A simple pendulum with mass m = 1.2 kg and length L = 2.46 m hangs from the ceil
ID: 1493782 • Letter: A
Question
A simple pendulum with mass m = 1.2 kg and length L = 2.46 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?
2.
What is the magnitude of the force on the pendulum bob perpendicular to the string at t=0?
3.
What is the maximum speed of the pendulum?
4.
What is the angular displacement at t = 3.52 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?
6.
What is the magnitude of the radial acceleration as the pendulum passes through the equilibrium position?
7.
Which of the following would change the frequency of oscillation of this simple pendulum?
Explanation / Answer
1) T = 2pi*sqrt(2.46/9.8) = 3.15 sec
2) The magnitude of the force perpendicular to the string is given by the equation
F = mg*sin(angle) = 1.2(9.81)(sin11) = 2.25 N.
3) The tangential velocity is solved from knowing the kinetic energy comes from the potential energy.
Thus mgh = 0.5mv^2.
v = sqrt(2*9.81*(2.46 - 2.46*cos(11))) = 0.942 m/s.
The angular velocity = v/2.46 = 0.383 m/s
4) The angle position is found from the equation, angle = initial angle*cos(sqrt(g/L)*t) where g is 9.81, L is 2.46, the initial angle is 11 and t is 3.52.
Angle = 11*cos[sqrt(9.81/2.46)*3.52] = 8.08
8.08 degrees from the vertical is the pendulum's position at t = 3.52 s. This angle is positive if at t = 0 the pendulum's 11 degree angle was to the right of the vertical, otherwise it is negative.
5) the tangential acceleration as the pendulum passes through the equilibrium position will be ZERO.
6) The radial (=centripetal) acceleration is:
Ar = v² / L = 0.942^2/2.46 = 0.36 m/s^2
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