How Do You Know Whether to Use Cos or Sin in Uniform Circular Motion Physics
half dozen.3: Uniform circular motion
- Folio ID
- 19400
As we saw in Chapter four, "uniform round motion" is defined to exist motion along a circumvolve with constant speed. This may be a practiced time to review Section iv.4 for the kinematics of movement along a circle. In particular, for the uniform round motion of an object around a circumvolve of radius \(R\), you should call up that:
- The velocity vector, \(\vec v\), is always tangent to the circumvolve.
- The acceleration vector, \(\vec a\), is always perpendicular to the velocity vector, considering the magnitude of the velocity vector does not change.
- The acceleration vector, \(\vec a\), always points towards the eye of the circumvolve.
- The acceleration vector has magnitude \(a=5^2/R\).
- The angular velocity, \(\omega\), is related to the magnitude of the velocity vector by \(v=\omega R\) and is constant.
- The angular acceleration, \(\alpha\), is zero for uniform round move, since the angular velocity does not change.
In particular, y'all should recall that even if the speed is constant, the acceleration vector is always non-zero in uniform circular movement because the velocity changes direction. According to Newton's 2nd Law, this implies that there must be a cyberspace force on the object that is directed towards the heart of the circle 1 (parallel to the dispatch): \[\begin{aligned} \sum \vec F = chiliad\vec a\end{aligned}\] where the acceleration has a magnitude \(a=five^2/R\). Because the dispatch is directed towards the middle of the circle, nosotros sometimes telephone call it a "radial" acceleration (parallel to the radius), \(a_R\), or a "centripetal" dispatch (directed towards the eye), \(a_c\).
Consider an object in uniform circular motion in a horizontal plane on a frictionless surface, as depicted in Effigy \(\PageIndex{1}\).
The simply style for the object to undergo uniform round motion as depicted is if the net forcefulness on the object is directed towards the middle of the circle. One mode to have a force that is directed towards the center of the circle is to adhere a cord between the center of the circumvolve and the object, as shown in Figure \(\PageIndex{1}\). If the string is under tension, the forcefulness of tension will always exist towards the center of the circumvolve. The forces on the object are thus:
- \(\vec F_g\), its weight with magnitude \(mg\).
- \(\vec Due north\), a normal forced exerted past the surface.
- \(\vec T\), a force of tension exerted by the string.
The forces are depicted in the free-trunk diagram shown in Figure \(\PageIndex{2}\) (equally viewed from the side), where nosotros also drew the acceleration vector. Annotation that this complimentary-body diagram is only "valid" at a particular instant in time since the acceleration vector continuously changes direction and would not always exist lined up with the \(x\) axis.
Writing out the \(x\) and \(y\) components of Newton's Second Constabulary: \[\brainstorm{aligned} \sum F_x &= T = ma_R\\ \sum F_y &= Northward - F_g =0\end{aligned}\] The \(y\) component but tells us that the normal force must accept the same magnitude as the weight because the object is not accelerating in the vertical direction. The \(x\) component tells us the relation between the magnitudes of the tension in the string and the radial acceleration. Using the speed of the object, we tin can also write the relation between the tension and the speed: \[\begin{aligned} T &= ma_R=m\frac{v^ii}{R}\\\end{aligned}\] Thus, we find that the tension in the string increases with the foursquare of the speed, and decreases with the radius of the circumvolve.
Exercise \(\PageIndex{i}\)
An object is undergoing uniform circular motion in the horizontal plane, when the string connecting the object to the center of rotation suddenly breaks. What path will the cake accept after the string bankrupt?
- A
- B
- C
- D
- Answer
Example \(\PageIndex{one}\)
A car goes effectually a curve which can be approximated as the arc of a circle of radius \(R\), as shown in Effigy \(\PageIndex{4}\). The coefficient of static friction between the tires of the car and the route is \(µ_{s}\). What is the maximum speed with which the car can go effectually the curve without skidding?
Solution:
If the car is going at constant speed around a circle, then the sum of the forces on the car must be directed towards the center of the circle. The merely forcefulness on the car that could be directed towards the center of the circumvolve is the force of friction between the tires and the road. If the road were perfectly slick (think driving in icy weather), it would not exist possible to drive around a curve since at that place could be no force of friction. The forces on the car are:
- \(\vec F_g\), its weight with magnitude \(mg\).
- \(\vec North\), a normal force exerted upwards by the road.
- \(\vec f_s\), a force of static friction between the tires and the route. This is static friction, because the surface of the tire does not move relative to the surface of the route if the motorcar is not skidding. The force of static friction has a magnitude that is at most \(f_s\leq\mu_sN\).
