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- A conical pendulum consists of a bob of mass ‘m’ revolving in a horizontal circle with constant speed ‘v’ at the end of a string of length ‘l’. In this case, the string makes a constant angle with the vertical. The bob of pendulum describes a horizontal circle and the string describes a cone.

#### Expression for Period of Conical Pendulum:

- Let us consider a conical pendulum consists of a bob of mass ‘m’ revolving in a horizontal circle with constant speed ‘v’ at the end of a string of length ‘l’. Let the string makes a constant angle ‘θ’ with the vertical. let ‘h’ be the depth of the bob below the support.

- The tension ‘F’ in the string can be resolved into two components. Horizontal ‘Fsin θ’ and vertical ‘Fcos θ’.

The vertical component (F cos θ) balances the weight mg of the vehicle.

F cosθ = mg ………….. (1)

The horizontal component (F sin θ) provides the necessary centripetal force.

F sin θ = mv^{2}/r ………… (2)

Dividing equation (2) by (1) we get,

But v = rω

Where ω is angular speed and T is the period of the pendulum.

From figure tan θ = r/h

Substituting in equation (4)

This is an expression for the time period of a conical pendulum.

- The time taken by the bob of a conical pendulum to complete one horizontal circle is called time period of the conical pendulum

#### Notes:

- The semi vertical angle θ or angle made by the string with vertical depends on the length and period of the conical pendulum. If time period decreases, then the quantity cosθ decreases. in turn θ increases. θ can never be 90°, because for this period T = 0.

#### Expression for Tension in the String of Conical Pendulum:

- Let us consider a conical pendulum consists of a bob of mass ‘m’ revolving in a horizontal circle with constant speed ‘v’ at the end of a string of length ‘l’. Let the string makes a constant angle ‘θ’ with the vertical. let ‘h’ be the depth of the bob below the support.

- The tension ‘F’ in the string can be resolved into two components. Horizontal ‘Fsin θ’ and vertical ‘Fcos θ’.

The vertical component (F cos θ) balances the weight mg of the vehicle.

F cosθ = mg ………….. (1)

The horizontal component (F sin θ) provides the necessary centripetal force.

F sin θ = mv^{2}/r ………… (2)

Dividing equation (2) by (1) we get,

Squaring equations (1) and (2) and adding

Substituting equation (5)

This is an expression for the tension in the string of a conical pendulum.

#### Notes:

- A simple pendulum is a special case of a conical pendulum in which angle made by the string with vertical is zero i.e. θ = 0°.
- The period of the simple pendulum is given by

- For conical pendulum θ < 10° time period obtained is almost same as the time period for simple pendulum having same length as that of conical pendulum. Hence when performing experiment on simple pendulum, teacher advise not to increase θ beyond 10°.

#### Example – 1:

- A cord 5.0 m long is fixed at one end and to its their end is attached a weight which describes a horizontal circle of radius 1.2 m. Compute the speed of the weight in the circular path. Find its time period.
**Solution:****Given:**Length of conical pendulum = l = 5.0 m, radius of circular path of bob = r = 1.2 m, g = 9.8 m/s^{2},**To find:**velocity of weight = v = ?, Period = T = ?

By Pythagoras theorem

**Ans: **The speed of the weight is 1.7 m/s, The period of the motion of the weight is 4.41 s.

#### Example – 2:

- A stone of mass 1 kg is whirled in a horizontal circle attached at the end of 1 m long string making an angle of 30° with the vertical. Find the period and centripetal force if g = 9.8m/s
^{2}. **Solution:****Given:**Length of pendulum = l = 1 m, angle with vertical = θ = 30°, g = 9.8 m/s^{2},**To find:**Period = T = ?, Centripetal force = F = ?

**Ans: **Period of motion = 1.867 s, Centripetal force = 5.657 N

#### Example – 3:

- A string of length 0.5 m carries a bob of mass 0.1 kg with a period 1.41 s. Calculate the angle of inclination of string with vertical and tension in the string.
**Solution:****Given:**Length of pendulum = l = 0.5 m, mass of bob = m = 0.1 kg, Period = T = 1.41 s, g = 9.8 m/s^{2},**To find:**The angle with vertical = θ = ?, Tension = F = ?

For equilibrium, Total upward force = Total downward force

**Ans: **Angle of string with vertical = 8°52’, The tension in string = 0.992 N

#### Example – 4:

- A conical pendulum has a length of 50 cm. Its bob of mass 100 g performs a uniform circular motion in a horizontal plane in a circle of radius 30 cm. Find a) the angle made by the string with the vertical b) the tension in the string c) the period d) the speed of the bob e) centripetal acceleration of the bob f) centripetal and centrifugal force acting on the bob.
**Solution:****Given:**length of pendulum = l = 50 cm = 0.5 m, mass of bob = m = 100 g = 0.1 kg, radius of circle = r = 30 cm = 0.3 m, g = 9.8 m/s^{2},**To find:**Angle made with vertical = θ = ?, Tension = T = ?, Period, Centripetal force = F = ?, centrifugal force = F = ?

Calculation of tension in string (In diagram F = T)

For equilibrium, Total upward force = Total downward force

Calculation of time period

**Ans: **The angle of string with vertical = 36°52’, The tension in the string = 1.225 N, Period = 1.27 s

Velocity of a bob = 1.48 m/s, Centripetal force = 0.73 N radially inward, Centrifugal force = 0.73 N radially outward

**Note: **Above explanations and problems are based on assumption that reference frame is inertial.

#### Period of a simple pendulum in Non-Inertial Reference Frame:

- Reference frames which are at rest or moving with constant velocity with respect to the earth are called inertial reference frames.
- Reference frames which are moving with acceleration with respect to the earth are called non-inertial reference frames. In case of the non-inertial reference frame, we have to consider the pseudo force acting on the bob of the pendulum and corresponding changes should be done in the formula.
- For a simple pendulum oscillating in a vehicle moving horizontally with acceleration (a) the time period is

The pendulum will make an angle θ = tan^{-1}(g/a) with the vertical

- For a simple pendulum oscillating in a lift moving upward with acceleration (a) the time period of oscillation is

- For a simple pendulum oscillating in a lift moving downward with acceleration (a) the time period of oscillation is

- For a conical pendulum oscillating in a vehicle moving horizontally with acceleration (a) the time period is

The pendulum will make an angle θ + tan-1(g/a) with the vertical

- For a conical pendulum oscillating in a lift moving upward with acceleration (a) the time period of oscillation is

- For a conical pendulum oscillating in a lift moving downward with acceleration (a) the time period of oscillation is

Science > Physics > Circular Motion > You are Here |

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