Brake mechanism in a car and pressurised oil

[brotherbobby]

Homework Statement

Disk brakes, such as those in your car, operate by using pressurized oil to push outward on a piston. The piston, in turn, presses brake pads against a spinning rotor or wheel, as seen in the figure I paste below. Consider a 15 kg industrial grinding wheel, 26 cm in diameter, spinning at 900 rpm. The brake pads are actuated by 2.0-cm-diameter pistons, and they contact the wheel an average distance 12 cm from the axis. If the coefficient of kinetic friction between the brake pad and the wheel is 0.60, what oil pressure is needed to stop the wheel in 5.0 s?

Relevant Equations

(1) Kinetic frictional force : $f_K = \mu_K N$
(2) Frictional torque : $\tau_f = I_{\text{CM}} \alpha$ and Newton's (second) law : $\Sigma F = ma$. The torque is to be found first using $\tau = F\times R$, where $R$ is the radius of the cylinder. These can then be used to compute the angular retardation $\alpha$
(3) Moment of inertial of a cylinder via an axis passing through its center and perpendicular to its two ends : $I_{\text{CM}} = \frac{1}{2}MR^2$ (symbols having their usual meanings)
(4) Force due to a pressurised liquid : $F = p A$, where $p$ is its pressure and $A$ is the area of contact.

(I must confess I couldn't get far, owing mostly to the term "an average distance (of 12 cm) from the axis". Axis of the (peached coloured) wheel at the center with the arrow pointing down? Is the arrow the location of the "axis" of the wheel? Let me see.)

Objective : To find the oil pressure, let it be $P$.

Attempt at a solution : Force exerted by the pressurised oil on the piston : $F = PA$ where $A = \pi d^2_p/4$, $d_p$ being the diameter of the piston. Hence $F = \frac{\pi P d_p^2}{4}$.
This force will be exerted normally on the brake pad through which, I am assuming, it will be transmitted on to the rotating wheel.
Frictional force acting on the wheel : $f= \mu_K F = \frac{\mu_K \pi P d_p^2}{4}$.
The torque due to this (frictional) force $\tau = f\times x_{PW} = \frac{\mu_K \pi P d_p^2}{4} x_{PW}$ where $x_{PW}$ is the distance between the piston and the axis of the wheel.
This torque would lead to an angular retardation : $\tau = I_{\text{CM}}\alpha \Rightarrow \frac{\mu_K \pi P d_p^2}{4} x_{PW} = I_{\text{CM}}\alpha \Rightarrow \boldsymbol{P = \frac{4I_{\text{CM}}\alpha}{\mu_K \pi d_p^2 x_{PW}}}$.

Calculations : Initial angular velocity of the wheel $\omega_0 = 900\; \text{rpm}\; = \frac{900\times 2\pi}{60} = 30\pi\; \text{rad/s}$.
The moment of inertia of the (cylindrical) wheel $I_{\text{CM}} = \frac{1}{2} MR^2 = \frac{1}{2}\times 15\;\text{kg}\;\times (0.13\;\text{m})^2 = 0.13\; \text{kg-m}^2$.
Angular acceleration $\alpha = \frac{\omega_F - \omega_0}{t} = \frac{0-30\pi}{5} = -6\pi\; \text{rad/s}^2$. Or angular retardation : $\alpha = 6\pi \; \text{rad/s}^2$.
Putting these in the equation for pressure of oil (see last equation in paragraph above in bold) : $P = \frac{4\times 0.13\times 6\pi}{0.6\pi \times (0.02\;\text{m})^2\; \times 0.12\;\text{m}}\Rightarrow \boxed{P = 1.08\times 10^5\; \text{Pa}}$ .

Of course this is the gauge pressure.

But it doesn't agree with the answer in the book. $\boxed{P = 53\; \text{kPa}}$.

A help would be welcome.

 

[Steve4Physics]

I haven't checked your working in detail but (from a quick inspection) the basic method looks OK. But note:

1) The disc is sandwiched between two pads which means friction acts on both sides of the disc. Each pad provides half the required total torque.

2) You have rounded too much in intermediate steps. For example you have:
$\frac{1}{2}\times 15\;\text{kg}\;\times (0.13\;\text{m})^2 = 0.13\; \text{kg-m}^2$
but the unrounded value is
$\frac{1}{2}\times 15\;\text{kg}\;\times (0.13\;\text{m})^2 = 0.12675\; \text{kg-m}^2$

Work to a reasonably high precision for intermediate steps and round the final answer.

3) The diagram is a bit misleading as it looks like the brake cylinders are on the disc’s axis. The actual set-up is more clearly shown here:
https://www.researchgate.net/profile/Jens-Wahlstroem/publication/257821335/figure/fig3/AS:667689492181000@1536200988485/Disc-brake-assembly-with-a-single-piston-floating-caliper-and-a-ventilated-disc.png
Note that the pistons/pads are positioned near the outer edge of the disc in order that the friction provides maximum torque. In this question we take the distance from the disc's axis to the cylinder/pads ($x_{PW}$) as 12cm which you have done correctly.

Mainly due to item 1), your final answer is twice as big as it should be.

Edit - typo' corrected.

 

[BvU]

A factor of two (*)

So perhaps the braking force is nicely split between two brake pads, one on each side of the wheel $-$ as the drawing suggests !

(*) Your result should be 1.06 $\times$ 10⁵ Pa. Avoid errors like this by not rounding off too early: the $I_{CM}$ is less than 0.13 kg.m².

[edit] hey, @Steve4Physics was faster. Fortunately we agree

$\$

 

[brotherbobby]

Thank you all, I solved the problem. Yes I should have considered both left and right side pistons exerting frictional torques on the rotating wheel to make it stop. Each of those torques is half the value I considered.

But regarding precision - should we not keep to two places after the decimal in order to keep significant digits from not becoming more than what we began with?

 

[Steve4Physics]

But regarding precision - should we not keep to two places after the decimal in order to keep significant digits from not becoming more than what we began with?

If you round at different internal stages of a calculation, rounding errors tend to accumulate. You don't want this!

Also (in case you are not already aware) note that decimal places are not generally equivalent to significant figures.

Here’s an example which should help if you work through it...

a = 5.12 (2 decimal places, 3 significant figures)
b = 341 (no decimal places, 3 significant figures)

In part 1 of a problem you are asked to calculate c, where c = ab:
c = ab = 5.12*341 = 1745.92 (unrounded)
Correct answer: c = 1750 (or if preferred 1.75*10³)

In part 2 (final part) of the problem you are asked to find c³:
Correct answer: c³ = 1745.92³ = 5.32*10⁹
Incorrect answer: c³ = 1750³ = 5.36*10⁹

I hope it’s clear that 5.32*10⁹ is the correct answer. (Just pretend that the only part of the problem was to find c³.)

Using the rounded value for c introduced an error of 0.04*10⁹