In this article, the topic, “centripetal acceleration in pendulum” with 5 important several Facts will be discuss in a brief manner. The acceleration in pendulum is not constant.

**The bob of the pendulum is worked by the center seeking force which is known as in technical form as, centrifugal force. The pendulum is follows a path in a circular way. For the tension of the string in the pendulum’s bob appears to obey the circular way. In the case of pendulum the acceleration will be in pick point at the last portion of the string.**

**What is the centripetal acceleration in pendulum?**

**What is the centripetal acceleration in pendulum?**

**Centripetal acceleration**** in pendulum can be defined as, the characteristics of the motion of the pendulum traversing a path in circular way, and centripetal acceleration always notified according to the centre of the path in circular way.**

From the second law of the Newton’s we can know that, a centrifugal force is present notified according to the centre of the path in circular way which is promise bond for the motion of the circular.

The derivation of **centripetal force** will be very clear when the concept of second law of the Newton’s and centripetal acceleration will be known.

For a matter the centripetal acceleration will be followed a constant speed of v in a circular way which radius is r. Then the expression can be written as,

For a matter the angular speed of will be ω in a circular way which radius is r. Then the expression can be written as,

a = ω^{2}r

When the matter goes from centripetal force to centripetal acceleration, the relation is follow second law of the Newton’s.

F = ma

The centripetal acceleration for a body is F and the mass will be carrying by the matter will be m, in that case the expression can be written as,

F =mv^{2}

and,

F = mω^{2}r

__Problem:-__

__Problem:-__

**A small sized solid ball which contains mass about 0.7 kilogram is connected with a string. The ball is continuously whirled in a horizontal circle at an unchanged speed. The radius will be for the horizontal circle 0.5 meter. The motion of the circular for the solid ball frequency will be 1.8 Hz.**

**1. Determine the value of the centripetal force.**

**2. Determine the amount of force would be required to move the solid ball with twice of the speed in the same circle.**

__Solution:-__

Given data are,

Mass (m) = 0.7 kg

Frequency (f) = 1.8 Hz

Radius (r) = 0.5 meter

We know that, the expression for the centripetal force is,

We also know that,

So,

F = 45 Newton

The centripetal acceleration for a body is F and the mass will be carrying by the matter will be m, in that case the expression can be written as,

For this reason, the value of speed is doubled then the value of acceleration force will be increase the factor of, 2^{2} = 4

So,

F = 4*45

f = 180 Newton

A small sized solid ball which contains mass about 0.7 kilogram is connected with a string. The ball is continuously whirled in a horizontal circle at an unchanged speed. The radius will be for the horizontal circle 0.5 meter. The motion of the circular for the solid ball frequency will be 1.8 Hz.

1. The value of the centripetal force is 45 Newton.

2. The amount of force would be required to move the solid ball with twice of the speed in the same circle is, 180 Newton.

**Is the centripetal acceleration in pendulum a constant?**

**No, the centripetal acceleration in pendulum is not unchanged.**

The tension and the centripetal acceleration are both directed toward the center of the circle. The net tangential force leads to a tangential acceleration. The centripetal acceleration is never constant, but if the radius of the orbit that the object is moving in is very large and the speed of the object is relatively less than for a fraction of a second or so, the centripetal acceleration might be considered as a constant value.

**How to find the centripetal acceleration in pendulum?**

**The way of finding the centripetal acceleration in pendulum is discuss below,**

**For a bob of a pendulum the mass will be m which is connected with the end portion of the string. The length of the string will be L and the centripetal acceleration will be followed a constant speed in a circular way which radius is r. The angle will be made by the string [latex]\Theta[/latex].**

**The bob of the pendulum is moves towards horizontal. The vertical motion is not observed in the bob for this reason the vertical force must be equality. Then the expression can be written as,**

