Physics, 13.11.2019 02:31 jetblackcap

Asolid disk of mass m1 = 9.8 kg and radius r = 0.19 m is rotating with a constant angular velocity of ? = 32 rad/s. a thin rectangular rod with mass m2 = 3 kg and length l = 2r = 0.38 m begins at rest above the disk and is dropped on the disk where it begins to spin with the disk.

1)what is the initial angular momentum of the rod and disk system? kg-m2/s

2)what is the initial rotational energy of the rod and disk system? j

3)what is the final angular velocity of the disk? rad/s

4)what is the final angular momentum of the rod and disk system? kg-m2/s

5)what is the final rotational energy of the rod and disk system? j

6)the rod took t = 5.8 s to accelerate to its final angular speed with the disk. what average torque was exerted on the rod by the disk?

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Answer from: oreo543

1) The initial angular momentum of the rod and disk system is; L_total,i = 5.664 kg•m²/s

2) The initial rotational energy of the rod and disk system is; K_total = 90.624 J

3) The final angular velocity of the disk is; ω_d,f = 26.58 rad/s

4) The final angular momentum of the rod and disk system is; L_total,f = 5.664 kg•m²/s

5) The final rotational energy of the rod and disk system is; E_total = 75.277 J

6) The average torque exerted on the rod by the disk is; τ = 0.165 N.m

We are given;

Mass of disk; m1 = 9.8 kg

Mass of rod; m2 = 3 kg

Radius of disk; r = 0.19 m

Angular velocity of disk; ω_d = 32 rad/s

Length of rod; L = 0.38 m

1) Moment of inertia of disk is;

I_disk = ½m1•r²

I_disk = ½ × 9.8 × 0.19²

I_disk = 0.177 kg.m²

Moment of inertia of rod;

I_rod = m2•L²/12

I_rod = (3 × 0.38²)/12

I_rod = 0.0361 kg.m²

Formula for angular momentum is; L = Iω

Thus, initial angular momentum of the disk and rod system is;

L_total = (0.177 × 32) + (0.0361 × 0)

L_total,i = 5.664 kg•m²/s

2) Initial rotational kinetic energy of the disk and rod system is;

K_total = ½I_disk*(ω_d)² + ½I_rod*ω²

K_total = ½(0.177 × 32²) + 0

K_total = 90.624 J

3) To get the final angular velocity of the disk, we will use the formula;

I_disk*ω_d,i = (I_disk + I_rod)ω_d,f

Plugging in the relevant values;

0.177 × 32 = (0.177 + 0.0361)ω_d,f

ω_d,f = (0.177 × 32)/(0.177 + 0.0361)

ω_d,f = 26.58 rad/s

4) From conservation of momentum, we can say that final angular momentum. Thus;

L_total,f = 5.664 kg•m²/s

5) Final total energy is;

E_total = ½(I_disk + I_rod)(ω_d,f)²

E_total = ½(0.177 + 0.0361)(26.58)²

E_total = 75.277 J

6) Average torque was exerted on the rod by the disk is;

τ = I_rod × α

Where;

α = (ω_d,f - ω)/t

α = (26.58 - 0)/5.8

α = 4.58 rad/s²

Thus;

τ = 0.0361 × 4.58

τ = 0.165 N.m

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Answer from: whatsittoya4261

Part a)

L = 5.76 kg m^2/s

Part b)

K = 92.16 J

Part c)

$\omega = 26.65 rad/s$

Part d)

L = 5.76 kg m^2/s

Part e)

E = 76.74 J

Part f)

$\tau = 0.16 Nm$

Explanation:

Part a)

Initial angular momentum of the system of rod + disc is given as

$L = I\omega$

here we know that

$I_{disc} = \frac{1}{2}mR^2$

$I_{disc} = \frac{1}{2}(9.8)(0.19)^2$

$I_{disc} = 0.18 kg m^2$

$I_{rod} = \frac{1}{12}mL^2$

$I_{rod} = \frac{1}{12}(3)(0.38)^2$

$I_{rod} = 0.0361 kg m^2$

so angular momentum is given as

L = 0.18(32) + 0.0361(0)

L = 5.76 kg m^2/s

Part b)

Initial rotational energy is given as

$K = \frac{1}{2}I\omega^2$

$K = \frac{1}{2}(0.18)(32^2) + \frac{1}{2}(0.0361)0$

K = 92.16 J

Part c)

As we know that angular momentum is conserved so we will have

$I_1\omega_1 = (I_1 + I_2)\omega_2$

$0.18(32) = (0.18 + 0.0361)\omega$

$\omega = 26.65 rad/s$

Part d)

Angular momentum is conserved so final angular momentum is same as initial angular momentum

L = 5.76 kg m^2/s

Part e)

Final total energy

$E = \frac{1}{2}(I_1 + I_2)\omega^2$

$E = \frac{1}{2}(0.18 + 0.0361)26.65^2$

E = 76.74 J

Part f)

Angular acceleration of the rod is given as

$\alpha = \frac{\omega_f - \omega_i}{t}$

$\alpha = \frac{26.65 - 0}{5.8}$

$\alpha = 4.6 rad/s^2$

$\tau = I\alpha$

$\tau = 0.0361(4.6)$

$\tau = 0.16 Nm$

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Asolid disk of mass m1 = 9.8 kg and radius r = 0.19 m is rotating with a constant angular velocity o...