Centripetal Force Question?
912301
Well, for centripetal force:\n\nForce = (mass) * (velocity squared) / radius\n\nand from there?  I'm lost.  I've tried plugging in all your information and it feels like we're missing a known variable.  My first assumption was to put in 7.18 as the force, but then we don' t the mass.   And if we plug in (7.17*mass), we're left with force as a variable.   \n\nOye, good luck.  It's been 10 years since I've taken physics.
904251
Let's say F1 and F2 are the tensions (forces) in the inner and outer sections of the rod respectively, w is the angular velocity, and r the distance to the center (half the length of the rod). Equating the radial force to (mass x centripetal acceleration) for each mass gives:\n\nr*w^2*m1 = F1 - F2 = 2F2\n(2r)*w^2*m2 = F2\n\nDividing the second equation by the first all variables cancel except for the ratio of the masses:\n\nm2/m1 = 1/4
1093836
Centrifugal force and gravity.
897942
You should only ask one question at a time.\n\nI will answer the first:\n\nConsider the forces on the box:\n\nGravity vertically downward\nfriction pulling backwards\nThe force of gravity parallel to the ramp \nminus the force of friction\n=m*a=m*1.22\nso\nm*sin(34)*g-f=m*1.22\nfriction can be expressed as\nf=m*u*cos(34)*g\nnote that mass divides out\nsin(34)*g-u*cos(34)*g=1.22\n\nu=.52\n\nj
371595
approx 19 miles/second.\n\nAny force from centripetal acceleration you might feel would be more than compensated by the gravitational attractions from the Sun and Earth themselves.\n\nIt might be a measureable amount, but just barely.
770512
The total force on the block is mg sin (theta). The mg cos (theta) and force normal cancel out. Since there is no friction, the only force, again, is only mg sin (theta). \n\nThis is obvious if you consider that there is a force mg, and break it down into components, the solution should seem obvious.
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! :)
784770
Consider the forces and directions of the forces on the car:\n\nThe force of gravity pulling the car directly downward\nThe centrifugal force pulling the car directly horizontal\nThe force of the car on the road, which is banked\nThe friction of the road on the tires.\n\nSince it is icy, let's set the friction to zero\n\nThen the forces parallel to the road are the parallel component of the centrifugal force:\n\nm*cos(3.4)*r*w^2\n\nand the force of gravity towards the center\nm*g*sin(3.4)\n\nWhen these force balance, there is no force of friction parallel to the road surface.  So \n\nm*cos(3.4)*r*w^2=m*g*sin(3.4)\nw=sqrt(g*sin(3.4)/(cos(3.4)*r))\nThe linear velocity, or what the driver will read on the speedometer\nis w*r\n\n8 m/s\n\nj
643953
depends on the direction of the force
938772
Of cause, help.\nthe force of gravity is mM/R^2 = 550 N\nThis is for R = 3400km\n\nIn case we take the object up for 12 km, the distance from the centre of the planet will be 3412 N.\n\nand its weight will be mM/(R+12km)^2 = Wn\nWe can find  Wn   just from these two equations:\nfrom the first equation we shall express \nmM = 550(N)xR^2\nthan  Wn = 550 (N) x R^2/(R+12)^2 = 550(N)x3400^2/3412^2 = \n550 (N) x 0.996483^2 = 550(N)x0.9929783 = 546.138 N\nThat is the answer
998322
Lets make an enormous assumption to fill in the gaps.\nYou have a thing on a string of length L and spin it round at a certain rate of rotation. Keeping the rate of roataion what would the tangential velocity of the thing be if the string was shortened to L/2?\nIn this case the tangential velocity (= the radius x the angular velocity) will also halve.
57539
Well, there are two components of acceleration here: there's centripetal acceleration, which is toward the center of the Ferris wheel, and tangential acceleration, whose direction is tangential to the wheel (in this problem, straight forward).  In order to calculate the total acceleration, you need both these components.\n\nYour centripetal acceleration is given by the velocity squared over the radius of the circle:\n\na_c = 3^2/14 = 9/14 m/s^2 = .64\n\nFrom the info in the problem, your tangential acceleration would be the .5 m/s^2.  \n\nNow draw a triangle with those two accelerations as sides.  Solve for the hypontenuse with Pythagoras to get .81 m/s^2.  Finally, the arctan of .64/.5 is 65 degrees (more or less), so your acceleration is .81 m/s^2 at a 65 degree angle.\n\nNow go back and check my work, because I just got up and I'm a bit rusty in the mornings.  AND I didn't get my newspaper this morning.  Can you believe that?
845902
Let’s consider from viewpoint of physics. F=G*M*m/r^2 is gravitation force holding the sat on its orbit. On the other hand F=m*a – centripetal force. Thus a=G*M/r^2, where M is mass of Earth (I don’t remember), r=R+h, R=6.4*10^6m is radius of Earth, h=0.615*10^6m is height of the sat, a is gravitational acceleration at h (like that ‘g’ on the ground), G is gravitational constant  (?? 6.67*10^-11 ??  I don’t remember!), m was mass of the satellite. Another formula for centripetal acceleration for circular track a=r*w^2, w is angular velocity or w=2pi/T, T being period. G*M/r^2=r*(2pi/T)^2. Now T=2pi*r*sqrt(r/(G*M)) seconds
984186
accn = v^2 / r or [(angular velocity)^2] * r\ntherefore,\naccn of hammer = [(2*pi)^2] * 1.8\n\nas F= ma\nTension = 7.0 * [(2*pi)^2] * 1.8\nthe tension in the chain is the force used to keep the hammer in centripetal accn.
742316
Torqu equals force applied times radius, so the equation would be T=Fr    or    T=mar    so, T=(4)(9.8)(6), or 235.2 n/m\n\nMoment of inertia is not needed to solve this problem unless it later asks for another different answer.
574012
I've alread answered two of your questiions in my last answer....\n\nIn this one, the guys above are correct - however, I'd try working it out for myself if I was you. Your tutor will soon realise something is wrong if you are giving answers to a subject you don't understand. Read the chapter, ask you teacher if you need more help. Learn Newtonian mechanics for yourself.
398654
In the special case where centrifugal force exactly balances gravity, the orbit IS circular.  But say you have a planet in a circular orbit.  Give it a little extra speed in the direction it's going.  Now it has too much centrifugal force, and it begins to gain distance from the star -- but that means it's gaining potential energy and so must be losing kinetic energy.  So it slows down.  The detailed math shows that by the time it's gone 180 degrees around it has lost so much speed that, even though it is farther from the star, now it doesn't have enough centrifugal force to counteract gravity, so it begins to speed up.  Again, if you do the math, you see that it will return to the exact position it was at when you gave it the extra push, and with the exact same velocity it had just after that push.  So it will just keep repeating that orbit, which is highest (and slowest) at one point and lowest (and fastest) 180 degrees around; in other words, an elliptical orbit.