A 0.300-kg object is swung in a circular path and in a vertical plane on a 0.500-m length string. If the angular speed at the bottom is 8.00 rad/s, what is the tension in the string when the object is at the bottom of the circle?Select one:a.23.52 Nb.25.08 Nc.41.34 Nd.9.6 Ne.12.54 N
Question
A 0.300-kg object is swung in a circular path and in a vertical plane on a 0.500-m length string. If the angular speed at the bottom is 8.00 rad/s, what is the tension in the string when the object is at the bottom of the circle?Select one:a.23.52 Nb.25.08 Nc.41.34 Nd.9.6 Ne.12.54 N
Solution
To solve this problem, we need to use the formula for the tension in a string of a pendulum at the bottom of its swing, which is:
T = mg + mω²r
where:
- m is the mass of the object (0.300 kg),
- g is the acceleration due to gravity (9.8 m/s²),
- ω is the angular speed (8.00 rad/s), and
- r is the radius of the circle, which is the length of the string (0.500 m).
First, calculate the gravitational force on the object:
mg = 0.300 kg * 9.8 m/s² = 2.94 N
Then, calculate the centripetal force on the object:
mω²r = 0.300 kg * (8.00 rad/s)² * 0.500 m = 9.6 N
Finally, add these two forces to find the tension in the string:
T = mg + mω²r = 2.94 N + 9.6 N = 12.54 N
So, the tension in the string when the object is at the bottom of the circle is 12.54 N. Therefore, the correct answer is e. 12.54 N.
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