2- A pendulum of mass M and length L is released from rest from an angle of 90°

Question

2- A pendulum of mass M and length L is released from rest from an angle of 90° with respect to the vertical. The gravitational potential energy is zero at the lowest point. Ignore friction. Data: M=O.4 Kg, L=0.5 m, g=9.8 rm/s2•

 d) Find the tension at A, B and at the initial position.

e)Now consider the problem with friction. If the work of the friction force was one fifth of the initial energy, what would be the value of the speed at the lowest point of the trajectory? (be careful with the sign of W!)

Explanation / Answer

2- A pendulum of mass M and length L is released from rest from an angle of 90° with respect to the vertical. The gravitational potential energy is zero at the lowest point. Ignore friction. Data: M=O.4 Kg, L=0.5 m, g=9.8 rm/s2• d) Find the tension at A, B and at the initial position. We have equation of energy conservation. R*mg=R*mg*(1-cos theta) + mv^2/2. so R*mg*cos theta=mv^2/2. so mv^2/R=2mg*cos theta. =F_centripetal. Tension of the string. T=F_centripetal+mg*cos theta = 3mg*cos theta= so theta = 35 . T=9.6(N) theta = 0 . T=11.76(N). e)No consider the problem with friction. If the work of the friction force was one fifth of the initial energy, what would be the value of the speed at the lowest point of the trajectory? (be careful with the sign of W!) energy is conserved. so WE have. Mg*R-Mg*R/5=mv^2/2 so 4Mg*R/5=mv^2/2. so mv^2/R=8Mg/5=1.6Mg. so T=2.6Mg= 10.2(N) I really need help with this question. This picture should clarify the

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