QUESTION 1: Consider a hoop of mass 2.1 kg and radius 0.57 m that rolls without
ID: 2299042 • Letter: Q
Question
QUESTION 1:
Consider a hoop of mass 2.1 kg and radius 0.57 m that rolls without slipping down an incline. The hoop starts at rest from a height h = 2.6 m above the bottom of the incline and then rolls to the bottom.
a) What forces act on the hoop? (Select all that apply.)
Centripetal force, normal force, friction, and/or gravity.
b) Which of these forces do work on the hoop? (Select all that apply.)
Centripetal force, normal force, friction, and/or gravity.
c) What is the total initial mechanical energy of the hoop? Take the zero of potential energy to be at the bottom of the incline.
d) If the hoop's speed at the bottom of the incline is vf, what is its mechanical energy? Express your answer in terms of vf. (use variables only).
e) Solve for vf.
Explanation / Answer
a) Centripetal force, normal force, friction, and gravity. (Answer)
All of these forces will act on the hoop
b) Work is done by force of friction and gravity (Answer)
c) Initial Total Mechanical Energy = Potential Energy + Translational Kinetic Energy + Rotational Kinetic Energy
therefore
Initial Total Mechanical Energy = mgh = 2.1 x 9.8 x 2.6 = 53.508 Joules (Answer)
d) Hoops Total Mechanical Energy at the botttom = Potential Energy + Translational Kinetic Energy + Rotational Kinetic Energy
PE = 0
Total Mechanical Energy at the botttom = 1/2m(vf)^2 + 1/2 x mr^2 x (vf/r)^2 = m(vf)^2 (Answer)
e) Now Total Mechanical Energy at the bottom and at the top will be same
therefore on equating we get
m(vf)^2 = 53.508
therefore on solving vf = 5.0477 m/s (Answer)