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(1) Acceleration acting on the object undergoing uniform circular motion is called centripetal acceleration. (2) It always acts on the object along the radius towards the centre of the circular path.       (3) Magnitude of centripetal acceleration,   \[a\,=\frac{{{v}^{2}}}{r}={{\omega }^{2}}r=4{{\pi }^{2}}{{n}^{2}}r=\frac{4{{\pi }^{2}}}{{{T}^{2}}}r\]    (4) Direction of centripetal acceleration : It is always the same as that of \[\Delta \overrightarrow{\upsilon }\]. When \[\Delta t\] decreases, \[\Delta \theta \] also decreases. Due to which \[\Delta \vec{\upsilon }\] becomes more and more perpendicular to \[\overrightarrow{\upsilon }\]. When \[\Delta t\to 0,\] \[\Delta \overrightarrow{\upsilon }\] becomes perpendicular to the velocity vector. As the velocity vector of the particle at an instant acts along the tangent to the circular path, therefore \[\Delta \overrightarrow{\upsilon }\] and hence the centripetal acceleration vector acts along the radius of the circular path at that point and is directed towards the centre of the circular path.    

(1) Displacement and distance : When particle moves in a circular path describing an angle \[\theta \] during time t (as shown in the figure) from the position A to the position B, we see that the magnitude of the position vector \[\vec{r}\] (that is equal to the radius of the circle) remains constant. i.e., \[\left| {{{\vec{r}}}_{1}} \right|\,=\,\left| {{{\vec{r}}}_{2}} \right|=r\] and the direction of the position vector changes from time to time.     (i) Displacement : The change of position vector or the displacement \[\Delta \vec{r}\] of the particle from position A to the position B is given by referring the figure. \[\Delta \,\vec{r}={{\vec{r}}_{2}}-{{\vec{r}}_{1}}\] Þ \[\Delta r=\,\left| \Delta \vec{r} \right|\,=\,\left| {{{\vec{r}}}_{2}}-{{{\bar{r}}}_{1}} \right|\]   \[\Delta r=\,\sqrt{r_{1}^{2}+r_{2}^{2}-2{{r}_{1}}{{r}_{2}}\,\cos \theta }\] Putting \[{{r}_{1}}={{r}_{2}}=r\] we obtain \[\Delta r=\sqrt{{{r}^{2}}+{{r}^{2}}-2r.r\,\cos \theta }\] \[\Rightarrow \] \[\Delta r=\sqrt{2{{r}^{2}}\left( 1-\cos \theta  \right)}\,\] \[=\,\sqrt{2{{r}^{2}}\left( 2{{\sin }^{2}}\frac{\theta }{2} \right)}\] \[\Delta \,r=2r\,\sin \frac{\theta }{2}\]     (ii) Distance : The distanced covered by the particle during the time t is given as d = length of the arc \[AB=r\theta \] (iii) Ratio of distance and displacement :   \[\frac{d}{\Delta r}=\frac{r\theta }{2r\,\sin \theta /2}\]   \[=\frac{\theta }{2}\text{cosec}\,(\theta /2)\] (2) Angular displacement \[(\theta )\] : The angle turned by a body moving in a circle from some reference line is called angular displacement. (i) Dimension =\[[{{M}^{0}}{{L}^{0}}{{T}^{0}}]\] (as \[\theta \] = \[arc/\text{radius)}\]. (ii) Units =  Radian or Degree. It is some time also specified in terms of fraction or multiple of revolution. (iii) \[2\pi \,\text{rad}\,=\,{{360}^{o}}=1\,\text{Revolution}\] (iv) Angular displacement is a axial vector quantity. Its direction depends upon the sense of rotation of the object and can be given by Right Hand Rule; which states that if the curvature of the fingers of right hand represents the sense of rotation of the object, then the thumb, held perpendicular to the curvature of the fingers, represents the direction of angular displacement vector. (v) Relation between linear displacement and angular displacement \[\overrightarrow{s}=\overrightarrow{\theta }\times \overrightarrow{r}\] or                 \[s=r\theta \]     (3) Angular velocity \[(\omega )\] : Angular velocity of an object in circular motion is defined as the time rate of change of its angular displacement. (i) Angular velocity w =\[\frac{\text{angle}\,\text{ traced}}{\text{time}\,\text{ taken}}=\underset{\Delta t\to 0}{\mathop{Lt}}\,\frac{\Delta \theta }{\Delta t}=\frac{d\theta }{dt}\] \[\therefore \] \[\omega =\frac{d\theta }{dt}\]  (ii) Dimension : \[[{{M}^{0}}{{L}^{0}}{{T}^{-1}}]\] (iii) Units : Radians per second \[(rad.{{s}^{-1}})\] or Degree per second. (iv) Angular velocity is an axial vector. Its direction is the same as that of \[\Delta \theta \]. For anticlockwise rotation of the point object on the circular path, the direction of \[\omega ,\] according to Right hand rule is along the axis of circular path directed upwards.  For clockwise rotation of the point object on the circular path, the direction of \[\omega \] is along the axis of circular path directed downwards. (v) Relation between angular velocity and linear velocity \[\overrightarrow{v}=\overrightarrow{\omega }\times \overrightarrow{\,r}\] (vi) For uniform circular motion \[\omega \] remains constant where as for non-uniform motion w varies with more...

