Rotatory Motion
Angular Displacement ( dq) : The angle through which the radius vector rotates in a given time is called angular displacement. It is a vector quantity. This vector is positive if angle described is in anti-clockwise sense and negative if in clockwise sense. The direction of the vector is perpendicular to the plane of rotation as guided by the right hand thumb rule. The unit of angular displacement is redian ( rad ). Angular velocity (w) : The rate of angular displacement of a body is called angular velocity. The rate of linear displacement is called linear velocity ( v ) . Angular velocity , w = dq / dt If w is uniform or constant then w = q / t = 2p / T = 2pn rad / sec. Where q is the angular displacement in a time t. T is the period of revolution and n is the number of revolutions made by the body per sec. The unit of angular velocity is radian / sec . Its direction is the same as that of angular displacement. Angular acceleration ( a ) : The rate of change angular velocity of a rotating body is called angular acceleration . a = dw / dt = d2q / dt2 rad / s2 If a is constant , a = w - w0 / t where w0 is the initial angular velocity, and w is the final angular velocity after a time t. It is a vector. The unit of angular acceleration is radian sec-2Value of radian : The angle subtended at the center of a circle by an arc of length equal to its radius is called a radian. The angle in radian = arc / radius p radian = 180° 1 radian = 180° / p = 57° 18' Relation between w,v,a and a : If a particle rotates in a circle of radius r with uniform angular velocity w and linear velocity v , v = rw ………………………………..( 1 ) If the particle rotates with constant angular acceleration a and linear acceleration a in circular motion , a = ra ………………………………( 2 ) If a particle rotates in circular motion with constant angular acceleration, w = w0 + at …………………..( 1 )0 = w0t + ½ at2 ……………….( 2 )w2 - w02 = 2aq…………………( 3 ) where w0 is the initial angular velocity , w is the final angular velocity and q is the angular displacement after a time t . The no . of revolutions made by a particle. N = q / 2p . Centripetal Force : The centripetal force is that constant force, which acting at right angles to the motion of the body causes it to move in a circular path with constant angular velocity. The centripetal acceleration is acting towards the center O of the circle . This must be due to the action of a constant force F on the body. According to Newton's second law F = ma = mv2 / r = mrw2 Centrifugal Reaction : The centrifugal reaction is that radial force which acts outwards on the agency making the body to move in a circular path. Its magnitude is equal to that of the centripetal force but it is directed away from the center. The centrifugal reaction = mv2 / r = mrw2 Centrifugal Force : The centrifugal force is that constant force in the system acting away from the center of the circle along the radius which always equals the centripetal force . It may be noted that the centripetal fore and centrifugal force not form action reaction pair. This is always equal and opposite to the centripetal force. This is directed away from the center along the radius of the circle .
Centrifugal force = mv2 / r = mrw2 Angle of Banking : The angle of banking is the angle made by the elevated path with the horizontal . 1Expression for Angle of Banking : q = Tan-1 ( v2 / rg ) Height of Banking :
H = l * v2 / rgCondition for overturning of a vehicle ( on unbanked road ) : Let a vehicle move along a curved path on an unbanked road . Let G be the center of gravity of the vehicle at a vertical height h, above the ground , Let 2a be the distance between the front two wheels, Let N1 and N2 be the normal reactions of the ground on the inner and outer wheels respectively when the vehicle travels along a curved path of radius r with a speed v . The frictional force F between the wheels and the ground supplies the necessary centripetal force. Since there is no net vertical motion . N1 + N2 = mg ………………………….( 1 ) If there is to be no turning effect of the vehicle at G.Fn + N1a = N2a ………………………..( 2 )( N2 - N1) a + Fh = mv2h / r or N2 - N1 = mv2h / ar …………………………( 3 ) Addings eqs (1 ) and (3 ) 2 N2 = mg + mv2h / ar orN2 = ½ m ( g + v2 h / ar )………………………..( 4 ) And N1 = mg - N2 = ½ m ( g - v2h / ar )……………………….( 5 ) From eq (4) it follows that the reaction, N2 on the outer wheel is always positive.From eq (5) if v increases, N1 decreases and when g = v2h / ar…………….( 6 )N1 = 0 . Thus the inner wheel is no longer in contact with the ground , and the car overturns outwards. Rigid body : A body which does not undergo any change in its shape and volume by the application of force is called a rigid body . There are no perfect rigid bodies in nature. Ex : doors , pulleys , vehicles Translatory motion : A body is said to have translatory motion if all particles in it move with the same velocity and their paths are parallel. Ex : Motion of train , running boy Rotatory Motion : A body is said to have rotatory motion when it moves in a circular path about a fixed point or about an axis . Rotatory motion is caused and altered by a single couple , the angular acceleration of each particle is the same. Ex : The rotatory motion of earth about its own axis. The motion of a fly wheel. In general most of the bodies will have both translatory and rotatory motions. Non - uniform circular motion : In uniform circular motion the linear acceleration is along the radial direction but in non-uniform circular motion, the net linear acceleration is not along the radial direction and is inclined to it. In non - uniform circular motion , the rotating particle has a) radial centripetal acceleration component ar = v2 / r which only changes the direction of the velocity and b) tangential acceleration component at = ra , which changes the magnitude of the velocity. The resultant acceleration,
