Online Tutoring Is The Easiest, Most Cost-Effective Way For Students To Get The Help They Need Whenever They Need It.
Centripetal acceleration is experienced by a body moving around a circular path. It is always directed towards the center of the circle. Equation of centripetal acceleration (ac) is given as ac = v2/ r
Formula for calculating centripetal acceleration can be derived as follows.
X and Y are the two points in the circular path where the velocities are ‘v1’ and ‘v2’ respectively. The radius
as shown above is ‘r’. ‘dr’ is the change in positions. Consider here the motion is uniform i.e. velocities are
constant. From the definition of acceleration we can write a = (v2 – v1)/ (t2 – t1) ---------- (1)
It appears that the above expression will result zero. But, due to different direction the expression 1 will result
some value. We can write equation 1 as
a = dv/dt ------- (2)
Representing the velocity vectors by vector diagram we get.
Example1: A stone is rotating with a velocity of 2 rad/s around a radius of 1m. What is the centripetal acceleration of the stone?
Solution: Angular velocity of the stone is given 2 rev/s. Linear velocity ‘v’ is expressed in terms of angular velocity as
ω is the angular velocity and r is the radius of the circle. Centripetal acceleration in terms of angular velocity is written as
ac = ω2r
Placing the values in the above equation
ac = 22/1
ac = 4 m/s2
Example: What is the centripetal acceleration experienced by a car that is taking a turn of diameter 6m with a velocity of 25 m/s?
Solution: Centripetal acceleration is expressed in terms of velocity as
ac = v2/r -------------(1)
v = 25 m/s
Radius of the turn is half of the diameter value given in the question.
r = d/2
r = 6/2
r = 3m
Placing the values in equation 1
ac = 252 /3
ac = 208.33 m/s2
HAVE A QUESTION?
Chat With Our Tutoring Experts Now
I just want you to know how great Brinda has been in tutoring me on probability. She is patient and very knowledgeable in all concepts. I look forward to many more sessions with her. I appreciate her help and the great work that Tutor Pace is doing.
9th Grade, from Barbara
Thank you for your excellent tutoring services. It has surely made a lot of difference in our child's academic performance.
9th Grader, from Venkat (father of Sai)
I have enjoyed my session with Aparna and it was helpful. I will hope to see her as my Math tutor in future sessions as well.
10th Grade, from Senia Revayo
Your tutor has been such an inspiration to me as I have difficulty getting some of the physics concepts down and truly understanding them. She takes the time to fully explain each and every aspect of any problem I might ask for help and always with encouragement and humor.