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Q5.21  A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev./min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?

Answers (1)


We are given,    frequency :

                                                   n\ =\ \frac{40}{60}\ =\ \frac{2}{3}

So, the angular velocity becomes :

                                                   \omega \ =\ 2\Pi n

By Newton's law of motion, we can write :

                                                  T\ =\ F_{centripetal }

or                                                     =\ \frac{mv^2}{r}\ =\ mw^2r

or                                                     =\ (0.25)\left ( 2\times\Pi \times \frac{2}{3} \right )^2(1.5)

or                                                     =\ 6.57\ N


Now, we are given maximum tension and we need to find maximum velocity for that :

                                                    T_{max}\ =\ \frac{mv_{max}^2}{r}

Thus,                                            v_{max}\ =\ \sqrt{\frac{T_{max}.r}{m}}

or                                                               =\ \sqrt{\frac{200\times 1.5}{0.25}}

or                                                                =\ 34.64\ m/s 

Posted by

Devendra Khairwa

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