A small rock with mass 0.12kg is fastened to a massless stringwith length 0.8m t

Question

A small rock with mass 0.12kg is fastened to a massless stringwith length 0.8m to form a pendulum. The pendulum is swingingso as to make a maximum angle of 45degree with the vetical. Air resistance is negligible. a) what is the speed of therock when the string passes through the vertical position? b) Whatis the tension in the string when it makes an angle of 45degreewith the vertical? c) What is the tension in the string as itpasses through the vertical? Can someone help me solve thisquestion, please explain it in detail, i will give top score forthe complete answer. Thank you, A small rock with mass 0.12kg is fastened to a massless stringwith length 0.8m to form a pendulum. The pendulum is swingingso as to make a maximum angle of 45degree with the vetical. Air resistance is negligible. a) what is the speed of therock when the string passes through the vertical position? b) Whatis the tension in the string when it makes an angle of 45degreewith the vertical? c) What is the tension in the string as itpasses through the vertical? Can someone help me solve thisquestion, please explain it in detail, i will give top score forthe complete answer. Thank you,

Explanation / Answer

m = 0.12 kg, L = 0.8 m, = 45o, a) what is the speed of the rock when the string passes through thevertical position? mgL(1 - cos) = mv2/2 v = [2gL(1 - cos)] = 2.14 m/s b) What is the tension in the string when it makes an angle of45degree with the vertical? T - mgcos = 0 T = mgcos = 0.832 N c) What is the tension in the string as it passes through thevertical? T - mg = mv2/L T = m(g + v2/L) = 1.86 N

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