=w*sinx/m+f*w*cosx/m\na=g*sinx+f*g*cosx\na=g(sinx+f*cosx)----(1)\n0.5*m*v^2=m*a*s\ns=v^2/(2*a)---(2)\nInsert (1) in (2) and solve for s\ns=v^2/(2*g(sinx+f*cosx))\ns=3.1^2/(2*9.81(sin26+0.5*cos26))\ns=0.55 meter