The 40 mm diameter shaft AB rotates with a constant angular

The 40 mm diameter shaft AB rotates with a constant angular velocity of 20 rad/s. Ring C has an inner diameter of 150 mm. Assuming that no a slipping occurs between shaft AB and ring C, . determine the angular velocity of ring C and the acceleration vectors of the points of shaft AB and ring C which are in contact.

Solution

Given quantities,

Let Diameter of shaft AB be D₁ = 40mm

so Radius( r₁) = 20mm

and Diameter of ring C be D₂ = 150mm

so radius (r₂) =75mm

Angular Velocity (w₁) of shaft AB = 20 rad/s

To find:

Step 1: The linear velocity (v) of the shaft at the point of contact with the ring is given by

  v = rw , where w represents angular velocity

Step 2: Since there is no slippage so linear velocity for both shaft and ring at the point of contact would be same

so , v₁=v₂   

   or r₁w₁=r₂w₂

or 20x20 = 75xw₂

   or w₂= 0.53 rad/s

So angular velocity of ring C os 0.53 rad/s

Step 1: The tangential acceleration of both the shaft and ring will be zero as linear velocity is constant as a_t=dv/dt

Step 2: Radial acceleration of shaft AB at point of contact is given by a_n= w₁²r₁

=20² x 20

= 8000 rad/s²

Similarly, the radial acceleration of Ring C at point of contact is given by a_n=w₂²r₂

=0.53² X 75

= 21.06 rad/s²