Given that the disk shown rolls without sliding RWOS across
Solution
1) since the disc moves without slipping, the velocity of the centre A equals the tangential velocity at the end of the radius.
Since Vc=5 in the horiz direction, the velocity of the rod resolved along the horiz direction must equal 5 ft/sec
Now the angle at the ground made with the rod is arcsin( 11/24) = 27.28 deg
sin ( ground angle) = .889
If the velocity of the rod is Vo, then Vo *cos (27.28) =5
V0= 5.624 fps
the angle made by the tangent to the radius with the top of the rod is (90-27.28- 60) =2.72 deg
Resolving, the tangential velocity Vt at the point B is Vt cos (2.72) = 5.624
Hence Vt = 5.63 fps
The velocity of the tip of the radius would be 8/6* 5.63 = 7.507 fps.
By non slip, this is the velocit of the centre of the disc, in the horizontal dirn.
To find the angular velocity of BC, (about the contact pt C)
at B normal to BC, or at C, the angular velocity should be same.
Evaluating at B, the tangential velocty there was 5.63 ftps
Making this staionary means applying a velocity in the opposite direction of 7.507 fps to keep the rotn centre static
Knowing the radius at this point ( 6in), the angular velocity can be found
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