A rock with mass .12 kg is fastened to a massless string withlength .80m to form

Question

A rock with mass .12 kg is fastened to a massless string withlength .80m to form a pendulum. The pendulum is swinging so as tomake a max. angle of 45 degrees with the vertical. Air resistanceis negligable. a. what is the speed of the rock when the srongpasses through the vertical position?
b. what is the tension in the string when it makes an angle of45deg with vertical c. what is the tension in the string as it passes through thevertical?
b. what is the tension in the string when it makes an angle of45deg with vertical c. what is the tension in the string as it passes through thevertical?

Explanation / Answer

   mass, m = 0.12 kg    lenth of the pendulum , L = 0.8 m    radius of circular path, r = 0.8 m    angle, = 45 o (a)   Velocity, V = [ 2gL ( 1 - cos ) ] = [ 2 * 9.8 * 0.8 * 0.293 ] = 2.14 m/s (b)           Velocity at the maxangle position, V = 0       Tension at 45o,T1 = mV 2 / r + mg cos = 0 +0.12 * 9.8 * 0.707 = 0.8314N (c)   Tension at lowest point, T2 = mV2 / r + mg  = ( 0.12 * 2.142 / 0.8) + 0.12 * 9.8                                                                            = 0.6869 + 1.176 = 1.8629 N (a)   Velocity, V = [ 2gL ( 1 - cos ) ] = [ 2 * 9.8 * 0.8 * 0.293 ] = 2.14 m/s (b)           Velocity at the maxangle position, V = 0       Tension at 45o,T1 = mV 2 / r + mg cos = 0 +0.12 * 9.8 * 0.707 = 0.8314N (c)   Tension at lowest point, T2 = mV2 / r + mg  = ( 0.12 * 2.142 / 0.8) + 0.12 * 9.8                                                                            = 0.6869 + 1.176 = 1.8629 N

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