<< Chapter < Page Chapter >> Page >

The normal component of acceleration is also called the centripetal component of acceleration or sometimes the radial component of acceleration . To understand centripetal acceleration, suppose you are traveling in a car on a circular track at a constant speed. Then, as we saw earlier, the acceleration vector points toward the center of the track at all times. As a rider in the car, you feel a pull toward the outside of the track because you are constantly turning. This sensation acts in the opposite direction of centripetal acceleration. The same holds true for noncircular paths. The reason is that your body tends to travel in a straight line and resists the force resulting from acceleration that push it toward the side. Note that at point B in [link] the acceleration vector is pointing backward. This is because the car is decelerating as it goes into the curve.

This figure has a curve representing the path of a car. The curve decreases and increases. There are two circles along the path The first circle has point A where the curve meets the circle. At point A there are three vectors. The first vector is asubt and is tangent to the curve at A. The second vector is asubr and is orthogonal to vector asubt. In between these vectors is vector a. The second circle has point B where the curve meets the circle. At point A there are three vectors. The first vector is asubt and is tangent to the curve at A. The second vector is asubr and is orthogonal to vector asubt. In between these vectors is vector a.
The tangential and normal components of acceleration can be used to describe the acceleration vector.

The tangential and normal unit vectors at any given point on the curve provide a frame of reference at that point. The tangential and normal components of acceleration are the projections of the acceleration vector onto T and N , respectively.

Finding components of acceleration

A particle moves in a path defined by the vector-valued function r ( t ) = t 2 i + ( 2 t 3 ) j + ( 3 t 2 3 t ) k , where t measures time in seconds and distance is measured in feet.

  1. Find a T and a N as functions of t .
  2. Find a T and a N at time t = 2 .
  1. Let’s start with [link] :
    v ( t ) = r ( t ) = 2 t i + 2 j + ( 6 t 3 ) k a ( t ) = v ( t ) = 2 i + 6 k a T = v · a v = ( 2 t i + 2 j + ( 6 t 3 ) k ) · ( 2 i + 6 k ) 2 t i + 2 j + ( 6 t 3 ) k = 4 t + 6 ( 6 t 3 ) ( 2 t ) 2 + 2 2 + ( 6 t 3 ) 2 = 40 t 18 40 t 2 36 t + 13 .

    Then we apply [link] :
    a N = a 2 a T 2 = 2 i + 6 k 2 ( 40 t 18 40 t 2 36 t + 13 ) 2 = 4 + 36 ( 40 t 18 ) 2 40 t 2 36 t + 13 = 40 ( 40 t 2 36 t + 13 ) ( 1600 t 2 1440 t + 324 ) 40 t 2 36 t + 13 = 196 40 t 2 36 t + 13 = 14 40 t 2 36 t + 13 .
  2. We must evaluate each of the answers from part a. at t = 2 :
    a T ( 2 ) = 40 ( 2 ) 18 40 ( 2 ) 2 36 ( 2 ) + 13 = 80 18 160 72 + 13 = 62 101 a N ( 2 ) = 14 40 ( 2 ) 2 36 ( 2 ) + 13 = 14 160 72 + 13 = 14 101 .

    The units of acceleration are feet per second squared, as are the units of the normal and tangential components of acceleration.
Got questions? Get instant answers now!
Got questions? Get instant answers now!

An object moves in a path defined by the vector-valued function r ( t ) = 4 t i + t 2 j , where t measures time in seconds.

  1. Find a T and a N as functions of t .
  2. Find a T and a N at time t = −3 .
  1. v ( t ) = r ( t ) = 4 i + 2 t j a ( t ) = v ( t ) = 2 j a T = 2 t t 2 + 4 , a N = 2 t 2 + 4
  2. a T ( −3 ) = 6 13 13 , a N ( −3 ) = 2 13 13
Got questions? Get instant answers now!

Projectile motion

Now let’s look at an application of vector functions. In particular, let’s consider the effect of gravity on the motion of an object as it travels through the air, and how it determines the resulting trajectory of that object. In the following, we ignore the effect of air resistance. This situation, with an object moving with an initial velocity but with no forces acting on it other than gravity, is known as projectile motion    . It describes the motion of objects from golf balls to baseballs, and from arrows to cannonballs.

First we need to choose a coordinate system. If we are standing at the origin of this coordinate system, then we choose the positive y -axis to be up, the negative y- axis to be down, and the positive x- axis to be forward (i.e., away from the thrower of the object). The effect of gravity is in a downward direction, so Newton’s second law tells us that the force on the object resulting from gravity is equal to the mass of the object times the acceleration resulting from to gravity, or F g = m g , where F g represents the force from gravity and g represents the acceleration resulting from gravity at Earth’s surface. The value of g in the English system of measurement is approximately 32 ft/sec 2 and it is approximately 9.8 m/sec 2 in the metric system. This is the only force acting on the object. Since gravity acts in a downward direction, we can write the force resulting from gravity in the form F g = m g j , as shown in the following figure.

Practice Key Terms 6

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Calculus volume 3' conversation and receive update notifications?

Ask