Angular Momentum worksheet

1. Angular momentum: 17,280,000 g·cm²/s

1. Find how hard the solid sphere is to spin (its moment of inertia) with the formula for that shape: I = 2/5 × m × r²

   $\frac{2}{5}\times 6\times {0.3}^2$$=0.216\text{\hspace{0.17em}kg·m²}$

2. Multiply how hard it is to spin by how fast it spins. That gives the angular momentum: L = I × ω

   $0.216\times 8$$=1.728\text{\hspace{0.17em}kg·m²/s}$

3. Change the answer to g·cm²/s, the unit you picked: divide by 0.0000001

   $\frac{1.728}{0.0000001}$$=\mathrm{17,280,000}\text{\hspace{0.17em}g·cm²/s}$

4. For reference, the energy stored in the spin (rotational kinetic energy): E = ½ × L × ω, the same as ½ × I × ω²

   $0.5\times 1.728\times 8$$=6.912\text{\hspace{0.17em}J}$

2. Angular momentum: 14 J·s

1. Multiply how hard it is to spin by how fast it spins. That gives the angular momentum: L = I × ω

   $2\times 7$$=14\text{\hspace{0.17em}kg·m²/s}$

2. For reference, the energy stored in the spin (rotational kinetic energy): E = ½ × L × ω, the same as ½ × I × ω²

   $0.5\times 14\times 7$$=49\text{\hspace{0.17em}J}$

3. Angular momentum: 27.5 J·s

1. For a small mass moving around a circle, multiply mass × speed × distance from the center: L = m × v × r

   $5\times 5\times 1.1$$=27.5\text{\hspace{0.17em}kg·m²/s}$

2. For reference, the energy stored in the spin (rotational kinetic energy): E = ½ × L × ω, the same as ½ × I × ω²

   $0.5\times 27.5\times 4.545455$$=62.5\text{\hspace{0.17em}J}$

4. Angular momentum: 12 J·s

1. Multiply how hard it is to spin by how fast it spins. That gives the angular momentum: L = I × ω

   $2\times 6$$=12\text{\hspace{0.17em}kg·m²/s}$

2. For reference, the energy stored in the spin (rotational kinetic energy): E = ½ × L × ω, the same as ½ × I × ω²

   $0.5\times 12\times 6$$=36\text{\hspace{0.17em}J}$

5. Angular momentum: 64.8 kg·m²/s

1. For a small mass moving around a circle, multiply mass × speed × distance from the center: L = m × v × r

   $6\times 6\times 1.8$$=64.8\text{\hspace{0.17em}kg·m²/s}$

2. For reference, the energy stored in the spin (rotational kinetic energy): E = ½ × L × ω, the same as ½ × I × ω²

   $0.5\times 64.8\times 3.333333$$=108\text{\hspace{0.17em}J}$

6. Angular momentum: 474.61 lb·ft²/s

1. Multiply how hard it is to spin by how fast it spins. That gives the angular momentum: L = I × ω

   $2\times 10$$=20\text{\hspace{0.17em}kg·m²/s}$

2. Change the answer to lb·ft²/s, the unit you picked: divide by 0.042140110094

   $\frac{20}{0.042140110094}$$=474.607208\text{\hspace{0.17em}lb·ft²/s}$

3. For reference, the energy stored in the spin (rotational kinetic energy): E = ½ × L × ω, the same as ½ × I × ω²

   $0.5\times 20\times 10$$=100\text{\hspace{0.17em}J}$

7. Angular momentum: 237.3 lb·ft²/s

1. Multiply how hard it is to spin by how fast it spins. That gives the angular momentum: L = I × ω

   $1\times 10$$=10\text{\hspace{0.17em}kg·m²/s}$

2. Change the answer to lb·ft²/s, the unit you picked: divide by 0.042140110094

   $\frac{10}{0.042140110094}$$=237.303604\text{\hspace{0.17em}lb·ft²/s}$

3. For reference, the energy stored in the spin (rotational kinetic energy): E = ½ × L × ω, the same as ½ × I × ω²

   $0.5\times 10\times 10$$=50\text{\hspace{0.17em}J}$

8. Angular momentum: 427.15 lb·ft²/s

1. Multiply how hard it is to spin by how fast it spins. That gives the angular momentum: L = I × ω

   $2\times 9$$=18\text{\hspace{0.17em}kg·m²/s}$

2. Change the answer to lb·ft²/s, the unit you picked: divide by 0.042140110094

   $\frac{18}{0.042140110094}$$=427.146487\text{\hspace{0.17em}lb·ft²/s}$

3. For reference, the energy stored in the spin (rotational kinetic energy): E = ½ × L × ω, the same as ½ × I × ω²

   $0.5\times 18\times 9$$=81\text{\hspace{0.17em}J}$

9. Angular momentum: 0.72 kg·m²/s

1. Find how hard the thin hollow sphere is to spin (its moment of inertia) with the formula for that shape: I = 2/3 × m × r²

   $\frac{2}{3}\times 3\times {0.2}^2$$=0.08\text{\hspace{0.17em}kg·m²}$

2. Multiply how hard it is to spin by how fast it spins. That gives the angular momentum: L = I × ω

   $0.08\times 9$$=0.72\text{\hspace{0.17em}kg·m²/s}$

3. For reference, the energy stored in the spin (rotational kinetic energy): E = ½ × L × ω, the same as ½ × I × ω²

   $0.5\times 0.72\times 9$$=3.24\text{\hspace{0.17em}J}$

10. Angular momentum: 498.34 lb·ft²/s

1. For a small mass moving around a circle, multiply mass × speed × distance from the center: L = m × v × r

   $3\times 5\times 1.4$$=21\text{\hspace{0.17em}kg·m²/s}$

2. Change the answer to lb·ft²/s, the unit you picked: divide by 0.042140110094

   $\frac{21}{0.042140110094}$$=498.337568\text{\hspace{0.17em}lb·ft²/s}$

3. For reference, the energy stored in the spin (rotational kinetic energy): E = ½ × L × ω, the same as ½ × I × ω²

   $0.5\times 21\times 3.571429$$=37.5\text{\hspace{0.17em}J}$