Circular motion is another example of motion in two dimensions. To create circular motion in a body it must be given some initial velocity and a force must then act on the body which is always directed at right angles to instantaneous velocity.                              Since this force is always at right angles to the displacement therefore no work is done by the force on the particle. Hence, its kinetic energy and thus speed is unaffected. But due to simultaneous action of the force and the velocity the particle follows resultant path, which in this case is a circle. Circular motion can be classified into two types - Uniform circular motion and non-uniform circular motion.  

Let a particle be projected up with a speed u from an inclined plane which makes an angle \[\alpha \] with the horizontal and velocity of projection makes an angle \[\theta \] with the inclined plane. We have taken reference x-axis in the direction of plane. Hence the component of initial velocity parallel and perpendicular to the plane are equal to \[u\cos \theta \] and \[u\sin \theta \] respectively i.e. \[{{u}_{||}}=u\cos \theta \] and  \[{{u}_{\bot }}=u\sin \theta \]. The component of g along the plane is \[g\sin \alpha \] and perpendicular to the plane is \[g\cos \alpha \] as shown in the figure i.e. \[{{a}_{||}}=-g\sin \alpha \] and \[{{a}_{\bot }}=g\cos \alpha \].   Therefore the particle decelerates at a rate of \[g\sin \alpha \] as it moves from O to P.   (1) Time of flight : We know for oblique projectile motion \[T=\frac{2u\sin \theta }{g}\]or    we can say  \[T=\frac{2{{u}_{\bot }}}{{{a}_{\bot }}}\] \[\therefore \] Time of flight on an inclined plane \[T=\frac{2u\sin \theta }{g\cos \alpha }\] (2) Maximum height : We know for oblique projectile motion \[H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\]      or  we can say \[H=\frac{u_{\bot }^{2}}{2{{a}_{\bot }}}\] \[\therefore \] Maximum height on an inclined plane \[H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g\cos \alpha }\] (3) Horizontal range : For one dimensional motion \[s=ut+\frac{1}{2}a{{t}^{2}}\] Horizontal range on an inclined plane \[R={{u}_{||}}T+\frac{1}{2}{{a}_{||}}{{T}^{2}}\] \[R=u\cos \theta \,T-\frac{1}{2}g\sin \alpha \,\,{{T}^{2}}\]            \[R=u\cos \theta \,\left( \frac{2u\sin \theta }{g\cos \alpha } \right)-\frac{1}{2}g\sin \alpha \,{{\left( \frac{2u\sin \theta }{g\cos \alpha } \right)}^{2}}\] By solving   \[R=\frac{2{{u}^{2}}}{g}\,\frac{\sin \theta \,\cos (\theta +\alpha )}{{{\cos }^{2}}\alpha }\] (i) Maximum range occurs when \[\theta =\frac{\pi }{4}-\frac{\alpha }{2}\] (ii) The maximum range along the inclined plane when the projectile is thrown upwards is given by            \[{{R}_{\max }}=\frac{{{u}^{2}}}{g\,(1+\sin \alpha )}\] (iii) The maximum range along the inclined plane when the projectile is thrown downwards is given by \[{{R}_{\max }}=\frac{{{u}^{2}}}{g\,(1-\sin \alpha )}\]    

