Physics- Springs and Swings

Question

A pendulum made of a string of length 10.7 m and a spherical bob of mass 9.1 kg and negligible raius swings in a vertical plane. The pendulum is released from an angular position 59 ◦ from vertical as shown in the figure below. The string hits a peg located a distance 3 m below the point of suspension and swings about the peg up to an angle α on the other side of the peg. Then, the bob proceeds to oscillate back and forth between these two angular positions. The acceleration of gravity is 9.8 m/s 2 .

What is the maximum angle α which the pendulum will swing to the right after hitting the peg as shown above? Answer in units of ◦

Answer 1

Let l=10.7m and l₁ = 7.7m. Let beta=59⁰. using conservation of energy gives the following equation: 0+ mgl(1-cosbeta) = 12mv² +mgl₁(1-cosalpha). To find the maximum value of alpha v =0 at that point. Equation becomes

l(1-cos 59⁰)= l₁(1-cos alpha). 10.7(1- .515) = 7.7(1-cos alpha), 10.7(.485) = 7.7(1-cos alpha). 5.1895= 7.7(1-cos alpha), (1-cosalpha)=.67396 so cos alpha = .326 and alpha= cos^-1(.326) = 70.97 degrees.