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

In summary: This applies to all points on the object, since the object rolls without slipping.In summary, we used the equations of motion, torque, and the relationship between linear and angular acceleration for a rolling object to show that the linear acceleration of the disc in this scenario is 2/5 g.
  • #1
markosheehan
136
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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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  • #2
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):

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

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):

\(\displaystyle \tau=Rf=I\alpha\)

Solving for $f$, there results:

\(\displaystyle f=\frac{I\alpha}{R}\)

And so we now have:

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

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

\(\displaystyle a=R\alpha\)

Since the disk rolls without slipping, we know:

\(\displaystyle a=a_{\text{CM}}\)

Hence:

\(\displaystyle \alpha=\frac{a_{\text{CM}}}{R}\)

And now we have:

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

Or:

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

So, what do you get when you solve for $a_{\text{CM}}$?
 
  • #3
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.
 
  • #4
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:

\(\displaystyle a=R\alpha\)
 

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

1) What is disc rotation?

Disc rotation is the movement of a disc or circular object around a fixed point or axis.

2) How does mass affect disc rotation?

The mass of a disc affects its rotation by determining the amount of force required to move or rotate the disc. A larger mass will require more force to rotate compared to a smaller mass.

3) What is the significance of the radius in disc rotation?

The radius of a disc is important in determining the rotational speed of the disc. A larger radius will result in a greater rotational speed compared to a smaller radius.

4) What is the meaning of "tan^-1 3/4" in disc rotation?

"Tan^-1 3/4" represents the angle of the slope that the disc is rolling down. This angle is measured in radians and is used in calculations to determine the disc's rotational speed.

5) How does slope affect disc rotation?

The slope of a surface can influence disc rotation by altering the amount of gravitational force acting on the disc. A steeper slope will result in a greater gravitational force and therefore a faster rotation speed.

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