ONLINE: MAHEMAICS EXENSION opic 6 MECHANICS 6.6 MOION IN A CICLE When a particle moes along a circular path (or cured path) its elocity must change een if its speed is constant, hence the particle must be accelerating. According to Newton s Second Law, the particle must be acted upon by a non-zero resultant force that produces the acceleration. (1) () d a instantaneous acceleration dt 1 a Fi Newton s Second Law m i Position, displacement and angular displacement he equation of the circle with centre at (0, 0) and radius is (3) x y At any instant, the position of the particle is specified by its x and y Cartesian coordinates. he position of the particle can also be specified in terms of its polar coordinates (, ) where is the radius of the circle and is the angular displacement of the particle. he angular displacement must be gien in radians and is measured in an anticlockwise sense from the X axis as shown in figure (1). Fig. 1. Particle moing in a circle. Its position is gien in Cartesian coordinates (x, y) and in polar coordinates (, ). physics.usyd.edu.au/teach_res/math/math.htm mec66 1
Velocity and angular elocity When the particle is moing in a circle its displacement ector s has a constant magnitude but its direction gien by increases with time. At time t = t 1 the particle is at the point P 1 and moes to a point P in a time interal t. x ( t ) cos( ) y ( t ) sin( ) 1 1 1 1 1 1 x ( t t) cos( ) y ( t t) sin( ) Fig.. Position of a particle at times t 1 and t 1 +t. he angular displacement changes by 1 in the time interal t. he rate of change of the angular displacement with time is called the angular elocity (Greek letter omega). (4) aerage aerage angular elocity [rad.s -1 radians/second] t In the limit t 0 we can define the instantaneous angular elocity as (5) d instantaneous angular elocity dt he instantaneous elocity is equal to the time rate of change of displacement. herefore, the X and Y components of the particle s elocity are (6) dx d d x cos sin dy d d y sin cos sin cos x y physics.usyd.edu.au/teach_res/math/math.htm mec66
he magnitude of the instantaneous elocity is (7) x y sin cos he direction of the instantaneous elocity is by the angle y cos 1 tan sin tan x But tan is the slope of the displacement ector, therefore the elocity ector at all times is perpendicular to the displacement ector. he instantaneous elocity is always in the direction of the tangent to the circle at the instantaneous position of the particle. his elocity is known as the tangential elocity. he is no radial motion, so the radial elocity is zero. 0 + Y x y y tan x x cos y sin (0, 0) s + X d dt Fig. 3. he instantaneous elocity is at all times perpendicular to the displacement ector. physics.usyd.edu.au/teach_res/math/math.htm mec66 3
Acceleration a and angular acceleration Forces he instantaneous acceleration is the time rate of change of the elocity and the time rate of change of the angular elocity is known as the angular acceleration. 1 d d d d d a d d he acceleration a can be resoled into its X and Y components or its radial a and tangential components a. a ax ay a a Fig. 4. he acceleration a can be resoled into its X and Y components or its radial a and tangential components a. he X and Y components of the acceleration are found by differentiating the X and Y components of the elocity gien in equation (6) with respect to time. (8) d d d ax x x d d d ay y y cos sin cos sin sin cos sin cos he radial and tangential components are easily found from equation (8) by taking 0 rad or / rad physics.usyd.edu.au/teach_res/math/math.htm mec66 4
Fig. 5. he radial and tangential components of the acceleration can be calculated from the X and Y components when = 0 rad or = / rad. d a a dt 0 rad x cos 0 sin 0 a d rad a ay sin cos dt a d 0 rad a a y sin 0 cos dt d a dt d rad a ax cos sin dt d a dt (9) a radial acceleration d (10) a tangential acceleration dt physics.usyd.edu.au/teach_res/math/math.htm mec66 5
he magnitude of the tangential elocity is (7) and so the time rate of change of the speed of the particle gies the tangential acceleration a d d d a which is the same result as gien in equation (10). he radial acceleration is also called the normal component a N since it is always at right angles to the elocity of the particle. In Physics, it is mostly called the centripetal acceleration a C. he centripetal acceleration represents the time rate of change which represents the change in direction of the elocity ector. a an ac he particle moing in a circle always has a centripetal acceleration directed towards the centre of the circle. If the speed of the particle changes it has a tangential component of acceleration which is tangent to the circle. his tangential acceleration represents the time rate of change of the magnitude of the elocity ector. he magnitude of the acceleration ector a is (11) a a a a a a x y and its direction can be determined from a (1) tan A a y x where A is the angle of the acceleration ector measured with respect to the X axis. As with any acceleration, there must be a resultant (net) force in the direction of the acceleration to produce it. For centripetal acceleration, this force is called the centripetal force. It is not a new kind of force, but merely a label for the force needed for circular motion. he centripetal force may be due to a spring, string, friction, graitation, etc or a combination of these forces. he centripetal force is always directed towards the centre of the circle. here is no arrow for centripetal force in a free-body diagram (force diagram) as it is just a label for the force acting towards the centre of the circle. he word centripetal is deried from two Greek words meaning seeking centre. physics.usyd.edu.au/teach_res/math/math.htm mec66 6