a = Ö ar2 + at2 Vertical circular motion : Let a heavy particle of mass m be suspended by a light inelastic thread from a fixed point.Let a velocity v1 be given to the particle at a point so that it undergoes vertical circular motion of radius r having a velocity v2 at the top.At any instant let the string make an angle q with the vertical and let v be its tangential velocity. Then T - mg cos q = mv2 / r ………………………( 1 ) Case (1) : If q = 0 and T1 is the tension in the thread at the bottom A , T1 - mg = mv12 / r ( or ) T1 = mv12 / r + mg ………………………………..(2) If q = 180° , and T2 is the tension at the top, T2 + mg = mv22 / r Or T2 = mv22 / r - mg Case (2) : Critical velocities at the top and bottom. If the velocity is to be minimum at the top or, if the string is not to slacken T2 = 0 And mv22 / r = mg . v2 = Ö gr …………………………………………(4) If v1 is the minimum velocity at the bottom so that the particle completes circular motion at the top, it follows from the law of conservation of energy. ½ mv12 = ½ mv22 + 2mgr. Or v12 = v22 + 4 gr. But v22 = gr v12 = 5 gr or v1 =Ö 5gr…………………………………………(5) Case (3) :If the tensions are so adjusted that v1 = v2 = v T1 = mv2 / r + mg………………………(6) And T2 = mv2 / r - mg ………………………..(7) Thus the tension in the string is maximum at the bottom and minimum at the top. Case (4) :If v1 < Ö 5gr , the particle either oscillates about the lowest point or leave the circular path. Conical pendulum : A conical pendulum consists of a heavy bob suspended by a light inelastic suspension thread from a fixed point. The bob is drawn away from its vertical position and set to describe a horizontal uniform circular motion about the vertical axis passing through the point of suspension. Then the suspension thread traces the lateral surface of a cone and hence it is called conical pendulum. Let m be mass of the bob of the conical pendulum .Let l be the length of the thread making an angle q with the vertical . Let the bob undergo uniform horizontal circular motion of radius r with linear velocity v and angular velocity w . Then T sin q = mv2 / r ……………………………(1) T cos q = mg ………………………………(2) tan q = v2 / rg and tan q = r / h = r / Ö l2 - r2 ………………………….(4) v = rw = r * 2pn = r * 2p / T…………………………….(5) where T is the period of revolution of the bob
Angular Displacement ( dq) : The angle through which the radius vector rotates in a given time is called angular displacement. It is a vector quantity. This vector is positive if angle described is in anti-clockwise sense and negative if in clockwise sense. The direction of the vector is perpendicular to the plane of rotation as guided by the right hand thumb rule. The unit of angular displacement is redian ( rad ). Angular velocity (w) : The rate of angular displacement of a body is called angular velocity. The rate of linear displacement is called linear velocity ( v ) . Angular velocity , w = dq / dt If w is uniform or constant then w = q / t = 2p / T = 2pn rad / sec. Where q is the angular displacement in a time t. T is the period of revolution and n is the number of revolutions made by the body per sec. The unit of angular velocity is radian / sec . Its direction is the same as that of angular displacement. Angular acceleration ( a ) : The rate of change angular velocity of a rotating body is called angular acceleration . a = dw / dt = d2q / dt2 rad / s2 If a is constant , a = w - w0 / t where w0 is the initial angular velocity, and w is the final angular velocity after a time t. It is a vector. The unit of angular acceleration is radian sec-2Value of radian : The angle subtended at the center of a circle by an arc of length equal to its radius is called a radian. The angle in radian = arc / radius p radian = 180° 1 radian = 180° / p = 57° 18' Relation between w,v,a and a : If a particle rotates in a circle of radius r with uniform angular velocity w and linear velocity v , v = rw ………………………………..( 1 ) If the particle rotates with constant angular acceleration a and linear acceleration a in circular motion , a = ra ………………………………( 2 ) If a particle rotates in circular motion with constant angular acceleration, w = w0 + at …………………..( 1 )0 = w0t + ½ at2 ……………….( 2 )w2 - w02 = 2aq…………………( 3 ) where w0 is the initial angular velocity , w is the final angular velocity and q is the angular displacement after a time t . The no . of revolutions made by a particle. N = q / 2p . Centripetal Force : The centripetal force is that constant force, which acting at right angles to the motion of the body causes it to move in a circular path with constant angular velocity. The centripetal acceleration is acting towards the center O of the circle . This must be due to the action of a constant force F on the body. According to Newton's second law F = ma = mv2 / r = mrw2 Centrifugal Reaction : The centrifugal reaction is that radial force which acts outwards on the agency making the body to move in a circular path. Its magnitude is equal to that of the centripetal force but it is directed away from the center. The centrifugal reaction = mv2 / r = mrw2 Centrifugal Force : The centrifugal force is that constant force in the system acting away from the center of the circle along the radius which always equals the centripetal force . It may be noted that the centripetal fore and centrifugal force not form action reaction pair. This is always equal and opposite to the centripetal force. This is directed away from the center along the radius of the circle .
