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 3.00 m, as shown in the figure below. When θ = 17.0°, the speed of the object is 9.00 m/s. An object is swinging to the right and upward from the end of a cord attached to a horizontal surface. The cord makes an angle θ with the vertical. An arrow labeled vector v points in the direction of motion.

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

(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 °_____

Answer 1

(a)
The tension in the supports the weight of the object and provides the centripetal force that keeps the object moving in a circular path.

The vertical component of tension = T * cos 17˚
The vertical component of the tension supports the weight of the object.
T * cos 17˚ = m * g = 0.550 * 9.8 = 5.39
T = 5.39 ÷ cos 17˚ = 5.636 N This is the tension if the object was not moving.

Centripetal force = m * v^2/r = 0.55 * 9^2 /3 = 14.85 N
Total tension in cord = 5.636 + 14.85 = 20.486 N

(b)

The 20.486 N tension force is directed toward the center of the circle. So, all of the tension is the force causing the radial acceleration.
The component of the weight that is directed toward the center = 0.55 * 9.8 * cos 17˚

The net force that is directed toward the center = 20.486 – 0.55 * 9.8 * cos 17˚
Radial acceleration = net radial force ÷ mass
Radial acceleration = (20.486 – 0.55 * 9.8 * cos 17˚) ÷ 0.55

= 27.875 m/s^2


The direction of the tangential acceleration is perpendicular to the tension force in the cord, so the tension has no component causing tangential acceleration. The weight has a component that is tangent to the circle.
The tangential force = m* g * sin θ = 0.55 * 9.8 * sin 17˚
Tangential acceleration = tangential force ÷ mass

= 9.8 * sin 17°

= 2.865 m/s^2

(C)
Total acceleration = (27.875^2 +2.865^2)^0.5 = 28.02 m/s^2

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