From Newton s Second Law (equation ) and the equations for the acceleration (equations 9 and 10), the components of the force acting on the particle executing circular motion are: (13) F m m radial or normal or centripetal force (14) d d d F m m m m m tangential force Fig. 6. Motion in a circle: elocity and acceleration ectors. he resultant force acting on the particle is in the same direction as the acceleration a. physics.usyd.edu.au/teach_res/math/math.htm mec66 7
UNIFOM CICULA MOION When the particle moes in a circle with constant speed, the motion is called uniform circular motion. he tangential acceleration is zero and the acceleration ector is directed towards the centre of the circle and perpendicular to the elocity ector as shown in figure (6). a 0 a a a a a a C C N (9) a C centripetal (radial or normal) acceleration In uniform circular motion, the period is the time interal taken for the particle to trael one reolution ( rad), that is, in the time interal the particle traels a distance equal to the circumference of the circle with a constant speed. Speed Angular speed s t t he number of reolutions per second is called the frequency f and the angular speed is often referred to as the angular frequency. Period and frequency 1 f Angular frequency f 4 Centripetal acceleration ac 4 f physics.usyd.edu.au/teach_res/math/math.htm mec66 8
We now consider an alternatie deriation of the formula relating the centripetal acceleration to the speed of the particle. Fig. 7. Finding the change in elocity and hence the centripetal acceleration. So from figure (7), the centripetal acceleration is gien in equation (9). he X and Y components of the acceleration are a C which is the same result as (8) d d d ax x x d d d ay y y cos sin cos sin sin cos sin cos d For uniform circular motion 0 hence the x and Y components reduce to dt (15) ax x x ay y y cos sin Equations (15) are simply the equations for simple harmonic motion. As the particles moes around the circle at a uniform speed he projection of the x position of the particle onto the X-axis moes with simple harmonic motion. he projection of the y position of the particle onto the Y-axis moes with simple harmonic motion. physics.usyd.edu.au/teach_res/math/math.htm mec66 9
Example A bucket of water is swung in a ertical circle of radius. he speed of the bucket at the top of the swing is and the speed of the bucket at the bottom of the swing is B. (a) (b) (c) (d) Find the force acting on the water by the bucket at the top of the swing. Find the minimum speed for the water to remain in the bucket. Find the force exerted by the bucket on the water at the bottom of the swing. Estimate the minimum speed at the top of the circle and the period of reolution that would keep you from getting wet if you swung the bucket of water in a ertical circle at a constant speed. Solution Draw free-body diagrams for the water at the top and bottom of the circle (a) op of swing: apply Newton s Second Law to the water Fx 0 Fy F m G FN FG m g Find the normal force acting on the water F N m m g physics.usyd.edu.au/teach_res/math/math.htm mec66 10
(b) he bucket cannot exert an upward force on the water at the top. he minimum force it can exert is zero F N = 0 (c) m 0 _ min _ min g mg Bottom of swing: apply Newton s Second Law to the water B Fx 0 Fy F m G FN FG m g Find the normal force acting on the water F N mb m g Note: here is no arrow for the centripetal force in the free-body diagrams. Centripetal force is not a kind of force but simply a label. When the whirling bucket is at the top of the circle, both graity and the contact force of the bucket on the water contribute to the centripetal force. When the water is moing at its minimum speed at the top of the swing, the water is in free fall (acceleration = g) and the only force acting on the water is graity (m g). At the bottom of the swing the normal force must be greater than the weight (m g) by enough to proide the necessary centripetal force to moe the water in a circle. (d) Assume = 1.0 m and g = 9.8 m.s - _ min g 9.8 m.s 3 m.s s _ min -1-1 physics.usyd.edu.au/teach_res/math/math.htm mec66 11
Example (Syllabus) A string is 0.500 m long and will break if an object of mass exceeding 40.0 kg is hung ertically from it. An object of mass.00 kg is attached to one end of the string and it reoles around a horizontal circle with a uniform speed. Find the greatest angular elocity which may be imparted to the object without breaking the string. L = 0.500 m m 1 = 40.0 kg g = 9.80 m.s - m =.00 kg max =? rad.s -1 radius of circular motion = L = 0.500 m Max force Fmax exerted on string before breaking F max = m 1 g = (40)(9.8) N = 39 N Max centripetal force F C exerted by string on object FC F max m m g max 1 mg m 0.5 40 9.8 rad.s 19.8 rad.s 1-1 -1 max physics.usyd.edu.au/teach_res/math/math.htm mec66 1