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

3.27 turns

Step-by-step 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 = (\tau)/(I)`

Where:

I: is the moment of inertia for the disk

τ: is the torque

The moment of inertia is:

`I = (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 = (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 = (-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 = -((\omega_(0))^(2))/(2\alpha) = -((71.150 (rev)/(min))^(2))/(2*(-1.35 (rad)/(s^(2)))*(1 rev)/(2\pi rad)*((60 s)^(2))/((1 min)^(2))) = 3.27 rev = 3.27 turns`

I hope it helps you!