A smooth circular hoop with a radius of 0.900 m is placed flat on the floor. A 0

Question

A smooth circular hoop with a radius of 0.900 m is placed flat on the floor. A 0.375-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 4.00 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

here,

r = 0.900 m
M = 0.375 Kg
v1 = 9.50 m/s
v2 = 4 m/s

A)
From Conservation of Energy we have :

Initial Ke = Final KE + Friction

0.5 * M * v1^2 = 0.5 m * v2^2 + Energy due to frictional Force
Eff = 0.5m(v1^2 - v2^2)
Eff = 0.5 * 0.375 *(9.5^2 - 4^2)
Eff = 0.1875 * 74.25
Eff = 13.92 J

B)
S distnace tarvelled by particle in one peroid
S = 2 * pi *r
S = 2 * 3.14 * 0.900
S = 5.652 m

a = (vf^2 - vi^2)/2s
a = (4^2 - 9.50^2)/(2*5.652)
a = -6.568 m/s^2

Therefore, Distance travelled by particle with this acceleration and intial velocity v1 :

S = v1^2 /2a
S = 9.5^2 / (2*6.568)
s = 6.870 m

Rev = S/(2*pi*r)
Rev = 6.870/(2*3.14*0.900)
Rev = 1.215

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