If an object of mass 10kg is whirled around a horizontal circle of rad...
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.