- #1

brotherbobby

- 651

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- 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.

**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.**