If a meteor struck Earth tangentially against its rotation and embedded, Earth's day length would

Answer: Increase slightly (slower spin) The impact reduces Earth's angular momentum. Constant approximate I ⇒ smaller \omega ⇒ longer day.

Angular momentum of a system is conserved when

Answer: The net external torque is zero Conservation requires \sum \vec\tau\text{ext} = 0.

Angular momentum of a system is conserved when

Answer: No net external torque acts. Conservation of angular momentum requires zero <strongnet external torque</strong (analogous to momentum conservation and net force).

AP Physics 1 · Topic 6.4

Conservation of Angular Momentum Practice

Part of Energy and Momentum of Rotating Systems.(TOP-6.D)

Practice questions

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Sample questions

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Sample 1difficulty 2/5
$If a meteor struck Earth tangentially against its rotation and embedded, Earth's day length would$
- A
  $Decrease (faster spin)$
- B
  $Increase slightly (slower spin)$
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- C
  $Reverse direction$
- D
  $Remain unchanged$
Why
$I$
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Sample 2difficulty 2/5
$Angular momentum of a system is conserved when$
- A
  $Velocity is constant$
- B
  $The net external torque is zero$
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- C
  $There is no friction$
- D
  $The system is at rest$
Why
$\sum \vec\tau_\text{ext} = 0$
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Sample 3difficulty 2/5
$Angular momentum of a system is conserved when$
- A
  $No net external torque acts.$
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- B
  $No external forces act.$
- C
  $The net external force is nonzero but constant.$
- D
  $Internal torques sum to zero.$
Why
$Conservation of angular momentum requires zero <strong> net external torque </strong> (analogous to momentum conservation and net force).$
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Sample 4difficulty 2/5
$A figure skater spinning with arms extended pulls her arms in, reducing her moment of inertia to one-third of its original value. By what factor does her angular velocity change?$
- A
  $3$
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- B
  $9$
- C
  $1/3$
- D
  $1$
Why
$L = I\omega$
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Sample 5difficulty 3/5
$I = 0.20~\text{kg·m}^2$
- A
  $4.2~\text{rad/s}$
- B
  $6.0~\text{rad/s}$
- C
  $4.9~\text{rad/s}$
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- D
  $3.6~\text{rad/s}$
Why
$I_\text{new} = 0.20 + (0.50)(0.30)^2 = 0.245$
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