At a time t = 2.70s , a point on the rim of a wheel with a radius of 0.230m has

Question

At a time t = 2.70s , a point on the rim of a wheel with a radius of 0.230m has a tangential speed of 53.0m/s as the wheel slows down with a tangential acceleration of constant magnitude 10.8m/s^2 . A) Calculate the wheel's constant angular acceleration. B) Calculate the angular velocity at t=2.7s. C) Calculate the angular velocity at t=0s. D) Through what angle did the wheel turn between t=0 and t=2.7s? E) At what time will the radial acceleration equal g=9.81m/s^2?

Explanation / Answer

ang accl.= a/r=46.95 rad/s2 ang. vel=v/r=230.43 rad/s ang vel at t=0 =230.43+46.95*2.7=357.19 rad/s D) 793.33 rad E)V^2/r=9.81 =>v=1.5 ang vel.= 6.5 t=357.19-6.5/46.95=7.47 s

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