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A pendulum consists of an object of mass 1.21kg tht hangs at the end of a massless bar a distance 1.28m from the pivot point calculate the magnitude of the torque due to gravity about the pivot point?
Wiki User
∙ 15y ago
The torque is the component of the weight that is perpendicular to the bar. So when the bar hangs vertically down, parallel to the force of gravity, there is no torque. If the bar makes an angle "A" with the vertical then the component of weight perpendicular to the bar would be mgSin(A) and the torque would be mgLSin(A) , where m= 1.21 kg, and L = 1.28m, g=9.8m/ss , so all you need is the angle "A".
Wiki User
∙ 15y ago
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A pendulum consists of an object of mass 1.21kg tht hangs at the end of a massless bar a distance 1.28m from the pivot point calculate the magnitude of the torque due to gravity about the pivot....?
The question says nothing about the motion of the pendulum, or about its position at the instant we're supposed to consider, only that it "hangs". If it's just hanging there limp, then the force is acting straight through the pivot, the distance between the line of the force and the pivot is zero, so the torque = R x F is zero.================================================= Did the question mean to say that the 1.28 m is the horizontaldistance between the pivot and the vertical, on account of the bar is slanted ??? If that's the case, why then the vertical force due to gravity is the weight of the bob = m x g = 1.21 x 9.8 = 11.858 N. The magnitude of the torque = |R x F| = 1.28 x 11.858 = 15.178 Newton-Meters.
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