Pendulum. A smallrock with mass 0.12 kg isfastened to a massless string with len

Question

Pendulum. A smallrock with mass 0.12 kg isfastened to a massless string with length 0.69 m to form a pendulum. The pendulum isswinging so as to make a maximum angle of 45° with thevertical. Air resistance is negligible. (a) What is the speed of the rock when the string passes throughthe vertical position?
m/s

(b) What is the tension in the string when it makes an angle of45° with the vertical?
N

(c) What is the tension in the string as it passes through thevertical?
N (a) What is the speed of the rock when the string passes throughthe vertical position?
m/s

(b) What is the tension in the string when it makes an angle of45° with the vertical?
N

(c) What is the tension in the string as it passes through thevertical?
N

Explanation / Answer

a, Let the potential at the vertical position is zero. At the highest point, the speed is zero, so the energy there isE1 = mgh = mgl (1-cos 45 ) = 0.24 J Conservation of energy: 1/2 m v^2 = mgl (1-cos 45) ==> v =(2gl(1-cos 45)) = 2 m/s b, At the highest point, there is no speed, so the acceleration isperpendicular to the string, i.e no acceleration in the directionparallel to the string. ==>T = mg cos 45 =0.85 N c, As it pass through the vertical position, there is alsocentrifugal force point downwards, but still no acceleration: T = mg + m v^2 / l = 1.9 N

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