A simple pendulum with mass m = 2 kg and length L = 2.67 m hangs from the ceilin

Question

A simple pendulum with mass m = 2 kg and length L = 2.67 m hangs from the ceiling. It is pulled back to an small angle of = 11° from the vertical and released at t = 0.

1)What is the period of oscillation?

s  

2)What is the magnitude of the force on the pendulum bob perpendicular to the string at t=0?

N  

3)What is the maximum speed of the pendulum?

m/s  

4)What is the angular displacement at t = 3.77 s? (give the answer as a negative angle if the angle is to the left of the vertical)

°  

5)What is the magnitude of the tangential acceleration as the pendulum passes through the equilibrium position?

m/s2

6)What is the magnitude of the radial acceleration as the pendulum passes through the equilibrium position?

m/s2

Explanation / Answer

m=2kg, L=2.67m theta =11 deg

1) time period T = 2pi sqrt(L/g)

T = 2*3.14 sqrt(2.67/9.8)

T = 3.27 S

2) the magnitude of the force on the pendulum
   bob perpendicular to the string at t=0 is

F = mgsin(theta)

F = 2 * 9.8 * sin(11)

F = 3.74 N

3) maximum speed v_max = sqrt[2gl(1-cos(theta))]

      
       v_max = sqrt[2*9.8*2.67(1-cos(11))]

       v_max = 0.96 m/s

4) angular displacement theta = 2pi*t/T

           theta = 2*3.14*3.77/3.27

           theta = 7.2 degree

5) The velocity is maximum at equilibrium position

    so the tangential acceleration is zero

6) radial acceleration a = 2g(1-cos(theta))

           a = 2*9.8(1-cos(11))
          
           a = 0.36 m/s^2

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