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

[markosheehan]

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

 

[MarkFL]

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 $a_{\text{CM}}$?

 

[markosheehan]

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.

 

[MarkFL]

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:

$$a=R\alpha$$