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SVEN [57.7K]
3 years ago
15

A(n) 0.2 kg object is swung in a vertical circular path on a string 0.1 m long. The acceleration of gravity is 9.8 m/s2 . If a c

onstant speed of 6.38 m/s is maintained, what is the tension in the string when the object is at the top of the circle.
Physics
1 answer:
Leya [2.2K]3 years ago
8 0

Answer:

T=83.37N

Explanation:

Since the object is under a circular motion, according to Newton's second law, when the object is at the top of the circle we have:

\sum F_y: T-mg=F_c

Where F_c is the centripetal force and is given by:

F_c=ma_c=m\frac{v^2}{r}

Replacing and solving for T:

T=m\frac{v^2}{r}+mg\\T=0.2kg\frac{(6.38\frac{m}{s})^2}{0.1m}+0.2kg(9.8\frac{m}{s^2})\\T=83.37N

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3 years ago
A merry-go-round rotates at the rate of 0.17 rev/s with an 79 kg man standing at a point 1.6 m from the axis of rotation.
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We can use the conservation of angular momentum to solve.

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2 years ago
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