A particle of mass 1 kg, tied to a 1.2 m long string is whirled to perform the vertical circular motion, under gravity. The minimum speed of a particle is 5 m/s. Consider the following statements. P) Maximum speed must be 5sqrt5 m/s. Q) Difference between maximum and minimum tensions along the string is 60 N. Select the correct option.

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KollegeApply · Accepted Answer

Only the statement Q is correct. Explanation: The minimum speed of the particle = v_A = 5 m/s The length of the string (h) = 2.4 m The radius (l) = 1.2 m The mass of the particle, m = 1 kg Conserving energy between A and B $\Rightarrow$ `$\frac{1}{2}$ mv_A^2 + mgh = $\frac{1}{2}$mv_B^2 + 0 $\Rightarrow$ v_A^2 + 2gh = v_B^2 $\Rightarrow$ 5^2 + 2 xx 9.8 xx 2.4 = v_B^2 $\Rightarrow$ v_B^2 = 72 $\Rightarrow$ v_B = sqrt72 m/s At point B, T_B - mg = (mv_B^2)/r T_B = (mv_B^2)/r + mg ...(1) At point A, T_A + mg = (mv_A^2)/r T_A = (mv_A^2)/r - mg ...(2) T_B - T_A = ((mv_B^2)/r + mg) - ((mv_A^2)/r - mg) = m/r (v_B^2 - v_A^2) + 2mg = $\frac{1}{1}$.2 (72 - 25) + 2(1)(9.8)` = 60 N

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