Acceleration is the velocity change with respect to time. When it is constant for a particular period of time then it is called **constant acceleration**. It can be of average acceleration, instantaneous acceleration, when an object undergoes circular motion then the acceleration is of three types that are centripetal acceleration, tangential, and angular acceleration.

The essential part of roller coaster is a curve or a circular path which undergoes**centripetal acceleration**. Its direction is in the direction of centre of circular path. When we sit in train then the force which feel by us pushing us toward the outer edge of the circular path, this is called **centrifugal force**. This is not an actual force while it is inertia of our body or resistance to direction of train. Here, we discuss about centripetal acceleration in which the direction is towards the centre of circular path and its two main factors that are tangential and radial and its mathematical formula.

The essential part of roller coaster is a curve or a circular path which undergoes

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When an object is moving with uniform acceleration in circular direction, it is said to be experiencing the centripetal acceleration.

To understand this type of acceleration clearly, lets understand three different types of acceleration properly :

2. Centripetal Acceleration :

3.

There are a few key points to be noted about the centripetal acceleration :

- The direction of the centripetal acceleration is always along the radius vector of the circular direction.
- The magnitude of the centripetal acceleration can be calculated by the tangential speed and angular velocity.
- A object traveling in a circular direction experiences both the centripetal and the angular acceleration.
- The centripetal acceleration and angular acceleration are always 90
^{o}to each other

If we consider that the object is moving in the curve with the radius ‘r’ with tangential velocity ‘v_{t}’ then the centripetal acceleration can be defined as:

** ****$a_{c}=\lim_{\Delta t\rightarrow 0}\frac{dv_{t}}{dt}$**

** ****$a_{c}$ = $\frac{v_{t}}{t}$**

where,

a_{c} = centripetal acceleration

v_{t} = tangential velocity

t = time

If we consider that the object is moving in the curve with the radius ‘r’ with tangential velocity ‘v_{t}’ then the centripetal acceleration is given by,

**$a_{c}$ =**** $\frac{v_{t}^{2}}{r}$**

Now, if we consider the angular acceleration (a_{t}) of the object also, then its total acceleration would be;

**$a_{total}=\sqrt{a_{c}^{2}+a_{t}^{2}}$**

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where,

a

v

t = time

If we consider that the object is moving in the curve with the radius ‘r’ with tangential velocity ‘v

Now, if we consider the angular acceleration (a

The equation of centripetal acceleration,

**$a_{c}=\frac{v_{t}}{t}$**

Put the units of for velocity and radius in the above equation, the unit of centripetal acceleration comes out to be, **m/sec**^{2}

So, the centripetal acceleration has same unit as that of acceleration, i.e.**m/sec**^{2}.

So, the centripetal acceleration has same unit as that of acceleration, i.e.

The following problems will help you learn more about the centripetal acceleration :

From the problem it is clear that we have following variables available with us,

V

r = 5 meters;

m = 5 kg

Now since object is moving in the circle so its acceleration is calculated by calculating the centripetal acceleration

$a_{c}$ = $\frac{v_{t}^{2}}{r}$

$a_{c}$ = $\frac{10^{2}}{5}$

$a_{c}$ = $\frac{100}{5}$

$a_{c}$ = 20m/sec

So, the acceleration of the object is 20 m/sec

$a_{c}$ = $\frac{v_{t}^{2}}{r}$

$v_{t}^{2}$ = $a_{c}\times r$

$v_{t}$ = $\sqrt{a_{c}\times r}$

Putting values from the question, we have

$v_{t}$ = $\sqrt{8.05\times48.2}$

$v_{t}$ = $\sqrt{388.01}$

So,

$v_{t}$ = 19.69m/sec

Using the above formula, we have;

$v_{c}=\sqrt{a_{c}\times r}$

$v_{c}=\sqrt{3\times 2.1}$

$v_{c}=\sqrt{6.3}$

$v_{c}=2.5m/sec$

So, the tangential speed of the girl is 2.5 m/sec.

$a_{c}$ = $\frac{v_{t}^{2}}{r}$

r = $\frac{v_{t}^{2}}{a_{c}}$

r = $\frac{30^{2}}{15}$

r = $\frac{900}{15}$

r = 60m

So, the distance between the car and the center of the track is 60 meters.

$a_{c}$ = $\frac{v_{t}^{2}}{r}$

$a_{c}$ = $\frac{3^{2}}{0.5}$

$a_{c}$ = $\frac{9}{0.5}$

$a_{c}$ = 18m/sec

So, the centripetal acceleration of the bus is 18 m/sec

$a_{c}$ = $\frac{v_{t}^{2}}{r}$

$a_{c}$ = $\frac{5^{2}}{4}$

$a_{c}$ = $\frac{25}{4}$

$a_{c}$ = 6.25 m/sec

So, the centripetal acceleration of the body is 6.25 m/sec

$a_{c}$ = $\frac{v_{t}^{2}}{r}$

Here we need to first find the vt, first and then we will be able to the centripetal acceleration. Since here the frequency of the rotation of the object is given so we can find the vt as;

$v_{t}$ = 2$\pi$ $\times$13$\times$0.15 = 12.25m/sec

so centripetal acceleration is,

$a_{c}$ = $\frac{12.25^{2}}{0.15}$

$a_{c}$ = $\frac{150.06}{0.15}$

$a_{c}$ = 1000.41m/sec

So, the centripetal acceleration of the object is 1000 m/sec

From the equation of centripetal acceleration it is clear that the centripetal acceleration is directly proportional to the square of the tangential velocity so if the tangential velocity is increased by a factor of ‘2’ the centripetal acceleration is increased by the factor of ’4’.

From the equation of centripetal acceleration it is clear that the centripetal acceleration is inversely proportional to the radius of the curvature so if the radius is decreased by a factor of ‘2’ the centripetal acceleration is increased by the factor of ’2’.

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