Ring C has an inside radius of 55 mm and an outside radius of 60 mm and is positioned between two wheels A and B, each of 24-mm outside radius. Knowing that wheel A rotates with a constant angular velocity of 300 rpm and that no slipping occurs, determine (a) the angular velocity of the ring C and of wheel B, (b) the acceleration of the points on A and B that are in contact with C.

Where [tex]a_A[/tex] is the acceleration of a point on the outer rim of wheel A and [tex]a_B[/tex] the acceleration of a point on the outer rim of wheel B.

Explanation:

a)

Because there are no specification on what units we should give the answer it would be educational to give the answer in both ways, rpm and rad/s.

Attached is a sketch of the situation.

Let's define the following variables:

[tex]r_A=24\,mm[/tex] which is the outer radius of wheel A.
[tex]r_{Co}=60\, mm[/tex] which is the outer radius of wheel C.
[tex]r_{Ci}=55\,mm[/tex] which is the inner radius of wheel C.
[tex]r_B=r_A=24\,mm[/tex] which is the outer radius of wheel B

Wheel A rotates with an angular velocity of [tex]\omega_A=300\, rpm=300\times\frac{2\pi}{60}=10\pi\approx31.42\,rad/s[/tex]

Reminding that the linear velocity [tex]v[/tex] is related to the angular velocity [tex]\omega[/tex] by [tex]v=r.\omega[/tex] where [tex]r[/tex] is the radius of the circular motion, we can determine the angular velocity of wheel B.

At point of contact 1 we have the same linear velocity for the outer radius of wheel A as the outer radius of wheel C. Thus we have:

[tex]v_1=r_a\omega_a=r_{Co}.\omega_C\implies\omega_c=\frac{r_A}{r_{Co}}.\omega_A[/tex]

[tex]\implies \omega_c=\frac{24}{60}.10\pi\approx12.56 \, rad/s \, or\, \frac{24}{60}.300=120\,rpm[/tex]

Now we move to point of contact 2:

[tex]v_2=r_B.\omega_B=r_{Ci}.\omega_C[/tex]

[tex]\implies \omega_b=\frac{r_{Ci}}{r_B}.\frac{r_A}{r_{Co}}.300\,rpm[/tex]

[tex]\implies \omega_b=\frac{r_{Ci}}{r_B}.\frac{r_B}{r_{Co}}.300\,rpm[/tex]

[tex]\implies \omega_b=\frac{55}{60}.300\,rpm[/tex]

[tex]\implies \omega_b=275\, rpm[/tex]

alternatively we can express this as rad/s per second by doing:

[tex]\omega_b=\frac{55}{60}.10\pi\,rad/s\approx 28.79\, rad/s[/tex]

b)

The acceleration of a point undergoing circular motion is

[tex]a=r\omega^2[/tex]

Thus for a point in the outer rim of A we have [tex]\omega_A\approx31.42\, rad/s[/tex] and [tex]r_A=0.024 \, m[/tex].

We obtain from the previous [tex]a_A=r_A\omega_A^2\approx23.69 \, m/s^2[/tex]

Similarly:

[tex]a_B=r_B\omega_B^2\approx19.89 \, m/s^2[/tex]