Question

1. integrated concepts

you have a grindstone (a disk) that is 90.0 kg, has a 0.340 - m radius, and is turning at 90.0 rpm, and you press a steel axe against it with a radial force of 20.0 n. (a) assuming the kinetic coefficient of friction between steel and stone is 0.20, calculate the angular acceleration of the grindstone. (b) how many turns will the stone make before coming to rest?

Explanation:

Step1: Calculate the frictional force

The frictional force $f$ is given by $f = \mu_kF_N$, where $\mu_k = 0.20$ is the coefficient of kinetic - friction and $F_N=20.0\ N$ is the normal force.
$f=\mu_kF_N=0.20\times20.0\ N = 4.0\ N$

Step2: Calculate the moment of inertia of the disk

The moment of inertia $I$ of a solid disk is $I=\frac{1}{2}mr^{2}$, where $m = 90.0\ kg$ and $r = 0.340\ m$.
$I=\frac{1}{2}\times90.0\ kg\times(0.340\ m)^{2}=\frac{1}{2}\times90.0\times0.1156\ kg\cdot m^{2}=5.202\ kg\cdot m^{2}$

Step3: Calculate the angular acceleration

The torque $\tau$ is related to the angular acceleration $\alpha$ by $\tau = I\alpha$. Also, $\tau=rf$ (where $r$ is the radius and $f$ is the frictional force). So, $\alpha=\frac{\tau}{I}=\frac{rf}{I}$.
$\alpha=\frac{0.340\ m\times4.0\ N}{5.202\ kg\cdot m^{2}}\approx - 0.261\ rad/s^{2}$ (negative because it is a decelerating torque)

Step4: Convert the initial angular velocity to rad/s

The initial angular velocity $\omega_0$ in rad/s: $\omega_0 = 90.0\ rpm$. Using the conversion factor $1\ rpm=\frac{2\pi}{60}\ rad/s$, we have $\omega_0=90.0\times\frac{2\pi}{60}\ rad/s = 3\pi\ rad/s\approx9.42\ rad/s$

Step5: Calculate the angular displacement before coming to rest

We use the equation $\omega^{2}=\omega_0^{2}+2\alpha\theta$, where $\omega = 0$ (comes to rest). So, $\theta=\frac{-\omega_0^{2}}{2\alpha}$.
$\theta=\frac{-(9.42\ rad/s)^{2}}{2\times(- 0.261\ rad/s^{2})}\approx170.5\ rad$

Step6: Convert the angular displacement to number of turns

The number of turns $n$ is given by $n=\frac{\theta}{2\pi}$.
$n=\frac{170.5\ rad}{2\pi\ rad}\approx27.2$ turns

Answer:

(a) $\alpha\approx - 0.261\ rad/s^{2}$
(b) $n\approx27.2$ turns