When a body is projected horizontally from a certain height 'y' vertically above the ground with initial velocity u. If friction is considered to be absent, then there is no other horizontal force which can affect the horizontal motion. The horizontal velocity therefore remains constant and so the object covers equal distance in horizontal direction in equal intervals of time. The horizontal velocity therefore remains constant and so the object covers equal distance in horizontal direction in equal intervals of time. (1) Trajectory of horizontal projectile : The horizontal displacement x is governed by the equation                 \[x=ut\Rightarrow t=\frac{x}{u}\]    ..(i)         The vertical displacement y is governed by    \[y=\frac{1}{2}g{{t}^{2}}\]            ...(ii) (since initial vertical velocity is zero) By substituting the value of t in equation (ii) \[y=\frac{1}{2}\frac{g\,{{x}^{2}}}{{{u}^{2}}}\]                 (2) Displacement of Projectile \[\mathbf{(}\vec{r}\mathbf{)}\] : After time t, horizontal displacement \[x=ut\] and vertical displacement \[y=\frac{1}{2}g{{t}^{2}}\]. So, the position vector \[\vec{r}=ut\,\hat{i}+\frac{1}{2}g{{t}^{2}}\,\hat{j}\] Therefore  \[r=ut\,\sqrt{1+{{\left( \frac{gt}{2u} \right)}^{2}}}\]        and     \[\alpha =\,{{\tan }^{-1}}\left( \frac{gt}{2u} \right)\]                              \[\alpha ={{\tan }^{-1}}\left( {\sqrt{\frac{gy}{2}}}/{u}\; \right)\]         \[\left( \text{as}\,\,t=\sqrt{\frac{2y}{g}} \right)\] (3) Instantaneous velocity : Throughout the motion, the horizontal component of the velocity is \[{{\upsilon }_{x}}=u\]. The vertical component of velocity increases with time and is given by \[{{\upsilon }_{y}}=0+gt=gt\]  (From \[\upsilon =u+gt\] So, \[\vec{v}={{v}_{x}}\hat{i}+{{v}_{y}}\hat{j}\]  = \[u\,\hat{i}+g\,t\,\hat{j}\]             i.e. \[v=\sqrt{{{u}^{2}}+{{\left( gt \right)}^{2}}}\] \[=\,u\sqrt{1+{{\left( \frac{gt}{u} \right)}^{2}}}\] Again \[\vec{v}=u\hat{i}+\sqrt{2gy}\,\hat{j}\] i.e. \[v=\sqrt{{{u}^{2}}+2gy}\]      Direction of instantaneous velocity : \[\tan \,\varphi =\frac{{{v}_{y}}}{{{v}_{x}}}\]  \[\Rightarrow \] \[\varphi ={{\tan }^{-1}}\left( \frac{{{v}_{y}}}{{{v}_{x}}} \right)\] \[={{\tan }^{-1}}\left( \frac{\sqrt{2gy}}{u} \right)\]  or \[\varphi ={{\tan }^{-1}}\left( \frac{gt}{u} \right)\] Where \[\phi \] is the angle of instantaneous velocity from the horizontal. (4) Time of flight : If a body is projected horizontally from a height h with velocity u and time taken by the body to reach the ground is T, then \[h=0+\frac{1}{2}g{{T}^{2}}\]          (for vertical motion)            \[T=\sqrt{\frac{2h}{g}}\] (5) Horizontal range : Let R is the horizontal distance travelled by the body            \[R=uT+\frac{1}{2}\,\,0\,{{T}^{2}}\] (for horizontal motion)            \[R=u\sqrt{\frac{2h}{g}}\]   (6) If projectiles A and B are projected horizontally with different initial velocity from same height and third particle C  is dropped from same point then (i) All three particles will take equal time to reach the ground. (ii) Their net velocity would be different but all three particle possess same vertical component of velocity.           (iii) The trajectory of projectiles A and B will be straight line w.r.t. particle C.       (7) If various particles thrown with same initial velocity but in different direction then               (i) They strike the ground with same speed at different times irrespective of their initial direction of velocities.   (ii) Time would be least for particle E which was thrown vertically downward. (iii) Time would be maximum for particle A which was thrown more...

