Angular acceleration and such

[warmfire540]

As you press down on the accelerator of your car you see the tachometer climb steadily from 1200 rpm to 5500 rpm in 2.5 s.
(a) What is the angular acceleration of the car?
(b) What is the tangential acceleration of a point on the edge of the engine’s 3.8 cm diameter crankshaft?
(c) How many revolutions does the engine make during this time?
(d) If the moment of inertia of the crankshaft is 0.009 kg.m2, what is the torque generated by the engine?
(e) What is the kinetic energy of the crankshaft at its maximum rpm? What is its angular momentum at that time?


I like to make sure I'm grasping this:

angular velocity intial and final (wi and wf)
a. wi=125.66 wf=575.96
wf=wi+at
575.96=125.66-a2.5
ang=180.12 rads-2

b.ang=a/r
ang*r=a
180.12*.019=a
a=3.42 ms-2

c.theta=wit+(1/2)at^2
theta=314.14+562.88
theta=877.02 rad
rev=theta/2pi
rev=877.02/2pi
rev=139.58

d. torque=I*ang
torque=.009*180.12
torque=1.62 mN

e. K=(1/2)Iw^2
K=(1/2)(.009)(575.98(^2
K=1492.78 J

 

[Raze2dust]

I can't spot any conceptual mistake... :)

 

[Moetasim]

L=Iw
L=.009*575.98
L=5.18 kg m2/s

I would like to clarify and confirm your calculations for each part of the question:

a) To find the angular acceleration, we can use the formula: angular acceleration (α) = (final angular velocity - initial angular velocity) / time. So, α = (575.96 rad/s - 125.66 rad/s) / 2.5 s = 180.12 rad/s^2.

b) The tangential acceleration is the linear acceleration at the edge of the crankshaft, which can be calculated using the formula: tangential acceleration (a) = angular acceleration (α) * radius (r). So, a = 180.12 rad/s^2 * 0.019 m = 3.42 m/s^2.

c) The number of revolutions can be calculated using the formula: number of revolutions = (final angular position - initial angular position) / (2π). In this case, the final angular position is 575.96 rad and the initial angular position is 125.66 rad. So, number of revolutions = (575.96 rad - 125.66 rad) / (2π) = 139.58 revolutions.

d) The torque generated by the engine can be calculated using the formula: torque (τ) = moment of inertia (I) * angular acceleration (α). So, τ = 0.009 kg.m^2 * 180.12 rad/s^2 = 1.62 Nm.

e) The kinetic energy of the crankshaft at its maximum rpm can be calculated using the formula: kinetic energy (K) = (1/2) * moment of inertia (I) * (angular velocity)^2. So, K = (1/2) * 0.009 kg.m^2 * (575.96 rad/s)^2 = 1492.78 J. The angular momentum at this time can be calculated using the formula: angular momentum (L) = moment of inertia (I) * angular velocity (ω). So, L = 0.009 kg.m^2 * 575.96 rad/s = 5.18 kg.m^2/s.