Question

At *t* = 2.55 s a point on the rim of a 0.230-m-radius
wheel has a tangential speed of 53.5 m/s as the wheel slows down
with a tangential acceleration of constant magnitude 12.0
m/s^{2}.

(a) Calculate the wheel's constant angular acceleration.

[-52 rad/s^{2}]

(b) Calculate the angular velocities at *t* = 2.55 s and
*t* = 0.

ω_{2.55 s} |
= [ ] rad/s |

ω_{0} |
= [ ] rad/s |

(c) Through what angle did the wheel turn between *t* = 0
and *t* = 2.55 s?

[ ] rad

(d) At what time will the radial acceleration equal
*g*?

[ ]s after *t* = 2.55 s

Answer #1

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Calculate the wheel's constant angular
acceleration. rad/s^2
Calculate the angular velocity at t = 3.10 s. rad/s
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