This is the second posting for this question. The last time I asked, I received

Question

This is the second posting for this question. The last time I asked, I received two wrong answers, and guy just posted a random solution to a problem that wasn't asked.

The rotation of rod OA about O is defined by the relation theta = t^3 - 4t, where theta and t are expressed in radians and seconds, respectively. Collar B slides along the rod so that its distance from O is r= 2.5t^3-t^2, where are and t are expressed in inches and seconds, respectivley. When t=1s, determine the radius of curvature of the path of the collar. Velocity is 3.536 ft/s or -2.5r+2.5theta, and acceleration is 12.5 ft/s^2 or 7.5r-10theta.

Explanation / Answer

V = -2.5r+2.5theta magnitude of velocity = 3.536 in /s unit vector in the direction of velocity(v) = (-2.5r+2.5theta)/3.536 a = 7.5r-10theta or 12.5 in/s^2 component of acceleration in the direction of velocity = a.v = (-7.5*2.5 -10*2.5)/3.536 = 12.37 in/s^2 component of acceleration perpendicular to the velocity (an) = sqrt( 12.5^2-12.37^2) = 1.77in/s^2 radius of carvature->r r= V^2/(an) = 3.536^2/1.77 in = 7.06 in [Ans]

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