The forces on the auto are shown in the gratis-body diagram in Figure half-dozen.3.5.
The \(y\) component of Newton'southward Second Law tells usa that the normal force exerted past the road must equal the weight of the motorcar: \[\begin{aligned} \sum F_y = N-F_g&=0\\ \therefore North &=mg\terminate{aligned}\] The \(10\) component relates the force of friction to the radial dispatch (and thus to the speed): \[\brainstorm{aligned} \sum F_x = f_s =ma_R&=m\frac{v^ii}{R}\\ \therefore f_s &= m\frac{v^ii}{R}\finish{aligned}\] The force of friction must be less than or equal to \(f_s\leq\mu_sN=\mu_smg\) (since \(N=mg\) from the \(y\) component of Newton'due south Second Law), which gives usa a condition on the speed: \[\begin{aligned} f_s = thousand\frac{v^2}{R}&\leq\mu_smg\\ v^2 &\leq \mu_s g R\\ \therefore five &\leq \sqrt{\mu_s g R}\end{aligned}\] Thus, if the speed is less than \(\sqrt{\mu_s chiliad R}\), the motorcar will non slip and the magnitude of the strength of static friction, which results in an acceleration towards the center of the circle, will be smaller or equal to its maximal possible value.
Give-and-take:
The model for the maximum speed that the automobile tin travel around the curve makes sense because:
- The dimension of \(\sqrt{\mu_s g R}\) is speed.
- The speed is larger if the radius of the curve is larger (ane can get faster effectually a wider curve without skidding).
- The speed is larger if the coefficient of friction is large (if the force of friction is larger, a larger radial acceleration can be sustained).
Example \(\PageIndex{2}\)
A ball is fastened to a mass-less string and executing circular motility along a circle of radius \(R\) that is in the vertical aeroplane, as depicted in Effigy \(\PageIndex{6}\). Can the speed of the ball be abiding? What is the minimum speed of the ball at the meridian of the circle if it is able to brand information technology around the circumvolve?
Solution:
The forces that are acting on the ball are:
- \(\vec F_g\), its weight with magnitude \(mg\).
- \(\vec T\), a force of tension exerted by the string.
Effigy \(\PageIndex{seven}\) shows the costless-body diagram for the forces on the brawl at three dissimilar locations forth the path of the circumvolve.
In social club for the ball to go around in a circle, there must be at least a component of the net force on the ball that is directed towards the center of the circle at all times. In the bottom half of the circle (positions one and 2), but the tension can have a component directed towards the center of the circle.
Consider in detail the position labeled ii, when the string is horizontal and the tension is equal to \(\vec T_2\). The free-torso diagram in Effigy \(\PageIndex{seven}\) too shows the vector sum of the weight and tension at position 2 (the carmine arrow labeled \(\sum \vec F\)), which points down and to the left. It is thus conspicuously impossible for the acceleration vector to point towards the center of the circle, and the acceleration volition have components that are both tangential (\(a_T\)) to the circle and radial (\(a_R\)), every bit shown past the vector \(\vec a_2\) in Effigy \(\PageIndex{7}\).
The radial component of the acceleration will change the direction of the velocity vector so that the ball remains on the circle, and the tangential component volition reduce the magnitude of the velocity vector. According to our model, information technology is thus impossible for the ball to go around the circle at abiding speed, and the speed must subtract as it goes from position 2 to position 3, no matter how one pulls on the string (yous can convince yourself of this by drawing the gratuitous-torso diagram at any point between points 2 and 3).
The minimum speed for the ball at the top of the circumvolve is given by the condition that the tension in the string is zero only at the acme of the trajectory (position 3). The ball can still get around the circle because, at position 3, gravity is towards the center of the circle and tin thus requite an acceleration that is radial, even with no tension. The \(y\) component of Newton's Second Law, at position 3 gives: \[\begin{aligned} \sum F_y = -F_g &= ma_y\\ \therefore a_y &=-g\stop{aligned}\] The magnitude of the acceleration is the radial dispatch, and is thus related to the speed at the top of the trajectory: \[\begin{aligned} a_R&=-a_y=yard = m\frac{five^2}{R}\\ \therefore v_{min}&=\sqrt{\frac{gR}{yard}}\end{aligned}\] which is the minimum speed at the superlative of the trajectory for the ball to be able to go along along the circle. The tension in the cord would change as the brawl moves around the circle, and volition be highest at the bottom of the trajectory, since the tension has to be bigger than gravity and so that the cyberspace strength at the bottom of the trajectory is upwards (towards the heart of the circumvolve).
