Question

1. For the rotational part of the experiment, a student measures a force of 1.7 0.2 N when the radius is 12.0 0.1 cm and the angular velocity 1 1.0 0.3 rads. What is the total mass of the object undergoing uniform circular motion (Equation 5)? The uncertainty is given by ơn-m/(?),(7), (2ジ force, including uncertainty (Section 1.4.2)? a. σω b. If the same mass (at the same radius) is rotated at 4.0 ± 0.1 rad/s, what is the expected 2. For the loop-the-loop of diameter is 45.0+0.5 cm: a. What is the expected minimum starting height hcale that the ball must be released from to make it around the loop (Equation 10), including uncertainty (Section 2.3)? b. If the height found by testing is hmeas130 cm above the base of the track, what fraction of the energy is lost to friction and wobble, including uncertainty (Section 2.4)?

Answer 1

Given that: - force, 'F' = 1.7 plusminus 0.2 N radius, 'r' = 12.0 plusminus 0.1 cm angular velocity, w = 11.0 plusminus 0.3 rad/sec therefore we have, F = mRW^2 m = F/RW^2 = 1.7/12.0 times 10^-2 times (11.0)^2 = 0.117 kg uncertainty sigma_m = squareroot (6f/F)^2 + (6r/r)^2 + (2 times 6w/w)^2 = 0.117 squareroot (0.2/17)^2 + (0.1/11.0)^2 + (2 times 0.3/11.0)^2 = 0.117 squareroot 0.02 + 0.26 + 0.003 = 0.117 squareroot 0.283 = 0.117 times 0.53 = 0.06 kg \r\n given data: radius(r) = 4.0 plusminus 0.1 rad/s F = mRW^2 = 0.117 times 4^2 times 11 times 10^-2 = 21.12 times 10^-2 kg F = 21.12 times 10^-2 kg sigma_F = F_ca/c squareroot (6m/m)^2 + (6r/r)^2 + (2 times 6w/w)^2 = 21.12 times 10^-2 squareroot (0.06/0.12) + (0.1/11)^2 + (2 times 0.1/4)^2 = 21.12 times 10^-2 squareroot 0.25 + 0.000081 + 0.0025 = 21.25 times 10^-2 times 0.5 = 10.56 times 10^-2 kg