(open image in new tab if it\'s too small to see...) One end of a cord is fixed

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One end of a cord is fixed and a small 0.650 -kg object is attached to the other end, where it swings in a section of a vertical circle of radius 3.00 m as shown in the figure below. When theta = 25.0 degree the speed of the object is 7.00 m/s. At this instant, find each of the following the tension in the string T = N the tangential and radial components acceleration a_r = m/s^2 inward a_t = m/s^2 downward tangent to the circle the total acceleration a_total = m/s^2 inward and below the cord at degree Is your answer hanged it the object is swinging down towards point instead of swinging up? Yes No Explain your answer to part (d).

Explanation / Answer

m = 0.65 kg ; L = 3 m ; theta = 25 deg ; v = 7 m/s

a)Tension will be :

T = mg cos(theta) + m v^2/R

T = m [ g cos(theta) + v^2/R ]

T = 0.65 [ 9.8 x cos25 + 7^2/3] = 16.4 N

Hence, T = 16.4 N

b)aT = g sin(theta)

aT = 9.8 x sin25 = 4.14 m/s^2

aR = v^2/R = 49/3 = 16.33 m/s^2

Hence, aT = 4.14 m/s^2 and aR = 16.4 N

c)a = sqrt (aT^2 + aR^2)

a = sqrt (4.14^2 + 16.33^2) = 16.85 m/s^2

Hence, a = 16.85 m/s^2

d)Yes

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