I have no idea Why I get wrong in question b A ring hangs from a horizontal rod
ID: 1849436 • Letter: I
Question
I have no idea Why I get wrong in question b
A ring hangs from a horizontal rod as shown. The rod rotates with a consant angular velocity, and there is no slip between the ring and the rod. It is observed that the angular velocity of the ring is a constant 1.6641 rad/s. The values of rl, r2 and r3 are 0.028, 0.139 and 0.099 m respectivley. Answer all questions to at least 3 sf. If necessary, you may enter answers in standard form (e.g. answer = 7.12e7, for an answer of 71.2 million). Enter your answers in SI units but do not enter the units. Determine the magnitude of the angular velocity of the rod. (rad/s) Determine the magnitude of the acceleration of the point on the rod which is in contact with the ring, (m/s2) Determine the magnitude of the velocity of a point on the outside of the ring, (m/s) Partially correct Marks for this submission: 1/3.Explanation / Answer
a) The velocity of the contact point relative to rod and ring will be same , so angular velocity of rod * radius of rod = angular velocity of ring * radius of ring => angular velocity of rod = 1.6641*0.139/0.028 = 8.261 b) acceleration = centripetal acceleration = (angular velocity)^2 *r = 8.261^2 * 0.028 = 1.91 c) Velocity at point on outer side = angular velocity of ring * outer radius = 1.6641*0.099 = 0.1647 m/s