A 0.5-kg collar is attached to a spring and slides without friction along a circ
ID: 1861375 • Letter: A
Question
A 0.5-kg collar is attached to a spring and slides without friction along a circular rod in a vertical plane. The spring has an undeformed length of 150 mm and a constant k = 200 N/m. Knowing that the collar has a speed of 3 m/s as it passes through point B, determine, at this instant, the tangential acceleration of the collar and the force of the rod on the collar.
Explanation / Answer
Spring length at B = sqrt [(175+125)^2 + 125^2] = 325 mm
Spriing deflection at B = 325-150 = 175 mm
Spring force at B = 200*0.175 = 35 N
Direction of spring force above horizontal = atan(125/(175+125)) = 22.62 deg
In Tangential direction,
m*a_t = (35*Cos22.62)
0.5*a_t = 32.3 N
a_t = 64.6 m/s^2
In normal direction, N = -35*Sin22.62 + mg + mv^2 /r
N = -13.46 + 0.5*9.81 + 0.5*3^2/0.125
N = 27.44 N
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