I pull back the string on a pendulum until it is at 60 degrees. I let go of it.
ID: 2171523 • Letter: I
Question
I pull back the string on a pendulum until it is at 60 degrees. I let go of it. When the pendulum is at its minimum point (theta=0), what is its speed?Possibly useful info: The string is 0.4 meters long.
so L=0.4 and m=0.05kg
Give your answer in units of meters per second, BUT DO NOT PUT THE UNITS IN YOUR ANSWER, JUST GIVE THE NUMBER THAT YOU GET IF YOU DO YOUR CALCULATION IN METERS PER SECOND!
part b
If I had asked you for the speed when it reaches theta=-30 degrees (i.e. it's on the left) what would your answer to the previous part be multiplied by? answer are either
0.86
1/2
3/4
1/sqrt(2)=0.707
None of these
part c
If we doubled the mass, what would your answer to part a be multiplied by?answers are
1, i.e. no change
1/sqrt(2)
1/2
sqrt(2)
None of these
2
Explanation / Answer
a)Potential energy of bob = mgl(1-cos 60) [l(1-cos60) is the change in height of the bob where l=0.4m] => PE =0.098 Now at theta=0 lall PE is converted into KE => 0.5*m*v^2 =0.098 =>0.5*0.05*v^2=0.098 =>v1=1.9799 [velocity of bob at theta =0] b)At theta =-30 PE=mgl(1-cos 30)=0.02625 By conservation of energy 0.098-0.02625=0.5*m*v^2 =>v2=1.6941m/sec v2/v1 =0.86 (approx) c)no change