1. A wheel with a 0.12 m radius is rotating at 37 rev/s. It then slows uniformly
ID: 2290284 • Letter: 1
Question
1. A wheel with a 0.12 m radius is rotating at 37 rev/s. It then slows uniformly to 13 rev/s over a 2.8 s interval. What is the angular acceleration of a point on the wheel?
2.A grindstone of radius 4.0 m is initially spinning with an angular speed of 2.8 rad/s. The angular speed is then increased to 14 rad/s over the next 7.2 seconds. Assume that the angular acceleration is constant. Through how many revolutions does the grindstone turn during the 7.2 second interval?
3. A spinning disc rotating at 140 rev/min slows and stops 42 s later. How many revolutions did the disc make during this time?
4. An amusement park ride, passengers are seated in a horizontal circle of radius 6.8 m. The seats begin from rest and are uniformly accelerated for 26 seconds to a maximum rotational speed of 1.6 rad/s. What is the tangential acceleration of the passengers during the first 13 s of the ride?
Explanation / Answer
1)angular accelration is rate of change of angular velocity
first findout the change in angularvelocity convert rps to rpm768
w=w1-w2
=224.196
alpha=224.196/2.8
=80.07
2) find change in angular speed
=w1-w2
=11.2rad/s
=107rpm
=1.783rev/s
3) in this question convert rev/minute to rev/s
140rpm=2.33rps
no of revoultions equal to 2.33x42
=97.86 revoultions
4)in this problem tangential accelration is rate of change of angular velocity
first find v=rw
=1.6x6.8
=10.88m/s
alpha=rate of change of velocity
=10.88/13
=0.836m/s2
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