Problem 2 A 0.5 kg rod 1 m in length has a bob of 1 kg attached to one end. The

Question

Problem 2 A 0.5 kg rod 1 m in length has a bob of 1 kg attached to one end. The other end of the rod is connected to a pivot so that the system can oscillate freely. The center of mass of the system is located at m from the pivot. The bob will be treated as a point mass. The moments of inertia for the bob and the rod about the pivot are given below Ibobmbobl Mlcmg (a)What is the angular frequency of the oscillations about the pivot (assuming the maximum angular displacement s small)? (b)Assuming that at t 0s the bob is released from a dis- placement of 0.1 radians, find the equation of motion for the position (t) (c)What is the magnitude of the maximum angular acceler ation experienced by the pendulum? (d)Why must we assume the maximum angular displacement is small?

Explanation / Answer

a] torque = 0.5mgL theta + MgL theta

i alpha = [0.5mgL theta + MgL theta]

alpha = [0.5mgL + MgL] theta/i = w^2 theta

w = sqrt(( 0.5mgL+MgL)/(mL^2/3 + ML^2))

= sqrt((0.5*0.5*9.8+1*9.8)/(0.5/3+1))

= 3.24 rad/s

b] theta = 0.1 sin(3.24t+pi/2) where theta is in radians

c] alpha = 0.1*w^2 = 0.1*3.24^2 = 1.05 rad/s^2

d] Because theta = sin theta approximation hold good at small theta only.

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