A smooth circular hoop with a radius of 0.600 m is placed flat on the floor. A 0.350-kg particle slides around the inside edge of the hoop. The particle is given an initial speed of 10.00 m/s. After one revolution, its speed has dropped to 5.50 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.
(b) What is the total number of revolutions the particle makes before stopping? Assume the friction force remains constant during the entire motion.
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Answer:
a. The energy transformed from mechanical to internal in the particle—hoop is 12.21 Joules
b. The total number of revolutions the particle makes before stopping is 1.43 revolutions
Explanation:
a.
Given
m = mass of particle = 0.350-kg
u = initial speed of 10.00 m/s
v = final speed = 5.50 m/s
r = radius = 0.600 m
We assume that the floor is horizontal;
This means that F = mg.
We also assume the rotational kinetic energy to be negligable.
Having listed the assumptions, we proceed as follows;
Let ∆E represent The energy transformed from mechanical to internal in the particle hoop.
This is given by
∆E = KE1 – KE2
Where KE1 = ½mu²
KE2 = ½mv²
So, ∆E = KE1 – KE2 becomes
∆E = ½mu² – ½mv²
∆E = ½m(u² – v²)
∆E = ½ * 0.350 * (10² – 5.5²)
∆E = 12.20625
∆E = 12.21J (Approximated)
Hence, the energy transformed from mechanical to internal in the particle—hoop is 12.21 Joules
b.
Let n = number of revolutions
The relationship between n and the energy is
1/n = (KE1 – KE2)/KE1
Make n the subject of formula
n = KE1 / (KE1 – KE2)
n = ½mu² / (½mu² – ½mv²) — Simplify
n = ½mu² / (½m(u² – v²)) —- Divide through by ½m
n = u² / (u² – v²)
n = 10² / (10² – 5.5²)
n = 1.433691756272401
n = 1.43 rev
Hence, the total number of revolutions the particle makes before stopping is 1.43 revolutions