1)An automobile engine slows down from 4580 rpm to 1140 rpm in 2.55 s. Calculate

Question

1)An automobile engine slows down from 4580 rpm to 1140 rpm in 2.55 s. Calculate its angular acceleration, assumed constant.


Calculate the total number of revolutions the engine makes in this time. Do not enter unit.


2)A wheel 30.7 cm in diameter accelerates uniformly from 246 rpm to 372 rpm in 6.02 s. How far will a point on the edge of the wheel have traveled in this time?



3)
The blades in a blender rotate at a rate of 6540 rpm. When the motor is turned off during operation, the blades slow to rest in 3.66 s. What is the angular acceleration as the blades slow down?
please do not answer 187.1 coz its wrong thanks

Explanation / Answer

1)

angular acce. = -(4580-1140)/(2.55/60) = -80941.17 rev/m^2

s = (v^2-u^2)/(2*a) = (4580^2-1140^2)/(2*80941.17) = 121.55 rev

2)

a = (372-246)/(6.02/60) = 1255.81 rev/m^2

s =  (372^2-246^2)/(2*1255.81) = 31.00 rev

so,

distance traveled by point on edge = 30.7*pi*31 = 2989.85 cm = 29.89 m

3)

angular acce. = (0-6540)/(3.66/60) = -107213.11 rev/m^2

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

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