Problem: A particle of mass m is attached to one end of a light slender rod whic

Question

Problem:

A particle of mass m is attached to one end of a light slender rod which pivots about a horizontal axis through point O. The spring constant k = 230 N/m and the distance b = 125 mm. If the system is released from rest in the position shown where the spring is unstretched, the bar is observed to deflect a maximum of 44 degree clockwise. Determine (a) the particle mass m and (b) the particle speed v after a displacement of 22 degree from the position shown. Neglect friction. m = kg When theta = 22 degree, v = m/s

Explanation / Answer

a)   intial energy of the system = pot. energy of spring + gravitational potential energy + kinetic energy of the rad and mass.

intial energy = 0 + 0 + 0 = 0 J

here... the axis OB is considered to be the reference for gravitiational potential energy ..


finally... height displacement of mass m = 2b sin 44 = 2*0.125 sin 44 = 0.1736646 m ..

so final gravitaitonal potentail energy = - m*9.8 * 0.1736646 = - 1.701913 m

vertical distance of the string from the pulley = ( b/2 + b sin 44 ) =

horizontal distance of the string from pulley = b(1- cos 44 )

so final length L = sqrt ( ( b/2 + b sin 44 )^2 + (b- bcos 44 ) ^2)
L = sqrt ( b^2 /4 + b2 sin^2 44 + b^2 sin 44 + b^2 + b^2 cos^2 44 - 2b^2 cos 44 )

L = b* sqrt ( 2 + 0.25 + sin 44 - 2 cos 44 ) = 1.2272 * b

intial length = b/2 = 0.5 b ..

so extension in spring = (1.2272 - 0.5 ) * b = 0.7272 *b = 0.7272 * 0.125 = 0.0909 m
so final potential energy of spring = 0.5 * 230 * 0.0909^2 = 0.95022315 J

final kinetic energy = 0 ..

so final energy = 0.95022315 - 1.701913 m + 0 = intial energy = 0

so mass m =   0.95022315 / 1.701913 = 0.5583265 kg ..

b)
at 22 degrees...


.. height displacement of mass m = 2b sin 22 = 2*0.125 sin 22 =0.093652 m ..

so final gravitaitonal potentail energy = - 0.5583265*9.8 * 0.093652 = - 0.5124263

vertical distance of the string from the pulley = ( b/2 + b sin22 )

horizontal distance of the string from pulley = b(1- cos 22 )

so final length L = sqrt ( ( b/2 + b sin 22 )^2 + (b- bcos 22 ) ^2)
L = sqrt ( b^2 /4 + b2 sin^2 22 + b^2 sin 22 + b^2 + b^2 cos^2 22 - 2b^2 cos 22 )

L = b* sqrt ( 2 + 0.25 + sin 22 - 2 cos 22 ) = 0.8776* b

intial length = b/2 = 0.5 b ..

so extension in spring = (0.8776 - 0.5 ) * b = 0.7272 *b = 0.7272 * 0.125 = 0.0472 m
so final potential energy of spring = 0.5 * 230 * 0.0472^2 =0.256246J

final kinetic energy = 0.5 * m * v^2 = 0.5 * 0.5583265 *v^2 = 0.27916325 * v^2

so final energy = 0.256246 - 0.5124263 + 0.27916325 v^2 = intial energy = 0

so velocity v =   0.957952 m/sec ..

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