MHB Disc Rotation: Mass m & Radius r Down Slope of tan^-1 3/4

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A disc of mass m and radius r rolls down a slope inclined at an angle of tan^-1 3/4 without slipping. The net force acting on the disc is derived from the gravitational component along the slope minus the frictional force. The torque caused by friction leads to a relationship between linear and angular acceleration, allowing the derivation of the linear acceleration formula. By applying the moment of inertia for the disc and solving the equations, the linear acceleration is determined to be 2/5 g. This analysis confirms the dynamics of rolling motion under the specified conditions.
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a disc of mass m and radius r rolls down a slope of incline tan^-1 3/4. the slope is rough enough to prevent slipping. the disc travels from rest a distance of s metres straight down the slope.
(i) show that the linear acceleration of the disc is 2/5 g.
what i did was i first revolved the forces and i got 2 equations
R=4/5 g and 3/5 mg-F=ma but i don't know where to from here
 
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Okay, if we recognize that the net force $F=Ma_{\text{CM}}$ on the disk parallel to the incline is the component of the disk's weight parallel to the incline less the force of friction $f$ keeping the disk from slipping, we may write (letting $\theta$ be the angle of the incline above the horizontal):

$$Mg\sin(\theta)-f=Ma_{\text{CM}}$$

The force of friction results in a torque about the center of mass, having a lever arm $R$, so we may write (where $I$ is the moment of inertia for the disk):

$$\tau=Rf=I\alpha$$

Solving for $f$, there results:

$$f=\frac{I\alpha}{R}$$

And so we now have:

$$Mg\sin(\theta)-\frac{I\alpha}{R}=Ma_{\text{CM}}$$

Now, we may relate the angular acceleration $\alpha$ of the disk to its linear acceleration$a$ as follows:

$$a=R\alpha$$

Since the disk rolls without slipping, we know:

$$a=a_{\text{CM}}$$

Hence:

$$\alpha=\frac{a_{\text{CM}}}{R}$$

And now we have:

$$Mg\sin(\theta)-\frac{I\dfrac{a_{\text{CM}}}{R}}{R}=Ma_{\text{CM}}$$

Or:

$$Mg\sin(\theta)-\frac{Ia_{\text{CM}}}{R^2}=Ma_{\text{CM}}$$

So, what do you get when you solve for $a_{\text{CM}}$?
 
i get 2/5 g when i solve for a but i don't get your full solution. what is the formula of a disc for angular acceleration what does a=Ra stand for.
 
For a rolling object (circular such as a cylinder, sphere, disk, hoop, etc.) of radius $R$, the linear acceleration $a$ and the angular acceleration $\alpha$ are related as follows:

$$a=R\alpha$$
 
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