Answer:
It will start to slip at [tex]t=\sqrt{\frac{\mu*g}{R}}*\frac{I_p+m*R^2}{\tau_0}[/tex]
Explanation:
We can find the angular acceleration from a sum of torque on the system:
[tex]\tau_0=\alpha *(I_p+m*R^2)[/tex] Solving for α
[tex]\alpha =\frac{\tau_0}{I_p+m*R^2}[/tex]
Now, on the object, we make a sum of forces on the centripetal-axis:
[tex]Ff=\mu*N=m*a=m*\frac{V^2}{R} =m*\omega^2*R[/tex] Solving for ω:
[tex]\omega=\sqrt{\frac{\mu*N}{m*R} }[/tex]
From a sum of forces on the axis perpendicular to the platform:
[tex]N-m*g=0[/tex] [tex]N=m*g[/tex]
Replacing this value into the ω equation:
[tex]\omega=\sqrt{\frac{\mu*g}{R} }[/tex] Now we have to find the amount of time that takes to the object to get this speed.
[tex]\omega=\alpha *t[/tex] Solving for t:
[tex]t=\frac{\omega}{\alpha} =\sqrt{\frac{\mu*g}{R}}*\frac{I_p+m*R^2}{\tau_0}[/tex]
