Question

A simple pendulum has a mass of 0.550 kg and a length of 2.00 m. It is displaced through an angle of 11.0

Answer 1

Given that the mass of pendulum is m = 0.550 kg

Length of pendulum is L = 7.00 m

Angle is ? = 11.0^o

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Amplitude of motion is

          A = L?

              = (7.00 m)(11^o*? /180 )    

              = 1.344 m

The angular frequency is ? = ?g/L

                                        = 1.18 rad/s

(a) Maximum velcoity is Vmax = A?

                                              = 1.59 m/s

(b) Maximum acceleration is a_max = A?²

                                                 = 1.87 m/s^2

maximum angular acceleration of the bob ? = a^max / L

                                                                = 0.267 rad/s^2

(c) Maximum restoring foce is F = ma

                                               = 1.0285 N

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Initial height of the bob is h = L - Lcos?

                                        = 0.129 m

(d) From conservation of energy

                      mgh = (1/2)mv²

                           v = ?2gh

                              = 1.59 m/s

Now the torque on the bob about point of suspension

                   ? = mgL sin?

                  I? = mgLsin?

                   ? = mgLsin? / I

                      = mgLsin? / mL²

                      = gsin? / L

                      = 0.267 rad/s²

^{Maximum force is F = mg sin?}

                              ^{= 1.028 N}