If the angular speed of an object changes from w1 to w2 in a time Dt, then its angular acceleration is a = Dw/Dt. The point P will initially have tangential speed vt1 = rw1, and end with tangential speed vt2 = rw2. Therefore Dvt = rDw. Dividing both sides by Dt gives: Dvt/Dt = rDw/Dt. In the limit that Dt goes to zero, Dvt/Dt is the tangential acceleration of P, at. at = ra.
A company called Sea Launch Corporation launched its first satellite into orbit on Sunday. The company launches rockets from a converted offshore oil platform which can be moved to a location on the equator. Why do they launch from the equator?
Because the earth is spinning, every point on the surface has a tangential speed proportional to its distance from the earth's axis. Rocket launchers can take advantage of this tangential speed to give their rockets a head start on their journey into orbit, and the greatest tangential speed is obtained at locations furthest from the earth's axis, i.e. on the equator.
A floppy disk in a computer rotates from rest up to an angular speed of 31.4 rad/s in a time of 0.892 s. The radius of the disk is 4.45 cm. What is the angular acceleration of the disk, the tangential acceleration of a point on the edge of the disk, and the final linear speed of this point.
Use a = Dw/Dt = (31.4 rad/s)/(0.892s) = 35.2 rad/s2. Then from the relationship between linear and angular acceleration we have at = ra = (0.0445 m)(35.2 rad/s2) = 1.57 m/s2. Finally, the relation between linear and angular speed gives vt = rw = (0.0445 m)(31.4 rad/s) = 1.40 m/s.
To any change in velocity there is a corresponding acceleration. Since velocity is a vector quantity, the changes can be in magnitude (what we were primarily concerned with in the preceding chapters) or direction. Circular motion involves constant changes in direction. A simple demonstration will easily verify that there is a force, and therefore an acceleration, for an object in circular motion, even when the speed is constant. Imagine a small weight attached to the end of a string, and spin it above your head. There is tension in the string, and thus a force on the weight. The force is directed inward, toward the center of rotation. Furthermore, if you release the string, the force will no longer be present, and instead of moving in a circle, the weight will fly off in a straight (neglecting gravity) line.
So, an object in circular motion experiences an acceleration towards the center of the circle. This acceleration is called centripetal (center seeking) acceleration, and has a magnitude of ac = vt2/r = rw2. A derivation of this equation can be found on p.186 of the text.
A rotating object always has centripetal acceleration. It may also have tangential acceleration if the rate of rotation (angular speed) is changing. Centripetal and tangential acceleration are always perpendicular. Neither of them has a fixed direction -- the direction changes as the object moves around in its rotation -- but they are always oriented perpendicularly to each other. Therefore, we find the total acceleration using Pythagorean's theorem: atot = Sqrt{ac2 + at2}.
A race car starts from rest on a circular track of radius 400 m. The car's speed increases at the constant rate of 0.500 m/s2. At the point at which the magnitudes of the centripetal and tangential accelerations are equal, determine (a) the speed of the race car, (b) the distance traveled, and © the elapsed time.
at = ra = 0.500 m/s2. a = at/r = 0.500 m/s2/400 m = 0.00125 rad/s2. (a) at = ac = v2/r, so v = Sqrt{r at} = Sqrt{(400 m)(0.500 m/s2} = 14.1 m/s. We can also find w = v/r = 0.0354 rad/s.
(b) Using w2 = w02 + 2aq, we see that q = w2/2a = (0.0354 rad/s)2 / 2(0.00125 rad/s2) = 0.50 rad = 0.080 rev. In terms of meters, this is a distance of s = qr = 200 m.
© Use q = w0t + (1/2)at2 to find the time. t = Sqrt{2q/a} = Sqrt{2(0.50 rad)/(0.00125 rad/s2)} = 28.3 s.
This can also be worked using tangential speed and acceleration.Consider a mass m tied to a string of length r, being swung in a horizontal circle. For the mass to remain in a circle, it experiences a centripetal acceleration, ac = vt2/r. To every acceleration there is associated a force, in this case it is called the centripetal force, Fc = mac = mvt2/r. If the centripetal force is removed, for instance by letting go of the string tied to the mass, then, following Newton's first law, the mass moves in a straight line.
It is the centripetal force, acting perpendicular to the instantaneous motion of the object, that keeps it moving in a circular path. The force must act perpendicular to the motion, since any component parallel to the motion will tend to speed up or slow down the object. Remember from chapter 5, an impulse FDt causes a change in momentum mDv.
When working problems involving a centripetal force, it is not convenient to think in terms of x and y components, since the direction of the centripetal force changes constantly. It is normally much more convenient to work with the components of the forces that are parallel (tangential) and perpendicular to the circular path.
