### Discussion :: Applied Mechanics - Section 3 (Q.No.21)

Rakesh Dabhi said: (Nov 14, 2014) | |

mv^2/r = (1*25)/1 = 25. |

Bret said: (Sep 27, 2016) | |

@Rakesh. Is it correct? |

Anc said: (Jun 18, 2018) | |

A one kg mass tied at the end of the string 0.5 m long is whirled in a horizontal circle, the other end of the string being fixed. The breaking tension in the string is 50 N. Find the greatest speed that can be given to the mass. Solution: Given: Mass of a body = m = 1 kg, radius of circular path = r = 0.5 m, Centripetal force = F = 50 N. To find: Greatest speed = v = ?, The necessary centripetal force acting on a body is given by tension in the string. Ans: The greatest speed can be given to mass = 5 m/s. |

Ihsan Ullah said: (Oct 13, 2020) | |

The tension in the string will be equal to the centrifugal force, so, T = Fc. T = (mv^2)/r or T=mw^2r As w=v/r. So put the values of each, T = (1)*(5^2)*(1), T = 25N. |

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