One end of a cord is fixed and a small 0.650-kg object is attached to the other

Question

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 = 16.0°, the speed of the object is 5.80 m/s.

(a) At this instant, find the tension in the string.

T = ________

(b) At this instant, find the tangential and radial components of acceleration.

ar =________ m/s2 inward

at =_______ m/s2 downward tangent to the circle

(c) At this instant, find the total acceleration. atotal = m/s2 inward and below the cord at ________ °

Explanation / Answer

the net force provides centripetal accleration of the ball

F_T - mg cos theta = mv^2/t

Tension the string

               F_T = mv^2 / r + mg cos theta

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

                     = 0.65 ( 5.8 ^2 / 3   + 9.8 cos 16 )

                      = 13.41 N

   radis acceleration is

a = v^2 / r   = 5.8 ^2 / 3 = 11.21 m/s^2

    tangential acceleration    

a= g sin thets = 9.8 sin 16 = 2.7 m/s^2  

total accleration is

a = root at^2 + ar^2 = root (11.21)^2 + (2.7)^2 = 11.53 m/s^2

angle is

theta = tan^-1 ( 2.7/11.21 ) = 13.54 degree

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