A pendulum bob with a mass 0.55kg is attached to a string with a length of 21.2m

Question

A pendulum bob with a mass 0.55kg is attached to a string with a length of 21.2m We choose the potential energy to be zero when the string makes an angle of 90 degrees with the vertical.

A) Finf the potential energy of system when the string makes an angle of 45 degrees with the vertical.

B) If released from this height (used in part A) what is the velocity of the bob when it reaches the low point at the bottom of its path.

C) If instead the bob is raised to an angle of 75 degrees and then released, how fast is the bob going when it reaches the low point at the bottom of its path.

Please show all work!! Thank you!!!!!

Explanation / Answer

a) U=-mg*R*cos(45 degrees)=-0.55*9.8*21.2*cos(45 degrees)=+80.799 J

b) U at lowest point=-mgR=- 0.55*9.8*21.2=-114.268

==>gain in kinetic energy=114.268+80.799=195.067 J

==> 0.5*0.55*v^2=195.067 J

==> v=26.63 m/s

c) simlarly mgR+mgR*sin(15)=0.5*mv^2

==> 2*9.8*21.2*(1+sin(15 degrees))=v^2

==> v=22.87 m/s

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