Question

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$FIGURECP14.74$. Consider a $15\mathrm{kg}$industrial grinding wheel, $26\mathrm{cm}$in diameter, spinning at $900\mathrm{rpm}$. The brake pads are actuated by $2.0-\mathrm{cm}-$diameter pistons, and they contact the wheel an average distance $12\mathrm{cm}$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 $5.0s$$5.0\mathrm{s}$?

Short answer

The oil pressure needed to stop the wheel is$53\mathrm{kPa}$

Step-by-step solution

Expression for rotational Kinetic energy 

The expression for rotational Kinetic energy of the grinding wheel is,

$E_{\mathrm{K}}=\frac{1}{2}I{\omega }^2$

Here, $I$is moment of inertia and$\omega$is angular velocity.

Replace $I=\frac{M{R}^2}{2}$in the equation of rotational kinetic energy.

$E_{\mathrm{K}}=\frac{1}{2}\left(\frac{M{R}^2}{2}\right){\omega }^2$

$=\frac{1}{4}M{R}^2{\omega }^2$

Here, $M$is mass of the wheel and $R$is radius of the wheel.

Calculation for rotational kinetic energy 

Calculate the rotational kinetic energy.

Substitute $15\mathrm{kg}$for $M$,$0.13\mathrm{m}$for $R$, and $900\mathrm{rpm}$for in the above equation.

$E_{\mathrm{K}}=\frac{1}{4}(15\mathrm{kg})(0.13\mathrm{m}{)}^2{\left[\left(900\mathrm{rpm}\right)\left(\frac{2\pi \mathrm{rad}}{1\mathrm{rev}}\right)\left(\frac{1\min }{60\mathrm{s}}\right)\right]}^2$

$=562,4\mathrm{J}$

Calculation of force

The Frictional energy$E_F$needed to stop the disk in a time $t$is,

$E_{\mathrm{F}}=\mu \mathrm{Fv}$

Here, $\mu$is coefficient of friction, $F$is force, $v$tangential velocity of the grinding wheel, and á $t$is the time taken by the wheel to come to rest.

Equate $E_{\mathrm{F}}=E_{\mathrm{K}}$and rewrite the equation for $F$.

$F=\frac{E_{\mathrm{K}}}{\mu vt}$

Replace $v=r\omega$in the above equation.

$F=\frac{E_{\mathrm{K}}}{\mu r\omega t}$

Substitute $562.4\mathrm{J}$for localid="1648197595292" $E_K$, $0.60$for $\mu$,$0.12\mathrm{m}$for role="math" localid="1648197792106" $r$,$94.2\mathrm{rad}/\mathrm{s}$for $\omega$,$5.0s$for $t$in the above equation.

$F=\frac{562.4\mathrm{J}}{0.60(0.12\mathrm{m})(94.2\mathrm{rad}/\sec )(5.0\mathrm{s})}$

$=16.5\mathrm{N}$

Calculation of oil pressure

Calculate the oil pressure needed to stop the wheel. The expression for the pressure is,

$P=\frac{F}{A}$

Here, $A$is area of cross-section.

Substitute $A=\pi r^2$in the above equation.

$P=\frac{F}{\pi r^2}$

Substitute$16.5\mathrm{N}$for $F$and $0.01\mathrm{m}$for $r$in the above equation.

$P=\frac{16.5\mathrm{N}}{\pi (0.01\mathrm{m}{)}^2}$

$=53000\mathrm{Pa}\left(\frac{1\mathrm{kPa}}{1000\mathrm{Pa}}\right)$

$P=53\mathrm{kPa}$