A mass m = 8.300kg is suspended from a string of length L = 1.590m. It revolves

Question

A mass m = 8.300kg is suspended from a string of length L = 1.590m. It revolves in a horizontal circle (see Figure). The tangential speed of the mass is 3.727m/s. What is the angle theta between the string and the vertical? Use "deg" as your units.

Hint: Be sure to use units of deg. Remember that tan is sin/cos and that sin^2 = 1 - cos^2. If you want to solve exactly, you will have to solve a quadratic equation in the variable cos(theta). Or you could just use trial-and-error to solve sinx tanx = constant. Pick an x(theta), plug it in your calculator, if the constant turns out too small, try a larger x, etc. A numerical solution, achieved by trial-and-error, is perfectly OK. In real life, most engineering problems are too complicated to solve analytically; only numerical solutions exist.

I dont understand the trial and error and have tried multiple methods ive found on line and nothing is working..ive gotten 41.7 degrees and its saying its wrong please help!

Explanation / Answer

centripetal force = mv^2/ r = 8.3 * 3.727^2 / 1.59 sin A = T sin A where T = tension in string; A = angle theta between the string and the vertical (in degrees) 8.3 * 3.727^2 / 1.59 sin A = T sin A 72.51 = T sin ^2 A ---(1) resolving vertically, T cos A = mg T = 8.3 * 9.81 / cos A sub into (1), 72.51 = [8.3 * 9.81 / cos A] * sin ^2 A 0.89cos A = sin^2 A 0.89 cos A = 1 - cos ^2 A cos^2 A + 0.89 cos A -1 =0 cos A = { -0.89 + sq root [ 4.7921] } /2 = 0.6495 A = 49.49 deg

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