Gnarls B.

Circular Motion: Finding the angle from the vertical

A .25 kg object is attached to a 0.80m string. The object is pulled to a point so that the string makes an angle θ from the vertical. It is then released. At the bottom of the object's path, the object reaches a speed of 3.0 m/s. What is angle θ?

I know T, the tension provides the centripetal force.

T=(mv2)/R

T=2.81 N

I don't know what to do next. The answer is 65° but I don't know how to get that answer

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Arturo O.

By the way, the centripetal force in this problem is a combination of the tension AND a component of weight.  If the motion was horizontal, then the centripetal force would be just the tension.
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03/23/18

Gnarls B.

Could you explain why this is the case? Gravity also supplying the centripetal force isn't making sense for me
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03/24/18

Arturo O.

Draw a free body diagram showing the forces acting on the mass when it is revolving in a vertical plane.  You will see that both the tension and a component of the weight act in the radial direction, and both have to be accounted for when calculating the centripetal force.  For example, at the top, both the tension and weight point down toward the center, so they both add up to the centripetal force.  At the bottom, the tension and weight point in opposite directions, so their difference is the centripetal force.  In between the top and bottom, you have to do a vector addition of tension and weight, accounting for the changing direction of the radius vectors.
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03/24/18

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