(A) Mass of the stone,$m = 0.25 \; kg$.Radius of the circle,$r = 1.5 \; m$.Frequency of revolution,$n = \frac{40}{60} = \frac{2}{3} \; rps$.Angular ve

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(A) Mass of the stone,$m = 0.25 \; kg$.Radius of the circle,$r = 1.5 \; m$.Frequency of revolution,$n = \frac{40}{60} = \frac{2}{3} \; rps$.Angular velocity,$\omega = 2 \pi n = 2 \times 3.1416 \times \frac{2}{3} \approx 4.189 \; rad/s$.The centripetal force is provided by the tension $T$ in the string: $T = m r \omega^2$.$T = 0.25 \times 1.5 \times (4.189)^2 \approx 6.58 \; N$.For the maximum speed,$T_{\max} = 200 \; N$.Using $T_{\max} = \frac{m v_{\max}^2}{r}$,we get $v_{\max} = \sqrt{\frac{T_{\max} \times r}{m}}$.$v_{\max} = \sqrt{\frac{200 \times 1.5}{0.25}} = \sqrt{1200} \approx 34.64 \; m/s$.

Class 11 Physics Newton's Laws of Motion and Friction Circular motion with Friction

Medium

$A$ 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 in $m/s$ with which the stone can be whirled around if the string can withstand a maximum tension of $200 \; N$ (in $; m/s$)?

A
$34.64$
B
$42.26$
C
$26.5$
D
$12.26$

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Newton's Laws of Motion and Friction

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Circular motion with Friction

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