Question

One end of a cord is fixed and a small 0.550-kg object is attached to the other end, where it swings in a section of a vertical circle of radius 1.00 m, as shown in the figure below. When θ = 17.0°, the speed of the object is 6.50 m/s. (a) At this instant, find the magnitude of the tension in the string. (b) At this instant, find the tangential and radial components of acceleration. at = m/s2 inward ac = m/s2 downward tangent to the circle (c) At this instant, find the total acceleration. inward and below the cord at ° (d) Is your answer changed if the object is swinging down toward its lowest point instead of swinging up? Yes No (e) Explain your answer to part (d).

Answer #1

Writing equations of motion

T = (m*v^2/r) + m*g*cos(theta)

T = (0.55*9.8*cos(17)) + (0.55*6.5^2/1) = 28.4 N is the tension in the string

b) a_tan = F_tan/m = (m*g*sin(17))/m = g*sin(17) = 9.8*sin(17) = 2.86 m/s^2

a_rad = v^2/r = (6.5^2)/1 = 42.25 m/s^2

C) total accelaration is a = sqrt(a_tan^2+a_rad^2) = sqrt(2.86^2+42.25^2) = 42.34 m/s^2

theta = tan^(-1)(a_tan/a_rad) = tan^(-1)(2.86/42.25) = 3.87 degrees

D) No

E) the radial and tangential components does not depend on the direction of motion

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