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:
[tex] \omega_{f}^{2} = \omega_{0}^{2} + 2\alpha*\theta [/tex]
Where:
[tex]\omega_{f}[/tex]: is the final angular velocity = 0
[tex]\omega_{0}[/tex]: 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:
[tex] \alpha = \frac{\tau}{I} [/tex]
Where:
I: is the moment of inertia for the disk
τ: is the torque
The moment of inertia is:
[tex] I = \frac{mr^{2}}{2} [/tex]
Where:
m: is the mass of the disk = 105.00 kg
r: is the radius of the disk = 0.297 m
[tex] I = \frac{105.00 kg*(0.297 m)^{2}}{2} = 4.63 kg*m^{2} [/tex]
Now, the torque is equal to:
[tex]\tau = -F x r = -\mu*F*r[/tex]
Where:
F: is the applied force = 46.650 N
μ: is the kinetic coefficient of friction = 0.451
[tex] \tau = -\mu*F*r = -0.451*46.650 N*0.297 m = -6.25 N*m [/tex]
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:
[tex] \alpha = \frac{-6.25 N*m}{4.63 kg*m^{2}} = -1.35 rad/s^{2} [/tex]
Finally, we can find the number of turns that the stone will make before coming to rest:
[tex] 0 = \omega_{0}^{2} + 2\alpha*\theta [/tex]
[tex]\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[/tex]
I hope it helps you!
