Position and velocity of a point on the disk:
\[ x = r \cos \theta \\ y = r \sin \theta \\ \dot{x} = r \sin \theta \, \dot{\theta} \\ \dot{y} = r \cos \theta \, \dot{\theta}\]
Since the friction force opposes the velocity, define a unit vector opposite the velocity.
$v = \sqrt{\dot{x}^2 + \dot{y}^2} \\ u_x =  \dot{x} / v \\ u_y =  \dot{y} / v$
Let the mass of the disk be $M$ and the radius of the disk be $R$. Derive the mass and normal force for a patch of the disk as follows:
$\rho = \frac{M}{\pi R^2} \\ dA = r \, dr \, d \theta \\ dm = \rho \, dA \\ dN = dm \, g$
If $x > 0$, the infinitesimal vector force on a patch of disk is:
$dF_x = \mu_1 \, dN \, u_x \\ dF_y = \mu_1 \, dN \, u_y$
If $x \leq 0$, the infinitesimal vector force on a patch of disk is:
$dF_x = \mu_2 \, dN \, u_x \\ dF_y = \mu_2 \, dN \, u_y$
The total force is then:
$F_x = \int_0^{2 \pi} \int_0^R dF_x \\ F_y = \int_0^{2 \pi} \int_0^R dF_y$
This could probably be done analytically, but I did it numerically instead. The answer is independent of the particular values of $M$, $R$, and $\dot{\theta}$. The initial acceleration of the center of mass is approximately $0.35$.
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Top Newest@Talulah Riley Here is my solution for Problem 23
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@Steven Chase Thank you so much for the solution . By the way check your last 8 hour notifications.
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@Talulah Riley Hey; try my latest problem.
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Tricky problem. Nice solution; I can't think of doing it by hand, however.
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Your line number 18 is wrong. The question says that we should use the approximation $\pi = \frac{22}7$, But you took $\pi$ as $3.141592653\ldots$.
The same issue arise in line number 31.
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Ah yes, old habits die hard
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