The normal acceleration is the centripetal acceleration given by,
Saturday, July 21, 2007
Irodov Problem 1.39
The normal acceleration is the centripetal acceleration given by,
Irodov Problem 1.38
(a) The normal acceleration is the centripetal acceleration of the particle and is given by, v2/R, v is the instantaneous tangential velocity of the particle. Since the tangential deceleration and normal acceleration are equal, this means that the particle is decelerating tangentially at a rate v2/R.
In other words,

To find the velocity v as a function of distance covered s we could use the relation between v and t to find the relation between s and t and so on... however, a simpler trick exists,
(b) Once (a) is done, we already know that the tangential and normal acceleration components are equal, thus, the net acceleration must be,
knowing the relation between v and s we can determine the total acceleration as a function of s as,
In other words,
To find the velocity v as a function of distance covered s we could use the relation between v and t to find the relation between s and t and so on... however, a simpler trick exists,
(b) Once (a) is done, we already know that the tangential and normal acceleration components are equal, thus, the net acceleration must be,
knowing the relation between v and s we can determine the total acceleration as a function of s as,
Irodov Problem 1.37
This is a good problem to demonstrate the origin of centripetal acceleration. So I will solve it using two methods, one using only math and the other on by using centripetal acceleration.
Solution Using Centripetal Acceleration:
As seen in the figure, the particle undergoes a tangential acceleration of a - this simply comes from the fact that the tangential velocity is given by v = at. In addition since the particle is turning around in a circle, it is also experience centripetal acceleration that is directed in the perpendicular (normal) direction towards the center of the circle and is equal to v2/r. The relation between the distance d traveled by the particle and the instantaneous tangential velocity is given by
, thus the centripetal acceleration is given by,
. When the particle travels n fraction of the circle, the distance d covered is
. Thus, the centripetal acceleration of the particle is equal to
. The total acceleration is thus given by,
Solution based on mathematics :
The instantaneous position of a particle rotating in a circle of radius r and a tangential velocity v
is given by,

Thus, for this problem the position vector of the particle is given by,

Knowing that,
we can reach the same result as obtain above. This solution shows the origin's of centripetal acceleration which comes from the basic fact that rotation itself by definition is acceleration!!!
Solution Using Centripetal Acceleration:
As seen in the figure, the particle undergoes a tangential acceleration of a - this simply comes from the fact that the tangential velocity is given by v = at. In addition since the particle is turning around in a circle, it is also experience centripetal acceleration that is directed in the perpendicular (normal) direction towards the center of the circle and is equal to v2/r. The relation between the distance d traveled by the particle and the instantaneous tangential velocity is given by
Solution based on mathematics :
The instantaneous position of a particle rotating in a circle of radius r and a tangential velocity v
is given by,
Thus, for this problem the position vector of the particle is given by,
Knowing that,
we can reach the same result as obtain above. This solution shows the origin's of centripetal acceleration which comes from the basic fact that rotation itself by definition is acceleration!!!
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