A smooth circular hoop with a radius of 0.600 m is placed flat on the floor. A 0
ID: 1461110 • Letter: A
Question
A smooth circular hoop with a radius of 0.600 m is placed flat on the floor. A 0.450-kg particle slides around the inside edge of the hoop. The particle is given an initial speed of 9.50 m/s. After one revolution, its speed has dropped to 4.50 m/s because of friction with the floor.
(a) Find the energy transformed from mechanical to internal in the particle—hoop—floor system as a result of friction in one revolution.
_____J
(b) What is the total number of revolutions the particle makes before stopping? Assume the friction force remains constant during the entire motion.
_____rev
Explanation / Answer
Q)A smooth circular hoop with a radius of 0.600 m is placed flat on the floor. A 0.450-kg particle slides around the inside edge of the hoop. The particle is given an initial speed of 9.50 m/s. After one revolution, its speed has dropped to 4.50 m/s because of friction with the floor.
PE = m * g * h
KE = ½ * m * v^2
As the particle moves upward, its kinetic energy decreases as its potential energy increases. As the particle moves downward, its kinetic energy increases as its potential energy decreases.
Since it moves the same distance in both directions, the increase of its kinetic energy should be equal to the decrease of its kinetic energy. This means the velocity of the particle after sliding around the inside edge of the hoop should still be 9.50 m/s.
Since velocity decreased from 10 m/s to 4 m/s, the kinetic energy decreased.
Initial KE = ½ * 0.450 * 10^2 = 22.5 J
Final KE = ½ * 0.450 * 4.5^2 = 4.55 J
Decrease of KE = 22.5 – 4.55 = 17.95 J
This is the amount that the kinetic decreased as the as the particle slid around the inside edge of the hoop one time.
The decrease of the kinetic energy is equal to the work done by the friction force.
Work of friction force = 17.95 N * m
This is the amount of work that the fiction force does as the particle slid around the inside edge of the hoop one time.
2) Number of revolutions = Decrease of KE ÷ 17.95 J/ revolution
When particle stops, it velocity is 0 m/s.
The velocity decreased from 9.5 m/s to 0 ms/. So, its kinetic energy decreased from 22.5 J to 0 J.
Decrease of KE = 22.5
Number of revolutions = 22.5 ÷ 17.95
In approximately 1.25 revolutions, the kinetic energy of the particle has decreased from 22.5 to 0.
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