In a popular amusement park ride, a rotating cylinder of radius 3.00 m is set in rotation at an angular speed of 5.00 rad/s. The floor then drops away, leaving the riders suspended against the wall in a vertical position. What minimum coefficient of friction between a rider's clothing and the wall is needed to keep the rider from slipping?
fs,max = μn, and in this case the normal force is the centripetal force exerted on the rider to keep her moving in a circle. That centripetal force is provided by the wall of the cylinder, and is Fc = mv2/r = mw2r. We are told that w = 5.00 rad/s, and r = 3.00 m. Since we don't know the mass, let's write Fc/m = w2r = (5.00 rad/s)2(3.00 m) = 75.0 m/s².
The force of static friction must balance the gravitational force on each rider, fs = μFc > mg. Therefore, μ > mg/Fc = (9.8 m/s²)/(75.0 m/s²) = 0.13. Since μ must be greater than or equal to 0.13, the minimum value is 0.13.
What happens if you're in a car that turns to the left at high speed? Assuming the car stays upright, you feel pulled to the right side of the car. This "force" that you feel is a fictitous force, this one being called a centrifugal force. It arises because for you to turn along with the car, you must be acted upon by a centripetal force, and this force comes from the car seat, steering wheel (if you are holding on to it), or whatever part of the car you are touching.
But this is not a real force in the sense we have been using. A person standing outside the car doesn't "see" this force, but instead sees you trying to continue moving in a straight line while the car turns to the left. You feel the force because your local environment (the car) is accelerating (a centripetal acceleration). Newton's laws don't work as is in an accelerating environment, additional corrections must be included to compensate for the acceleration. This is not something we will be studying in this course.
Consider a rollercoaster car going around a loop of radius R. What must be the speed at the bottom, as it enters the loop, so that it just makes it over the top of the loop without falling?
What we want is that at the top of the loop, the centripetal acceleration needed to keep the car moving on the circle just equals the gravitational force. This means that the rollercoaster track exerts no force on the car at the top of the loop, the car feels its own weight only. The we have mv2top/R = mg, or vtop = Sqrt{gR}.
Let us take the zero point of potential energy at the bottom of the loop. Then, at the top of the loop, the energy of the car is both kinetic and potential, while at the bottom it is all kinetic, and since we have only conservative forces (we neglect friction), the energy at the top equals the energy at the bottom.
What is the kinetic energy of something moving in a circle? It is still ½mv2, where v is the total velocity of the object. In the case of the rollercoaster car, v is just the tangential velocity, as we calculated above for the top of the loop. So, Etop = KEtop + PEtop = ½mv2top + mg2R = ½mgR + 2mgR = 2.5mgR. At the bottom of the loop, Ebot = KEbot = ½mv2bot. Using conservation of energy to equate this gives ½mv2bot = 2.5mgR, or vbot = Sqrt{5mgR}.
To have this amount of energy, the car must start at a height of at least 2.5R above the bottom of the loop. Note that this is more than 2R, the height of the loop. The additional starting height (meaning additional energy) is needed to have enough speed to stay on the loop as the car goes over the top.
| Every particle in the Universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. |
Notice some features of this force. First, it is proportional to 1/r2. We say that this is an inverse square law force, that is, the force is "inversely proportional to the square of the distance". If the distance between two objects is doubled, the force decreases by a factor of 4 (is 1/4 the original), or if the distance between two objects is halved, the force increases by a factor of 4.
Next, the force of m1 on m2, written F21, is equal in magnitude to the force of m2 on m1, written F12. The gravitational force is always attractive, so the directions of F21 and F12 are opposite. Written mathematically, F21 = -F12 as required by Newton's third law.
If one or both masses are extended spherical objects (the usual case for our work), then the same law holds, as long as the distance between the objects is measured between their centers, and the objects don't overlap. Pay attention to objects overlapping when, for instance, calculating what happens to something "launched" from the surface of the earth into space.
During a solar eclipse, the Moon, Earth, and Sun all lie on the same line, with the Moon between the Earth and the Sun. (a) What force is exerted on the Moon by the Sun? (b) What force is exerted on the Moon by the Earth? © What force is exerted on the Earth by the Sun?
From problem 31, the distance from the Earth to the Moon is 384,000 km = 3.84x108 m, and from table 7.3, the distance from the Earth to the Sun is 1.496x1011 m. During a solar eclipse, the distance from the Moon to the Sun is the difference of the above distances, or 1.492x1011 m. The mass of the Sun is 1.991x1030 kg, the Earth is 5.98x1024 kg, and the Moon is 7.36x1022 kg.
(a) Fmoon-sun = (6.673x10-11 Nm²/kg²)(1.991x1030 kg)(7.36x1022 kg)/(1.492x1011 m)2 = 4.39x1020 N
(b) Fmoon-earth = 1.99x1020 N
© Fearth-sun = 3.55x1022 N