whenever a particle rotate on circular path its velocity always changes why ? because its direction of motion changes hence it is clear that velocity will be always variable whether the magnitude of velocity is constant but velocity will be variable due to change in direction hence you know when velocity is variable the acceleration must be there hence centripetal acceleration comes in picture due to change in direction of velocity very important point since centripetal acceleration is a vector quantity hence its direction is always towards the center of circle along the radius as shown in above figure where its value is a? = v²/r where v is speed and r is radius now from previous post we have study v = r? for circular motion only see the first post where we can’t write v = r? so put value of in above equation we can also get a? = r²?²/r = r?² so we can also write a? =r?² where ? is angular velocity now i hope you have got how centripetal acceleration comes in picture now its formula derivation we discuss in next post don’t worry.
this acceleration arises due to change in speed of particle or change in magnitude of velocity hence if particle velocity magnitude is changing due to this a acceleration will come in picture which is called tangential acceleration this is also a vector quantity hence its will have definitely direction now direction depend upon the velocity magnitude if velocity magnitude is increasing then direction will be along the velocity direction and if velocity is decreasing then direction will be opposite to velocity direction tangent to the radius as shown in above picture now its value is define as a? = d|v|/dt remember this is change in speed upon change in time concept it is not change in velocity upon change time v = velocity and | v | = speed hence in tangential acceleration has a condition if velocity is constant then d|v|/dt = 0 because constant differentiation is zero so in this case a? = d|v|/dt =0 so only centripetal acceleration will be there no tangential acceleration sometime centripetal acceleration is also called normal acceleration or perpendicular acceleration because it is perpendicular to the surface and tangential acceleration is called parallel acceleration because it is parallel to surface as shown above figure .
Net acceleration(a???) this is due to total change in velocity due to direction and velocity magnitude both it is called net acceleration now its value is a??? = |dv/dt| now its direction will not be in velocity direction because a? and a? are always perpendicular hence a??? resultant will be a??? = √a?²+a?² now for direction use triangle law tan? = a?/a? now we will see some questions to clear concepts.
Q if tangential velocity v= 2t and radius r =9m find a?,a? and a??? at t =3s.
Ans for velocity v= 2t put t=3 v =2*3 =6m/s
centripetal acceleration a? = v²/r = 6*6/9 = 4m/s²
tangential acceleration a? = d| v |/dt = 2m/s²
net acceleration a??? = √a?²+a?² = √4²+2² = √20
Q2 if ? = 2t and radius r =2 find a?,a? and a???
now you can solve this problem yourself using v = r? .
Angular acceleration(? ) change in angular velocity divided by change in time ?avg = ∆?/∆t or ?inst = d?/dt
Q if ? = t² +1 find ? in 0 to 2s and at 2s.
Ans angular acceleration in 0 to 2s put t=0 ?0 = 1 now put t=2s ?2 = 2*2+1 = 5 ∆? = 5-1 = 4 hence ?avg = ∆?/∆t = 4/2 =2 rad/s²
now for ? at 2s ?inst = d?/dt = 2t = 2*2 = 4 rad/s²
now one relation in angular acceleration and linear acceleration
a? = r? where a? is tangential acceleration .
Q if | v | = constant in a circular motion then find ?, a?,a? which will be constant ?
Ans since | v | is constant then ? = v/r here both v and r constant then ? will be constant hence ? = dw/dt = 0 which is constant now for a? we know a? = r? = 0 which is also constant now for a? = v²/r here v and r are constant then a? magnitude will be constant but a? is a vector quantity and its direction is variable all the time in circular motion hence it will not be constant it will be variable remember this is important concept this is all about today topic we will continue in next post i hope you have enjoyed learning centripetal acceleration and tangential acceleration thanks for reading.
dated 16th sep 2018