Centrifugal force = mv2 / r = mrw2 Angle of Banking : The angle of banking is the angle made by the elevated path with the horizontal . 1Expression for Angle of Banking : q = Tan-1 ( v2 / rg ) Height of Banking :
H = l * v2 / rgCondition for overturning of a vehicle ( on unbanked road ) : Let a vehicle move along a curved path on an unbanked road . Let G be the center of gravity of the vehicle at a vertical height h, above the ground , Let 2a be the distance between the front two wheels, Let N1 and N2 be the normal reactions of the ground on the inner and outer wheels respectively when the vehicle travels along a curved path of radius r with a speed v . The frictional force F between the wheels and the ground supplies the necessary centripetal force. Since there is no net vertical motion . N1 + N2 = mg ………………………….( 1 ) If there is to be no turning effect of the vehicle at G.Fn + N1a = N2a ………………………..( 2 )( N2 - N1) a + Fh = mv2h / r or N2 - N1 = mv2h / ar …………………………( 3 ) Addings eqs (1 ) and (3 ) 2 N2 = mg + mv2h / ar orN2 = ½ m ( g + v2 h / ar )………………………..( 4 ) And N1 = mg - N2 = ½ m ( g - v2h / ar )……………………….( 5 ) From eq (4) it follows that the reaction, N2 on the outer wheel is always positive.From eq (5) if v increases, N1 decreases and when g = v2h / ar…………….( 6 )N1 = 0 . Thus the inner wheel is no longer in contact with the ground , and the car overturns outwards. Rigid body : A body which does not undergo any change in its shape and volume by the application of force is called a rigid body . There are no perfect rigid bodies in nature. Ex : doors , pulleys , vehicles Translatory motion : A body is said to have translatory motion if all particles in it move with the same velocity and their paths are parallel. Ex : Motion of train , running boy Rotatory Motion : A body is said to have rotatory motion when it moves in a circular path about a fixed point or about an axis . Rotatory motion is caused and altered by a single couple , the angular acceleration of each particle is the same. Ex : The rotatory motion of earth about its own axis. The motion of a fly wheel. In general most of the bodies will have both translatory and rotatory motions. Non - uniform circular motion : In uniform circular motion the linear acceleration is along the radial direction but in non-uniform circular motion, the net linear acceleration is not along the radial direction and is inclined to it. In non - uniform circular motion , the rotating particle has a) radial centripetal acceleration component ar = v2 / r which only changes the direction of the velocity and b) tangential acceleration component at = ra , which changes the magnitude of the velocity. The resultant acceleration,
a = Ö ar2 + at2 Vertical circular motion : Let a heavy particle of mass m be suspended by a light inelastic thread from a fixed point.Let a velocity v1 be given to the particle at a point so that it undergoes vertical circular motion of radius r having a velocity v2 at the top.At any instant let the string make an angle q with the vertical and let v be its tangential velocity. Then T - mg cos q = mv2 / r ………………………( 1 ) Case (1) : If q = 0 and T1 is the tension in the thread at the bottom A , T1 - mg = mv12 / r ( or ) T1 = mv12 / r + mg ………………………………..(2) If q = 180° , and T2 is the tension at the top, T2 + mg = mv22 / r Or T2 = mv22 / r - mg Case (2) : Critical velocities at the top and bottom. If the velocity is to be minimum at the top or, if the string is not to slacken T2 = 0 And mv22 / r = mg . v2 = Ö gr …………………………………………(4) If v1 is the minimum velocity at the bottom so that the particle completes circular motion at the top, it follows from the law of conservation of energy. ½ mv12 = ½ mv22 + 2mgr. Or v12 = v22 + 4 gr. But v22 = gr v12 = 5 gr or v1 =Ö 5gr…………………………………………(5) Case (3) :If the tensions are so adjusted that v1 = v2 = v T1 = mv2 / r + mg………………………(6) And T2 = mv2 / r - mg ………………………..(7) Thus the tension in the string is maximum at the bottom and minimum at the top. Case (4) :If v1 < Ö 5gr , the particle either oscillates about the lowest point or leave the circular path. Conical pendulum : A conical pendulum consists of a heavy bob suspended by a light inelastic suspension thread from a fixed point. The bob is drawn away from its vertical position and set to describe a horizontal uniform circular motion about the vertical axis passing through the point of suspension. Then the suspension thread traces the lateral surface of a cone and hence it is called conical pendulum. Let m be mass of the bob of the conical pendulum .Let l be the length of the thread making an angle q with the vertical . Let the bob undergo uniform horizontal circular motion of radius r with linear velocity v and angular velocity w . Then T sin q = mv2 / r ……………………………(1) T cos q = mg ………………………………(2) tan q = v2 / rg and tan q = r / h = r / Ö l2 - r2 ………………………….(4) v = rw = r * 2pn = r * 2p / T…………………………….(5) where T is the period of revolution of the bob
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