Question

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.

Answer 1

a)

`\omega_c=120\,rpm\, or\, 12.56 rad/s`

`\omega_B=275\,rpm\, or\, 28.79 \, rad/s`

b)

`a_A\approx23.68\,m/s^2`

`a_B\approx 19.90\,m/s^2`

Where
`a_A` is the acceleration of a point on the outer rim of wheel A and
`a_B` the acceleration of a point on the outer rim of wheel B.

Step-by-step 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:

• 
  `r_A=24\,mm` which is the outer radius of wheel A.
• 
  `r_(Co)=60\, mm` which is the outer radius of wheel C.
• 
  `r_(Ci)=55\,mm` which is the inner radius of wheel C.
• 
  `r_B=r_A=24\,mm` which is the outer radius of wheel B

Wheel A rotates with an angular velocity of
`\omega_A=300\, rpm=300*(2\pi)/(60)=10\pi\approx31.42\,rad/s`

Reminding that the linear velocity
`v` is related to the angular velocity
`\omega` by
`v=r.\omega` where
`r` 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:

`v_1=r_a\omega_a=r_(Co).\omega_C\implies\omega_c=(r_A)/(r_(Co)).\omega_A`

`\implies \omega_c=(24)/(60).10\pi\approx12.56 \, rad/s \, or\, (24)/(60).300=120\,rpm`

Now we move to point of contact 2:

`v_2=r_B.\omega_B=r_(Ci).\omega_C`

`\implies \omega_b=(r_(Ci))/(r_B).(r_A)/(r_(Co)).300\,rpm`

`\implies \omega_b=(r_(Ci))/(r_B).(r_B)/(r_(Co)).300\,rpm`

`\implies \omega_b=(55)/(60).300\,rpm`

`\implies \omega_b=275\, rpm`

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

`\omega_b=(55)/(60).10\pi\,rad/s\approx 28.79\, rad/s`

b)

The acceleration of a point undergoing circular motion is

`a=r\omega^2`

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

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

Similarly:

`a_B=r_B\omega_B^2\approx19.89 \, m/s^2`