If an object of mass 10kg is whirled around a horizontal circle of radius 4m by revolving string inclined at 30 degree to vertical. If the uniform speed of the object is 5m/s,the tension of the string(approx) is.?

Question

Answer

Calculation of Tension in the String Whirled Around a Horizontal Circle

Given Data:

Mass of the object = 10 kg

Radius of the circle = 4 m

Inclination of revolving string to vertical = 30°

Uniform speed of the object = 5 m/s

Formulae Used:

Centripetal force = mass x acceleration

Centripetal force = tension in the string

Acceleration = (velocity)^2 / radius

Force due to gravity = mass x acceleration due to gravity

Calculation:

The force acting on the object to keep it in circular motion is provided by the tension in the string. The force acting on the object is the centripetal force, given as:

Centripetal force = mass x acceleration

Acceleration of the object is given as:

Acceleration = (velocity)^2 / radius

Putting the values in the above equation, we get:

Acceleration = (5 m/s)^2 / 4 m

Acceleration = 6.25 m/s^2

Substituting the values of mass and acceleration in the centripetal force equation, we get:

Centripetal force = 10 kg x 6.25 m/s^2

Centripetal force = 62.5 N

As the object is in equilibrium, the tension in the string is equal to the centripetal force acting on the object. Therefore, the tension in the string is also 62.5 N.

Additionally, the force due to gravity acting on the object is given as:

Force due to gravity = mass x acceleration due to gravity

Force due to gravity = 10 kg x 9.81 m/s^2

Force due to gravity = 98.1 N

As the object is in equilibrium, the tension in the string must be greater than the force due to gravity acting on the object to keep it in circular motion. Therefore, the tension in the string must be greater than 98.1 N.

Answer:

The tension in the string is approximately 62.5 N. However, it must be greater than 98.1 N to keep the object in circular motion.

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