Finding the angular acceleration of a flywheel

[cpgp]

Homework Statement

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, 5.00° slope, from rest, with the flywheel engaged and no power supplied from the motor. Neglecting friction and the rotational inertia of the wheels, find (a) the speed of the car at the bottom of the slope, and (b) the angular acceleration of the flywheel at the bottom of the slope.

Relevant Equations

T=Ia

I have solved part a using the conservation of energy, getting a (correct) answer of 47.9 km/h, but I am unable to make headway with part b. Based on the flywheel rotating at 237rev/s when the car is moving at 86.5 km/h, I obtained omega = (237*2pi)v/24=62v. Differentiating both sides should give alpha = 62a. But that will yield alpha = 62*g*sin5=53 rad/s^2,when the correct answer is 3.65 rad/s^2.

 

[BvU]

Hello cpgp, !

So clearly, the acceleration is not equal to the acceleration you use -- which would be correct if the flywheel was not engaged ...

 

[Cutter Ketch]

But that will yield alpha = 62*g*sin5=53 rad/s^2

You differentiated both sides. There is still a dv/dt (i.e. the acceleration a) on the right. Much as you devised a relation between the car’s linear velocity v and the flywheel’s angular velocity $\omega$, you have now derived a relation between the car’s linear acceleration a and the flywheel’s angular acceleration $\alpha$. I don’t know if that will prove to be useful, but it is correct. It just isn’t the answer.

 

[cpgp]

Hello cpgp, !

So clearly, the acceleration is not equal to the acceleration you use -- which would be correct if the flywheel was not engaged ...

So, then it should be given by ma = I*alpha/r + mg sin5?

This gives me alpha=(g/r)sin5/(1-I/(mr^2)). Substituting the values of I=2.83, r=0.54, m=19.4, I get alpha=3.17 rad/s^2, but even that is not the right answer.

 

[haruspex]

ma = I*alpha/r + mg sin5?

What is r here?
Anyway, this does not look right. It says the car's acceleration exceeds g sin5°.

 

[cpgp]

What is r here?
Anyway, this does not look right. It says the car's acceleration exceeds g sin5°.

R is the radius of the flywheel.

 

[Cutter Ketch]

R is the radius of the flywheel.

But is the radius of the flywheel the lever arm that relates torque to the applied force?

 

[BvU]

No.

ma = I*alpha/r + mg sin5

1. Has the wrong sign
2. ma + I*alpha/r = mg sin5?
means the rim of the flywheel is scraping over the ground

You have to connect a and αα somehow -- but you did already ! And correctly.

 

[cpgp]

I've typed up my work a little better.

Relation derived in part a:
$$\omega = \frac{237\times 2\pi}{24}\times v$$
$$\omega = 62v$$

For part b:
$$\Sigma F = ma$$
$$ma = \frac \tau r + mg\sin(5)$$
$$ma = \frac {I\alpha} r +mg\sin5$$
$$ma = \frac {Ia} {r^2} +mg\sin5$$
$$a(m-\frac I {r^2}) = mg\sin5$$
$$a=\frac{mg\sin5}{m-\frac I {r^2}}$$
$$a=\frac{702.8}{812.1}$$
$$a=0.8654 ms^{-2}$$

From the relation derived in part a:
$$\omega = 62v$$
$$\frac{d\omega}{dt} = \frac d{dt}(62v)$$
$$\alpha = 62a$$
$$\alpha = 62\times0.8654$$
$$\alpha = 53.7 rad /s^{2}$$

 

[cpgp]

No. You have to connect a and $\alpha$ somehow.

I have shown that in my post above

 

[BvU]

I have shown that in my post above

edited my post. What does it mean for a ?

 

[cpgp]

2. ma + I*alpha/r = mg sin5?

That makes sense, but I still only get alpha = 52.4 rad/s^2.

 

[BvU]

unfortunately r is unknown.

Summarizing: you have calculated that constant acceleration $a$ yields 13.31 m/s in 1500 m. With simple SUVAT you find $a$ (which is NOT $g\sin\theta$ !) and hence $\alpha$ !

 

[haruspex]

R is the radius of the flywheel.

Not according to your finding that

alpha = 62a

 

[cpgp]

unfortunately r is unknown.

Summarizing: you have calculated that constant acceleration $a$ yields 13.31 m/s in 1500 m. With simple SUVAT you find $a$ (which is NOT $g\sin\theta$ !) and hence $\alpha$ !

Thanks a lot, that has given me the answer!