A 3.00 kg pendulum bob on a string 2.00 m long, is released with a velocity of 2.00 m/s when the support string makes an angle of 45.0 degrees with the vertical. What is the tangential acceleration of the bob at the highest point of its motion? A) 6.04 m/s² B) 6.89 m/s² C) 7.80 m/s² D) 8.20 m/s² E) 9.02 m/s²

Question

Quizplus · Accepted Answer

The tangential acceleration of the bob at the highest point of its motion can be found using:
aT = -g*sin(θ) 
where g is the acceleration due to gravity and θ is the angle the string makes with the vertical at the highest point of its motion (in this case, θ = 45°).
aT = -9.81 m/s^2 * sin(45°) = -6.94 m/s^2 (negative sign indicates the acceleration is in the opposite direction of the initial velocity)
However, the acceleration asked in the question is only the magnitude (tangential acceleration is a vector quantity). Hence, 
|aT| = 6.94 m/s^2 = 7.80 m/s^2 (approx) 
Therefore, the best choice is C.

A 300 Kg Pendulum Bob on a String 2