COLLEGE PHYSICS LL W/ 6 MONTH ACCESS
COLLEGE PHYSICS LL W/ 6 MONTH ACCESS
2nd Edition
ISBN: 9781319414597
Author: Freedman
Publisher: MAC HIGHER
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Chapter 8, Problem 117QAP
To determine

(a)

Average angular speed of the diver when he fell towards water having body tucked in

Expert Solution
Check Mark

Answer to Problem 117QAP

Average angular speed = 14.18 rads-1

Explanation of Solution

Given info:

Turns made by the diver while he dives = 3.5

Maximum height reached by the diver = 2.0 m

Height to the platform from the water level = 10.0 m

Length of the modeled rod = 2.0 m

Diameter of the modeled rod = 0.75 m

Formula used:

  t=2hg

  t= Time taken for the complete the flight

  h= Height

  g= Gravitational acceleration

  ω=number of turns*2πt

  ω= Angular momentum

  t= Time taken for the complete the flight

Calculation:

  Total hiehgt that diver travels=10.0 m+2.0 m=12.0 m

  t= 2hgt= 2*12 m 10  ms -2 t=1.55 s

  ω=number of turns*2πtω=3.5*2*3.141.55 sω=14.18 rads-1

Conclusion:

Average angular speed = 14.18 rads-1

To determine

(b)

Angular speed just after the diver stretched out

Expert Solution
Check Mark

Answer to Problem 117QAP

Angular speed just after the diver stretched out= 2.99 rads-1

Explanation of Solution

Given info:

Turns made by the diver while he dives = 3.5

Maximum height reached by the diver = 2.0 m

Height to the platform from the water level = 10.0 m

Length of the modeled rod = 2.0 m

Diameter of the modeled cylinder = 0.75 m

Formula used:

  I1ω1=I2ω2

  I1= Initial moment of inertia

  ω1= Initial angular momentum

  I2= Final moment of inertia

  ω2= Final angular momentum

  I2=112ml2

  I1= Final moment of inertia

  m= Mass of the diver

  r= Length of the modeled rod

  I1=12mr2

  I1= Initial moment of inertia

  m= Mass of the diver

  r= Radius of the modeled cylinder

Calculation:

  I1ω1=I2ω2ω2=I1ω1I2ω2=12mr2*14.18  rads -11 12ml2ω2=12* (0.375 m)2*14.18  rads -11 12 (2.0 m)2ω2=2.99 rads-1

Conclusion:

Angular speed just after the diver stretched out= 2.99 rads-1

To determine

(c)

Rotational kinetic energy difference

Expert Solution
Check Mark

Answer to Problem 117QAP

Rotational kinetic energy difference= 418 J

Explanation of Solution

Given info:

Turns made by the diver while he dives = 3.5

Maximum height reached by the diver = 2.0 m

Height to the platform from the water level = 10.0 m

Length of the modeled rod = 2.0 m

Diameter of the modeled cylinder = 0.75 m

Formula used:

  E=12Iω2

  E= Rotational kinetic energy

  I= Moment of inertia

  ω= Angular momentum

  I2=112ml2

  I1= Final moment of inertia

  m= Mass of the diver

  r= Length of the modeled rod

  I1=12mr2

  I1= Initial moment of inertia

  m= Mass of the diver

  r= Radius of the modeled cylinder

Calculation:

  E=12Iω2

Kinetic energy difference in the dive,

  dE=12I2ω2212I1ω12dE=12*12*75 kg*(0.375 m)2*(14.18 rads-1)212*75 kg*112(2.0 m)2(2.99 rads-1)2dE=530J112JdE=418J

Conclusion:

Rotational kinetic energy difference= 418 J

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Chapter 8 Solutions

COLLEGE PHYSICS LL W/ 6 MONTH ACCESS

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Rotational Kinetic Energy; Author: AK LECTURES;https://www.youtube.com/watch?v=s5P3DGdyimI;License: Standard YouTube License, CC-BY