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.
Question
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