1. For the rotational part of the experiment, a student measures a force of 1.8
ID: 3281496 • Letter: 1
Question
1. For the rotational part of the experiment, a student measures a force of 1.8 N when the radius is 12 cm and the angular velocity 10 rad/s. a) What is the total mass of the object undergoing uniform circular motion? b) If the same mass (at the same radius) is rotated at 5 rad/s, what will be the measured force?2. For a loop-the-loop of diameter 50 cm, what is the expected minimum starting height h that the ball must be released from to make it around the loop? If the height found by testing with a ball is 70 cm above the base of the track, what percent of the energy is lost to friction and wobble? 1. For the rotational part of the experiment, a student measures a force of 1.8 N when the radius is 12 cm and the angular velocity 10 rad/s. a) What is the total mass of the object undergoing uniform circular motion? b) If the same mass (at the same radius) is rotated at 5 rad/s, what will be the measured force?
2. For a loop-the-loop of diameter 50 cm, what is the expected minimum starting height h that the ball must be released from to make it around the loop? If the height found by testing with a ball is 70 cm above the base of the track, what percent of the energy is lost to friction and wobble? 1. For the rotational part of the experiment, a student measures a force of 1.8 N when the radius is 12 cm and the angular velocity 10 rad/s. a) What is the total mass of the object undergoing uniform circular motion? b) If the same mass (at the same radius) is rotated at 5 rad/s, what will be the measured force?
2. For a loop-the-loop of diameter 50 cm, what is the expected minimum starting height h that the ball must be released from to make it around the loop? If the height found by testing with a ball is 70 cm above the base of the track, what percent of the energy is lost to friction and wobble?
Explanation / Answer
1.
a) Centripetal force F = mr²
m = F/r² = 1.8 N/(0.12 m)(10 rad/s)² = 0.15 kg
b) Centripetal force F = mr² = (0.15 kg)(0.12 m)(5 rad/s)² = 0.45 N
2.
Energy conservation: mgh = ½ mgr + mg(2r)
Finally, h = 2r + r/2 = 5r/2.
h = 5(25cm)/2 = 62.5cm
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