Discussion:
The model for the minimum speed of the ball at the meridian of the circle makes sense because:
- \(\sqrt{\frac{gR}{thou}}\) has the dimension of speed.
- The minimum velocity is larger if the circumvolve has a larger radius (endeavour this with a mass attached at the stop of a cord).
- The minimum velocity is larger if the mass is bigger (again, effort this at abode!).
Exercise \(\PageIndex{2}\)
Consider a ball attached to a cord, being spun in a vertical circumvolve (such as the i depicted in Figure \(\PageIndex{6}\)). If you shortened the string, how would the minimum angular velocity (measured at the elevation of the trajectory) required for the brawl to make it effectually the circle change?
- It would decrease
- It would stay the aforementioned
- Information technology would increase
- Respond
Banked curves
As nosotros saw in Instance vi.3.1, in that location is a maximum speed with which a car tin can go around a curve before information technology starts to sideslip. You may have noticed that roads, highways especially, are banked where in that location are curves. Racetracks for cars that go around an oval (the deadening kind of car races) likewise accept banked curves. As we volition see, this allows the speed of vehicles to be higher when going around the curve; or rather, information technology makes the curves safer as the speed at which vehicles would slip is higher. In Example six.3.1, nosotros saw that it was the forcefulness of static friction between the tires of the car and the road that provided the just force with a component towards the heart of the circle. The idea of using a banked curve is to change the direction of the normal force between the road and the automobile tires so that information technology, too, has a component in the direction towards the middle of the circle.
Consider the car depicted in Effigy \(\PageIndex{8}\) which is seen from behind making a left turn around a curve that is banked by an bending \(\theta\) with respect to the horizontal and can be modeled as an arc from a circle of radius \(R\).
The forces exerted on the motorcar are the same as in Example 6.3.1, except that they indicate in different directions. The forces are:
- \(\vec F_g\), its weight with magnitude \(mg\).
- \(\vec N\), a normal force exerted by the route, perpendicular to the surface of the road.
- \(\vec f_s\), a force of static friction between the tires and the road. This is static friction, because the surface of the tire does not motion relative to the surface of the road if the motorcar is not skidding. The strength of static friction has a magnitude that is at most \(f_s\leq\mu_sN\) and is perpendicular to the normal strength. The force could be either upward or downwards, depending on the other forces on the car.
A free-body diagram for the forces on the car is shown in Figure 6.3.ix, along with the acceleration (which is in the radial direction, towards the center of the circle), and our choice of coordinate system (choosing \(x\) parallel to the dispatch). The direction of the force of static friction is not known a priori and depends on the speed of the car:
- If the speed of the machine is zero, the force of static friction is upwards. With a speed of naught, the radial acceleration is zero, and the sum of the forces must thus be zero. The impeding movement of the auto would be to slide downwards the banked curve (just similar a block on an incline).
- If the speed of the motorcar is very large, the forcefulness of static friction is downward, every bit the impeding move of the car would be to slide up the banking company. The natural motion of the machine is to get in a straight line (Newton's First Law). If the components of the normal forcefulness and of the force of static friction directed towards the center of the circle are besides minor to allow the car to plough, then the motorcar would slide up the bank (and then the impeding motion is up the bank and the force of static friction is downwards).
There is thus an "ideal speed" at which the force of static friction is precisely cypher, and the \(ten\) component of the normal strength is responsible for the radial acceleration. At higher speeds, the forcefulness of static friction is downwardly and increases in magnitude to keep the machine'southward acceleration towards the heart of the circle. At some maximal speed, the force of friction will reach its maximal value, and no longer be able to continue the car's acceleration pointing towards the center of the circle. At speeds lower than the ideal speed, the strength of friction is directed upwards to foreclose the auto from sliding downwards the depository financial institution. If the coefficient of static friction is too depression, it is possible that at depression speeds, the car would start to slide down the bank (so there would exist a minimum speed below which the car would start to slide downwardly).