**[latex]T cos \Theta = mg ….eqn[/latex] (1)**

**The resultant horizontal force functions as the centripetal force. Then the expression can be written as,**

**[latex]T sin \Theta = \frac{mv^2}{r}[/latex]….eqn (2)**

**Now, dividing the eqn (2) by eqn (1)**

**[latex]\frac{T cos \Theta }{T sin \Theta } = \frac{\frac{mv^2}{r}}{mg}[/latex]**

**[latex]tan \Theta = \frac{v^2}{rg}[/latex]….eqn (3)**

**Now,**

**[latex]tan \Theta = \frac{Opposite}{Adjacent}[/latex]**

**[latex]tan \Theta = \frac{r}{h}[/latex]**

**[latex]tan \Theta = \frac{r}{\sqrt{L^2 -r^2}}[/latex] …. eqn (4)**

**Comparing the eqn (3) and eqn (4) we can write,**

**[latex]\frac{r}{h} = \frac{v^2}{rg}[/latex]**

**[latex]\frac{r}{\sqrt{L^2 – r^2}} = \frac{v^2}{rg}[/latex]**

**[latex]\frac{rg}{\sqrt{L^2 – r^2}} = \frac{v^2}{r}[/latex]**

**[latex]\frac{rg}{\sqrt{L^2 – r^2}} = a_c[/latex]**

**[latex]a_c = \frac{rg}{\sqrt{L^2 – r^2}}[/latex]**

**The centripetal acceleration in pendulum is [latex]\frac{rg}{\sqrt{L^2 – r^2}}[/latex]**

**When centripetal acceleration in pendulum is zero?**

**At equilibrium position, the pendulum has zero acceleration, it is where acceleration start to change sign, by which it means it begin to decelerate.**

**Problem:-**

**A boat is travelling from X named place to Y named placed at the speed of 30 meter per second in the section of banked in the river. Now the radius for the horizontal circular way is about 250 meter.**

**Determine the angle of the banking required to stick the boat at the speed of the moving without any friction in between the surface of the boat and water of the river.**

**Solution:-**

**Given data are,**

**Speed of the boat (v) = 30 meter per second**

**Radius (r) = 250 meter**

**g = 9.81 meter per second square**

**If friction is present it would pay subscription to centripetal force and boat would be capable to move at a higher speed. Nevertheless, we let that the friction is zero here.**

**We know that,**

**[latex]tan \Theta = \frac{v^2}{rg}[/latex]**

**[latex]\Theta = tan^-^1 \frac{v^2}{rg}[/latex]**

**[latex]\Theta = tan^-^1 ( \frac{30^2}{250 \times 9.81})[/latex]**

**[latex]\Theta[/latex] = 13.74** **degree**

A boat is travelling from X named place to Y named placed at the speed of 30 meter per second in the section of banked in the river. Now the radius for the horizontal circular way is about 250 meter.

The angle of the banking required sticking the boat at the speed of the moving without any friction in between the surface of the boat and water of the river is 13.74 degree.

**Problem:-**

**A car is travelling from Kolkata to Durgapur at the speed of 35 meter per second in the section of banked in the road. Now the radius for the horizontal circular way is about 350 meter.**

**Determine the angle of the banking required to stick the car at the speed of the moving without any friction in between the tyre of the car and the road.**

**Solution:-**

**Given data are,**

**Speed of the car (v) = 35 meter per second**

**Radius (r) = 350 meter**

**g = 9.81 meter per second square**

**If friction is present it would pay subscription to centripetal force and tyre of the car would be capable to move at a higher speed. Nevertheless, we let that the friction is zero here.**

**We know that,**

**[latex]tan \Theta = \frac{v^2}{rg}[/latex]**

**[latex]\Theta = tan^-^1 \frac{v^2}{rg}[/latex]**

**[latex]\Theta = tan^-^1 ( \frac{35^2}{350 \times 9.81})[/latex]**

**[latex]\Theta[/latex] = 19.59** **degree.**

**A car is travelling from Kolkata to Durgapur at the speed of 35 meter per second in the section of banked in the road. Now the radius for the horizontal circular way is about 350 meter.**

**The angle of the banking required to stick the car at the speed of the moving without any friction in between the Tyre of the car and the road** **is 19.59 degree.**

**Conclusion:**

**The weight of the bob of the pendulum is only can move towards the perpendicular to the string, the acceleration force will be perpendicular to the string**.