At t = 3.35 s a point on the rim of a 0.220-m-radius wheel has a tangential spee

Question

At t = 3.35 s a point on the rim of a 0.220-m-radius wheel has a tangential speed of 49.5 m/s as the wheel slows down with a tangential acceleration of constant magnitude 12.5 m/s2. (a) Calculate the wheel's constant angular acceleration. Correct: Your answer is correct. rad/s2 (b) Calculate the angular velocities at t = 3.35 s and t = 0. 3.35 s = Correct: Your answer is correct. rad/s 0 = Correct: Your answer is correct. rad/s (c) Through what angle did the wheel turn between t = 0 and t = 3.35 s? Incorrect: Your answer is incorrect. rad (d) At what time will the radial acceleration equal g? Incorrect: Your answer is incorrect. s after t = 3.35 s

Explanation / Answer

angular acceleration = linear acceleration / radius

angular acceleration = 12.5 / 0.22

angular acceleration = 56.82 rad/s^2

angular velocity = linear velocity / radius

angular velocity at 3.35 sec = 49.5 / 0.22

angular velocity at 3.35 sec = 225 rad/s

by first equation of motion

v = u + at

225 = u - 56.82 * 3.35

u = 415.347 rad/sec

angular velocity at 0 sec = 415.347 rad/sec

by third equation of motion

v^2 = u^2 + 2as

255^2 = 415.347^2 - 2 * 56.82 * s

s = 945.865 rad

angle wheel will turn = 945.865 rad

radial acceleration = w^2 * r

w^2 * 0.22 = 9.8

w = 6.674 rad/sec

w^2 * r = 9.8

v = u + at

6.674 = 415.347 - 56.82 * t

t = 7.1924 sec

after 7.1924 sec radia acceleration will be equal to g

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.