The angular position of a point on the rim of arotating wheel is given by = 2.8t
ID: 1681996 • Letter: T
Question
The angular position of a point on the rim of arotating wheel is given by = 2.8t + 4.4t2 +2.0t3, where is in radians if t is given inseconds. a. What is the angular speed at t =3.0 s? b. What is the angular speed att = 5.0 s? c. What is the average angularacceleration for the time interval that begins at t =3.0 s and ends at t =5.0 s? d. What is the instantaneousacceleration at t = 5.0 s?a. What is the angular speed at t =3.0 s? b. What is the angular speed att = 5.0 s? c. What is the average angularacceleration for the time interval that begins at t =3.0 s and ends at t =5.0 s? d. What is the instantaneousacceleration at t = 5.0 s?
Explanation / Answer
a)angular speed=d/dt=2.8+8.8t+6t2 at t=3sec ang speed=2.8+8.8*3+6*9=83.2rad/sec b) solve in the same way c)find out ang speed at t=3 and 5 sec now ang acc=change in speed /(5-3) rad/sec2 d)we have angular speed=d/dt=2.8+8.8t+6t2 ang acc=8.8+12t at t=5secang acc=8.8+60=68.8rad/sec2
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