Complete Question
The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s.
(a) What is its angular acceleration in revolutions per minute-squared
(b) How many revolutions does the engine make during this 20 s interval?
rev
Answer:
a
![\alpha = 6261 \ rev/minutes^2](https://tex.z-dn.net/?f=%5Calpha%20%3D%206261%20%5C%20%20rev%2Fminutes%5E2)
b
![\theta = 613 \ revolutions](https://tex.z-dn.net/?f=%5Ctheta%20%20%3D%20613%20%5C%20revolutions)
Explanation:
From the question we are told that
The initial angular speed is ![w_i = 1120 \ rev/minutes](https://tex.z-dn.net/?f=w_i%20%3D%20%201120%20%5C%20rev%2Fminutes)
The angular speed after
is ![w_f = 2560 \ rev/minutes](https://tex.z-dn.net/?f=w_f%20%3D%202560%20%5C%20rev%2Fminutes)
The time for revolution considered is
Generally the angular acceleration is mathematically represented as
![\alpha = \frac{w_f - w_i }{t}](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7Bw_f%20-%20w_i%20%7D%7Bt%7D)
=>
=> ![\alpha = 6261 \ rev/minutes^2](https://tex.z-dn.net/?f=%5Calpha%20%3D%206261%20%5C%20%20rev%2Fminutes%5E2)
Generally the number of revolution made is
is mathematically represented as
![\theta = \frac{1}{2} * (w_i + w_f)* t](https://tex.z-dn.net/?f=%5Ctheta%20%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%2A%20%28w_i%20%2B%20w_f%29%2A%20%20t)
=> ![\theta = \frac{1}{2} * (1120+ 2560 )* 0.333](https://tex.z-dn.net/?f=%5Ctheta%20%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%2A%20%281120%2B%202560%20%29%2A%20%200.333)
=> ![\theta = 613 \ revolutions](https://tex.z-dn.net/?f=%5Ctheta%20%20%3D%20613%20%5C%20revolutions)
Answer:
![9.5 kg m^2/s](https://tex.z-dn.net/?f=9.5%20kg%20m%5E2%2Fs)
Explanation:
The angular momentum of an object is given by:
![L=mvr](https://tex.z-dn.net/?f=L%3Dmvr)
where
m is the mass of the object
v is its velocity
r is the distance of the object from axis of rotation
Here we have:
m = 350 g = 0.35 kg is the mass of the ball
v = 9.0 m/s is the velocity
r = 3.0 m is the distance of the object from axis of rotation (if we take the ground as the centre of rotation)
Therefore, the angular momentum is:
![L=(0.35)(9.0)(3.0)=9.5 kg m^2/s](https://tex.z-dn.net/?f=L%3D%280.35%29%289.0%29%283.0%29%3D9.5%20kg%20m%5E2%2Fs)
Answer:
D
Explanation:
For this kind of problem, forces add. F = F1 + F2
F1 = 6 N
F2 = 10 N
F = 6N + 10N
F = 16N
Do you have a picture then I could determine 1 millimeter