Question

A car is fitted with an energy-conserving flywheel, which in operation is geared to the driveshaft so that it rotates at 237 rev/s when the car is traveling at 86.5 km/h. The total mass of the car is 822 kg, the flywheel weighs 194 N, and it is a uniform disk 1.08 m in diameter. The car descends a 1500-m long, 5 degrees slope, from rest, with the flywheel engaged and no power supplied from the motor. Neglecting friction and the rotational inertia of the wheels, find the angular acceleration of the flywheel at the bottom of the slope.

Answer 1

I found the angular velocity and speed of the car at the bottom of the slope, using energy approach
w = 825 rad/s
v = 13.3 m/s
but I am stuck with the angular acceleration. Does anybody know what method to use to find it in this case?

Answer 2

If the slope is constant, there will be no angular acceleration of the car.

Angular acceleration is defined as the rate of change of the angular velocity. Angular velocity is defined as the rate of change of the angle of the car, in this example. If the slope is a constant 5 degrees, then \[\theta = 5\]\[\dot \theta = 0\]\[\ddot \theta = 0\]

The angular acceleration of the flywheel is a different story.

Answer 3

@eashmore Oh sorry! I've made a mistake in question, originally we are asked of the angular acceleration of the flywheel. I've corrected the question.

Answer 4

Okay. That makes more sense. Let me walk you through it.

I like to begin dynamics problems by defining the knowns, then the kinematic relationships, then coming up with kinetic equations of motion.

We know from the problem statement that \[\dot \theta = 237 \rm {rev \over s}\]when\[\dot x = 86.5 \rm {km \over hr} = 24.03 \rm {m \over s}\]
We need to determine the moment of inertia of the flywheel. For a disk, \[I = {m_{f}r_f^2 \over 2}\]where \(m_f\) is the mass of the flywheel and \(r_f\) is the radius of the flywheel. The mass of the flywheel is\[m_f = {W_f \over g}\] where \(W_f\) is the weight of the flywheel and g is acceleration due to gravity.

Now, let's define the kinematic relationship that relates the angular velocity of the flywheel and the velocity of the car.
\[\dot \theta = {\dot x \over r}\]
and for acceleration\[\ddot \theta = {\ddot x \over r}\]
Therefore, \[r = {\dot x \over \dot \theta} = {\ddot x \over \ddot \theta}\]\(r\) is the gear ratio between the wheels and the flywheel.


Let's set up the equations of motion of the car. Assuming there is no rolling resistance or wind resistance, \[m_c \ddot x = F_g\]

And the equations of motion of the flywheel\[I \ddot \theta = \tau = F_g \cdot r\]

The force of gravity, which drives the car down the slope, is defined as\[F_g = m_c g \sin(5^\circ)\]\[\therefore m_c \ddot x = m_c g \sin(5^\circ) \rightarrow \ddot x = g \sin(5^\circ)\]Rewritting the EOM for the flywheel \[I \ddot \theta = m_c g \sin(5^\circ) \cdot \left [ \ddot x' \over \ddot \theta' \right]\]
Here \(\ddot x'\) and \(\ddot \theta '\) are the values given in the problem statement.\[I \ddot \theta = m_c g \sin(5^\circ) \cdot \left [ 24.03 \over 237 \cdot 2 \pi \right]\]

Answer 5

I get \ddot \theta = 3.94 rad/s^2
while the answer is 3.65 rad/s^2
What am I doing wrong?

Answer 6

@eashmore
There was a mistake in your considerations
\[m_c\ddot x= F_g\] is wrong
This equation doesn't take into account the motion of the flywheel, it describes the motion of the car as if it were moving without it.
So if to consider the motion of the flywheel we will get
\[\frac {I\alpha}{r_{gear}}+m_ca=F_g\]
Then I get exactly what my textbook demands
\[\alpha=3.65\frac{rad}{s^2}\]
Anyway thank you for the help! I really appreciate it!