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one meter long string can bear maximum of 0.5 kg mass . a mass of 0.05 kg is tied to one end and rotated in a horizontal circle , calculate the maximum number of revolution so that string does not brakes
Asked by pushpakumari291279 | 20 Aug, 2022, 08:15: AM
Maximum Tension force Tmax that is bearable by string = M g

where M is mass hanging to give tension force in the string and g is acceleration due to gravity

Tension force F experienced by string, when it is attached to mass m and is revolved horizontally is given as

F = m ω2 L

where ω is angular speed of revolution and L is length of string

For maximum revolution speed , we have ,  F = Tmax

m ω2 L = M g

Number of revolution per second = 9.9 / ( 2 π )  ≈ 1.57

Number of revolution per minute  = 1.57 × 60  ≈ 94
Answered by Thiyagarajan K | 20 Aug, 2022, 11:31: AM

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