centripetal acceleration and tangential acceleration

Today our most important topic is centripetal acceleration and tangential acceleration in previous topic we have covered basic concepts of circular motion and its parameter involved in circular motion like angular displacement, angular velocity angular acceleration time period frequency i will suggest you before going to study this topic first you study previous post basic concept of circular motion very interesting and hidden concept elaborated in depth and easy way to understand circular motion link hidden concepts circular motion today we will discuss acceleration in circular motion that is called centripetal acceleration today you will learn there is no only centripetal acceleration in circular motion there are four acceleration in circular motion we will study here all types of acceleration involved in circular motion so far we have studies 

 (linear displacement)  s = r๐›ณ(angular displacement)
(linear velocity)  v = r๐›š(angular velocity) r is radius of circle
(linear acceleration) a = r๐›‚(angular acceleration)
now here we will learn four acceleration step by step and its concepts very first centripetal acceleration meaning and its concepts
centripetal acceleration(a๐’ธ) 

centripetal force and centrifugal force,centripetal acceleration,centrifugal acceleration



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 a๐’ธ  = vยฒ/r where v is speed and r is radius now from previous post we have study v = r๐›š for circular motion only see the first post where we can’t write v = r๐›š so put value of in above equation we can also get a๐’ธ = rยฒ๐›šยฒ/r = r๐›šยฒ so we can also write  a๐’ธ  =r๐›šยฒ  where  ๐›š 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.
Tangential acceleration(a๐“‰) 

centripetal force and centrifugal force,centripetal acceleration,centrifugal acceleration

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 a๐“‰ = 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 a๐“‰ = 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(a๐“ท๐“ฎ๐“ฝ) this is due to total change in velocity due to direction and velocity magnitude both it is called net acceleration now its value is   a๐“ท๐“ฎ๐“ฝ = |dv/dt|  now its direction will not be in velocity direction because a๐’ธ and a๐“‰ are always perpendicular hence a๐“ท๐“ฎ๐“ฝ resultant will be  a๐“ท๐“ฎ๐“ฝ  = โˆša๐’ธยฒ+a๐“‰ยฒ now for direction use triangle law tan๐ฐ  = a๐’ธ/a๐“‰ now we will see some questions to clear concepts.
Q if tangential velocity v= 2t and radius r =9m find a๐’ธ,a๐“‰ and a๐“ท๐“ฎ๐“ฝ at t =3s.
Ans for velocity v= 2t put t=3 v =2*3 =6m/s
centripetal acceleration a๐’ธ = vยฒ/r = 6*6/9 = 4m/sยฒ
tangential acceleration  a๐“‰ = d| v |/dt = 2m/sยฒ
net acceleration a๐“ท๐“ฎ๐“ฝ = โˆša๐’ธยฒ+a๐“‰ยฒ = โˆš4ยฒ+2ยฒ = โˆš20
Q2 if  ๐›š = 2t and radius r =2 find a๐’ธ,a๐“‰ and a๐“ท๐“ฎ๐“ฝ 
now you can solve this problem yourself using v = r๐›š .
Angular acceleration(๐›‚ )  change in angular velocity divided by change in time ๐›‚avg  = โˆ†๐›š/โˆ†t  or  ๐›‚inst = d๐›š/dt 
Q if  ๐›š = tยฒ +1 find ๐›‚ in 0 to 2s and at 2s.
Ans angular acceleration in 0 to 2s put t=0 ๐›š0 = 1 now put t=2s ๐›š2 = 2*2+1 = 5 โˆ†๐›š = 5-1 = 4 hence ๐›‚avg  = โˆ†๐›š/โˆ†t = 4/2 =2 rad/sยฒ
now for ๐›‚ at 2s ๐›‚inst = d๐›š/dt  = 2t = 2*2 = 4 rad/sยฒ
now one relation in angular acceleration and linear acceleration 
a๐“‰  = r๐›‚  where a๐“‰ is tangential acceleration .
Q if | v | = constant in a circular motion then find ๐›‚, a๐“‰,a๐’ธ which will be constant ?
Ans since | v | is constant then ๐›š = v/r here both v and r constant then ๐›š will be constant hence ๐›‚ = dw/dt = 0 which is constant now for a๐“‰ we know a๐“‰  = r๐›‚ = 0 which is also constant now for a๐’ธ = vยฒ/r here v and r are constant then a๐’ธ  magnitude will be constant but a๐’ธ  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