a mass of 5 kg is whirled in a horizontal circle at one end of a string 50 cm long. the other end being fixed. if the string when hanging vertically will just support a load of 200 kg mass without breaking , find the maximum whirling speed in m/s

joj

To find the maximum whirling speed at which the 5 kg mass can be whirled without the string breaking, we need to consider the tension force in the string.

When the 5 kg mass is whirled in a horizontal circle, the tension force in the string provides the centripetal force required to keep the mass in circular motion. This tension force can be calculated using the equation:

T = (m * v^2) / r

Where:
- T is the tension force
- m is the mass of the object (5 kg)
- v is the velocity of the object in circular motion
- r is the radius of the circular path (half the length of the string)

We need to find the maximum velocity (v) at which the tension force does not exceed the breaking strength of the string.

The breaking strength of the string is given by the maximum load it can support when hanging vertically. In this case, the load is 200 kg. The force due to the load (F_load) is given by:

F_load = mass * g

Where g is the acceleration due to gravity (9.8 m/s^2).

In order to prevent the string from breaking, we need to ensure that the tension force (T) does not exceed the force due to the load (F_load).

Setting T = F_load, we have:

(m * v^2) / r = mass * g

Substituting the given values, we have:

(5 * v^2) / 0.5 = 200 * 9.8

Simplifying the equation:

v^2 = (200 * 9.8 * 0.5) / 5

v^2 = 196

Taking the square root of both sides:

v = sqrt(196) ≈ 14 m/s

Therefore, the maximum whirling speed at which the 5 kg mass can be whirled without the string breaking is approximately 14 m/s.