Question

Rod OA rotates counterclockwise with a constant angular velocity of = 5 rad/s. The double collar B is pin connected together such that one collar slides over the rotating rod and the other slides over the horizontal curved rod, of which the shape is described by the equation r = 1.5(2 - cos θ) ft. If both collars weigh 0.75 lb, determine the normal force which the curved rod exerts on one collar at the instant θ = 120^o. Neglect friction.

Answer 1

Ans) In order to calculate force which circular rod exerts on one of the collar and force that OA exerts on the other collar at instant = 120 degree we need to write equilibrium equation in the cylinderical coordinates , where

Fr = m = m ( - r ) ...................................(1)

We have to calculate r , , , , ,

For r = 1.5(2 - Cos )

=> = 1.5 Sin

=> = 1.5 Cos + 1.5 Sin

Substitute the values of = 120 , = 5 rad/s and = 0 in r ,   , equations

=> r = 2.25 ft

=> = 6.495 ft/s

=> = -18.75 ft/s2

Putting values in equation 1,

=> Fr = m ( - r ) = 0.75 (-18.75 - 2.25()) = -56.25 lb

Now,

Fr = - N Cos (120)

=> -56.25 = - N (-0.5)

=> N = -112.5 lb

Hence, normal force exerted by rod on collar is -112.5 lb