Permit united states model the situation where the force of static friction is identically zero then that we can make up one's mind the platonic speed for the banked curve. The only two forces on the machine are thus its weight and the normal force. The \(x\) and \(y\) component of Newton's Second Law give:
\[\brainstorm{aligned} \label{eq:applyingnewtonslaws:carbank_x} \sum F_x &= Due north\sin\theta = ma_R=m\frac{v^ii}{R}\nonumber\end{aligned}\]
\[\therefore Due north\sin\theta = thousand\frac{v^{2}}{R}\]
\[\brainstorm{aligned} \characterization{eq:applyingnewtonslaws:carbank_y} \sum F_y &= North\cos\theta-F_g = 0\nonumber\end{aligned}\]
\[\therefore N\cos\theta =mg\]
Nosotros tin divide Equation six.iii.1 by Equation 6.3.two, noting that \(\tan\theta=\sin\theta/\cos\theta\), to obtain: \[\begin{aligned} \tan\theta &= \frac{v^2}{gR}\\ \therefore v_{ideal} &=\sqrt{gR\tan\theta}\end{aligned}\] At this speed, the forcefulness of static friction is zero. In exercise, one would use this equation to decide which banking concern bending to use when designing a route, so that the ideal speed is effectually the speed limit or the boilerplate speed of traffic. We leave information technology as an exercise to decide the maximal speed that the automobile can go around the curve before sliding out.
Inertial forces in circular movement
Equally you lot sit in a machine that is going around a curve, you volition feel pushed outwards, abroad from the center of the circle that the car is going effectually. This is because of your inertia (Newton's Commencement Police force), and your body would get in a straight line if the car were non exerting a net forcefulness on y'all towards the middle of the circle. Y'all are non so much feeling a force that is pushing you outwards every bit y'all are feeling the effects of the auto seat pushing you in; if you lot were leaning against the side of the motorcar that is on the outside of the curve, you would experience the side of the car pushing y'all inwards towards the center of the bend, fifty-fifty if information technology "feels" like you are pushing outwards against the side of the car.
If we model your motion looking at yous from the ground, we would include a force of friction between the car seat (or the side of the car, or both) and you that is pointing towards the centre of the circle, so that the sum of the forces exerted on you is towards the center of the circle. We can also model your motility from the non-inertial frame of the machine. Every bit you lot call up, because this is a non-inertial frame of reference, we need to include an additional inertial force, \(\vec F_I\), that points opposite of the dispatch of the car, with magnitude \(F_I=ma_R\) (if the net acceleration of the auto is \(a_R\)). Inside the non-inertial frame of reference of the car, your acceleration (relative to the reference frame, i.e. the automobile) is zero. This is illustrated by the diagrams in Effigy \(\PageIndex{10}\).
The \(y\) component of Newton's Second Police force in both frames of reference is the same: \[\begin{aligned} \sum F_y&=N-F_g=0\\ \therefore North&=mg\end{aligned}\] and just tells u.s. that the normal force is equal to the weight. In the reference frame of the ground, the \(x\) component of Newton'southward Second Police gives: \[\begin{aligned} \sum F_x &= f_s = ma_R\\ \therefore f_s &= m\frac{v^2}{R}\end{aligned}\] In the frame of reference of the automobile, where your acceleration is nada and an inertial force of magnitude \(F_I=mv^two/R\) is exerted on you, the \(x\) component of Newton's Second Constabulary gives: \[\brainstorm{aligned} \sum F_x &= f_s-F_I = 0\\ \therefore f_s - m\frac{v^2}{R} &= 0\stop{aligned}\] which of course, mathematically, is exactly equivalent. The inertial force is not a real forcefulness in the sense that it is not exerted by anything. It simply comes into play because nosotros are trying to use Newton'southward Laws in a non-inertial frame of reference. However, it does provide a good model for describing the sensation that we have of being pushed outwards when the car goes around a curve. Sometimes, people volition refer to this forcefulness as a "centrifugal" strength, which means "a force that points away from the center". You lot should however remember that this is not a existent force exerted on the object, just is the result of modeling motility in a non-inertial frame of reference.
Exercise \(\PageIndex{three}\)
Jamie is driving his tricycle around a circular swimming. Jamie feels a centrifugal force with magnitude \(F_I\). If Jamie pedals twice equally fast, what will be the magnitude of the centrifugal force that he experiences?
- \(\sqrt{two}F_I\)
- \(\frac{ane}{two}F_I\)
- \(2F_I\)
- \(4F_I\)
- Respond
Footnotes
i. The sum of the forces is ofttimes called the "net strength" on an object, and in the specific case of uniform circular motion, that net force is sometimes called the "centripetal strength" - however, it is not a strength in and of itself and it is always the sum of the forces that points towards the center of the circle.
Source: https://phys.libretexts.org/Bookshelves/University_Physics/Book:_Introductory_Physics_-_Building_Models_to_Describe_Our_World_%28Martin_Neary_Rinaldo_and_Woodman%29/06:_Applying_Newtons_Laws/6.03:_Uniform_circular_motion
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