In projectile motion, horizontal component of velocity \[(u\cos \theta ),\] acceleration (g) and mechanical energy remains constant while, speed, velocity, vertical component of velocity \[(u\sin \theta ),\] momentum, kinetic energy and potential energy all changes. Velocity, and KE are maximum at the point of projection while minimum (but not zero) at highest point. (1) Equation of trajectory : A projectile is thrown with velocity u at an angle \[\theta \] with the horizontal. The velocity u can be resolved into two rectangular components.                 \[u\cos \theta \] component along X-axis and \[u\sin \theta \] component along Y-axis. For horizontal motion \[x=u\cos \theta \times t\Rightarrow t=\frac{x}{u\cos \theta }\] ...(i) For vertical motion    \[y=(u\sin \,\theta \,)\,t-\frac{1}{2}\,g{{t}^{2}}\]     ...(ii) From equation (i) and (ii)  \[y=u\sin \theta \left( \frac{x}{u\cos \,\theta } \right)\,-\frac{1}{2}g\,\left( \frac{{{x}^{2}}}{{{u}^{2}}{{\cos }^{2}}\,\theta } \right)\] \[y=x\,\tan \,\theta \,-\frac{1}{2}\frac{g{{x}^{2}}}{{{u}^{2}}{{\cos }^{2}}\,\theta }\]     This equation shows that the trajectory of projectile is parabolic because it is similar to equation of parabola \[y=ax-b{{x}^{2}}\] Note : 

• Equation of oblique projectile also can be written as

 \[y=x\tan \theta \,\left[ 1-\frac{x}{R} \right]\] (where R = horizontal range = \[\frac{{{u}^{2}}\sin 2\theta }{g}\]) (2) Displacement of projectile \[(\vec{r})\] : Let the particle acquires a position P having the coordinates (x, y) just after time t from the instant of projection. The corresponding position vector of the particle at time t is \[\vec{r}\] as shown in the figure.   \[\,\,\vec{r}=x\hat{i}+y\hat{j}\]                                               ...(i)  The horizontal distance covered during time t is given as \[x={{v}_{x}}\,t\Rightarrow x=u\cos \theta \,\,t\]                 ...(ii) The vertical velocity of the particle at time t is given as \[{{v}_{y}}={{({{v}_{0}})}_{y}}-gt,\]                                ...(iii) Now the vertical displacement y is given as \[y=u\sin \theta \,t-1/2\,g{{t}^{2}}\]                      ...(iv) Putting the values of x and y from equation (ii) and equation (iv) in equation (i) we obtain the position vector at any time t as \[\vec{r}=(u\cos \theta )\,t\,\hat{i}+\left( (u\sin \theta )\,t-\frac{1}{2}g{{t}^{2}} \right)\,\hat{j}\]   \[\Rightarrow \]\[r=\sqrt{{{(u\,t\cos \theta )}^{2}}+{{\left( (u\,t\sin \theta )-\frac{1}{2}g{{t}^{2}} \right)}^{2}}}\] \[r=u\,t\sqrt{1+{{\left( \frac{gt}{2u} \right)}^{2}}-\frac{gt\sin \theta }{u}}\] and \[\varphi ={{\tan }^{-1}}(y/x)\] \[={{\tan }^{-1}}\left( \frac{ut\sin \theta -\frac{1}{2}g{{t}^{2}}}{(u\,t\cos \theta )} \right)\] or \[\varphi ={{\tan }^{-1}}\left( \frac{2u\,\sin \theta -gt}{2u\,\cos \theta } \right)\] Note :  The angle of elevation f of the highest point of the projectile and the angle of projection q are related to each other as   \[\tan \varphi =\frac{1}{2}\tan \theta \]      (3) Instantaneous velocity \[\upsilon \] : In projectile motion, vertical component of velocity changes but  horizontal component of velocity remains always constant. Example : When a man jumps over the hurdle leaving behind its skateboard then vertical component of his velocity is changing, but more...