Exercise 10.20 A string is wrapped several times around the rim of a small hoop

Question

Exercise 10.20

A string is wrapped several times around the rim of a small hoop with radius 8.00 cm and mass 0.180 kg . The free end of the string is held in place and the hoop is released from rest (the figure (Figure 1) ). After the hoop has descended 60.0 cm , calculate

Part A

the angular speed of the rotating hoop and

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Part B

the speed of its center.

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Figure 1 of 1

Exercise 10.20

A string is wrapped several times around the rim of a small hoop with radius 8.00 cm and mass 0.180 kg . The free end of the string is held in place and the hoop is released from rest (the figure (Figure 1) ). After the hoop has descended 60.0 cm , calculate

Part A

the angular speed of the rotating hoop and

= rad/s

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Part B

the speed of its center.

v = m/s

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Figure 1 of 1

Explanation / Answer

B)

Apply conservation of energy

final kinetic energy = initial potentail energy

0.5*m*v^2 + 0.5*I*w^2 = m*g*h

0.5*m*v^2 + 0.5*m*r^2*w^2 = m*g*h

0.5*m*v^2 + 0.5*m*v^2 = m*g*h

m*v^2 = m*g*h

==> v = sqrt(g*h)

= sqrt(9.8*0.6)

= 2.42 m/s <<<<<<<-------Answer for part B)

A) we know, Vcm = r*w

==> e = Vcm/r

= 2.42/0.08

= 30.3 rad/s <<<<<<<-------Answer for part A)

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