Question

You have a grindstone (a disk) that is 105.00 kg, has a 0.297-m radius, and is turning at 71.150 rpm, and you press a steel axe against it with a radial force of 46.650 N. Assuming the kinetic coefficient of friction between steel and stone is 0.451. How many turns will the stone make before coming to rest?

Answer 1

Answer:

3.27 turns

Explanation:

To find how many turns (θ) will the stone make before coming to rest we will use the following equation:

$\omega_{f}^{2} = \omega_{0}^{2} + 2\alpha*\theta$

Where:

$\omega_{f}$: is the final angular velocity = 0

$\omega_{0}$: is the initial angular velocity = 71.150 rpm

α: is the angular acceleration

First, we need to calculate the angular acceleration (α). To do that, we can use the following equation:

$\alpha = \frac{\tau}{I}$

Where:

I: is the moment of inertia for the disk

τ: is the torque  

The moment of inertia is:

$I = \frac{mr^{2}}{2}$

Where:

m: is the mass of the disk = 105.00 kg

r: is the radius of the disk = 0.297 m

$I = \frac{105.00 kg*(0.297 m)^{2}}{2} = 4.63 kg*m^{2}$

Now, the torque is equal to:

$\tau = -F x r = -\mu*F*r$

Where:

F: is the applied force = 46.650 N      

μ: is the kinetic coefficient of friction = 0.451

$\tau = -\mu*F*r = -0.451*46.650 N*0.297 m = -6.25 N*m$

The minus sign is because the friction force is acting opposite to motion of grindstone.    

Having the moment of inertia and the torque, we can find the angular acceleration:

$\alpha = \frac{-6.25 N*m}{4.63 kg*m^{2}} = -1.35 rad/s^{2}$

Finally, we can find the number of turns that the stone will make before coming to rest:

$0 = \omega_{0}^{2} + 2\alpha*\theta$  

$\theta = -\frac{(\omega_{0})^{2}}{2\alpha} = -\frac{(71.150 \frac{rev}{min})^{2}}{2*(-1.35 \frac{rad}{s^{2}})*\frac{1 rev}{2\pi rad}*\frac{(60 s)^{2}}{(1 min)^{2}}} = 3.27 rev = 3.27 turns$                        

I hope it helps you!