A bullet with a mass m = 4.83 g and initial speed of 330 m/s passes through a wheel which is initially at rest as in the figure. The wheel is a solid disk with mass M = 2.29 kg and radius R = 18.6 cm The wheel rotates freely about an axis through its center and out of the plane shown in the figure. The bullet passes through the wheel at a perpendicular distance 14.8 cm from the center. After passing through the wheel it has a speed of 201 m/s.
1) What is the angular speed (rad/s) of the wheel just after the bullet leaves it?
2) How much kinetic energy (J) was lost in the collision?
Initial angular momentum of bullet around axis of wheel = I w = m r^2 w = m r v = .00483 * .148 * 330 = .2359
Final angular momentum of bullet around axis of wheel = .00483 * .148 * 201 = .1437
change in angular momentum of bullet = .2359 - .1437 = .0922
that is the angular momentum of the wheel
I w = .0922
I = (1/2) M R^2 =.5*2.29*.186^2 = .03961
so
w = .0922/.03961 = 2.328 rad/s
Ke original = (1/2) m v^2 of bullet
Ke final =(1/2)mvfinal^2 of bullet + (1/2)I w^2 of wheel
Awesome!
Thanks a bunch
1) Well, well, well, looks like this bullet really stirred things up for the wheel! To find the angular speed of the wheel just after the bullet leaves, we can use the conservation of angular momentum. The initial angular momentum of the system is zero since the wheel is at rest. But after the bullet passes through, the wheel starts spinning. So, we have:
Initial angular momentum = Final angular momentum
Since the bullet doesn't exert any torque on the wheel, angular momentum is conserved. The initial angular momentum is zero, so we set that equal to the final angular momentum:
0 = (moment of inertia of the wheel) * (angular speed of the wheel)
The moment of inertia of a solid disk about its axis of rotation is given by (1/2) * mass * radius^2. Substituting the given values:
0 = (1/2) * 2.29 kg * (0.186 m)^2 * (angular speed of the wheel)
Now it's time to do some number crunching. Solve for the angular speed of the wheel, and voila, you've got your answer!
2) Now let's tackle the second question. To find the kinetic energy lost in the collision, we need to calculate the total initial kinetic energy of the system and subtract the final kinetic energy. The total initial kinetic energy consists of the kinetic energy of the bullet and the kinetic energy of the wheel.
The kinetic energy of the bullet is given by (1/2) * mass * speed^2, so plug in the given values to get it.
The kinetic energy of the wheel can be calculated using the equation (1/2) * moment of inertia * (angular speed)^2. We already know the moment of inertia of the wheel and we just found the angular speed of the wheel, so plug away!
Now, add up the kinetic energies of the bullet and the wheel just before the collision to get the total initial kinetic energy. Subtract the kinetic energy of the bullet just after the collision (which can be calculated using the bullet's new speed) from this total initial kinetic energy.
And there you have it! The amount of kinetic energy lost in the collision.
To answer these questions, we need to apply the principle of conservation of linear momentum and the principle of conservation of angular momentum.
1) Angular Speed of the Wheel:
The principle of conservation of angular momentum states that the total angular momentum before an event is equal to the total angular momentum after the event if no external torques act on the system.
The initial angular momentum of the wheel-bullet system is zero since the wheel is initially at rest and the bullet is moving in a straight line. Therefore, the final angular momentum must also be zero after the bullet passes through the wheel.
We can calculate the initial angular momentum L_initial of the bullet and the final angular momentum L_final of the wheel using the formulas:
L_initial = m_bullet * r_bullet * v_bullet
L_final = I * ω
where m_bullet is the mass of the bullet, r_bullet is the perpendicular distance from the center of the wheel to the path of the bullet, v_bullet is the initial speed of the bullet, I is the moment of inertia of the wheel, and ω is the angular speed of the wheel just after the bullet leaves it.
Since the bullet passes through the wheel without affecting its mass distribution, the moment of inertia I remains constant. Hence, we can equate the initial and final angular momenta:
m_bullet * r_bullet * v_bullet = I * ω
Now we can solve for ω:
I * ω = m_bullet * r_bullet * v_bullet
ω = (m_bullet * r_bullet * v_bullet) / I
To find I, we use the formula for the moment of inertia of a solid disk rotating about its center:
I = (1/2) * M * R^2
where M is the mass of the disk and R is its radius. Substituting the given values, we can calculate the angular speed ω.
2) Kinetic Energy Lost in the Collision:
To determine the kinetic energy lost in the collision, we need to find the initial and final kinetic energies of the system.
The initial kinetic energy E_initial of the system is given by:
E_initial = (1/2) * m_bullet * v_bullet^2
The final kinetic energy E_final of the system is the sum of the kinetic energy of the bullet after passing through the wheel and the kinetic energy of the rotating wheel.
E_final = (1/2) * m_bullet * v_final^2 + (1/2) * I * ω^2
Substituting the given values and solving for the kinetic energy lost, we can calculate the answer.
Note: In these calculations, make sure to convert all the units to be consistent. For example, convert the mass of the bullet from grams to kg, speed from m/s to cm/s, and radius from cm to meters.
To solve this problem, we can use the principle of conservation of angular momentum and kinetic energy.
1) The angular momentum before the collision is equal to the angular momentum after the collision. The angular momentum due to the bullet is given by:
L_bullet = m_bullet * v_bullet * r_bullet
where m_bullet is the mass of the bullet, v_bullet is the initial speed of the bullet, and r_bullet is the perpendicular distance between the bullet and the center of the wheel. Substituting the given values:
L_bullet = (4.83 g) * (330 m/s) * (14.8 cm) = 2.2634 kg·m²/s
The angular momentum due to the wheel can be calculated as follows:
L_wheel = I * ω
where I is the moment of inertia of the wheel and ω is the angular speed of the wheel. The moment of inertia of a solid disk rotating about its central axis is given by:
I = (1/2) * M * R²
Substituting the given values:
I = (1/2) * (2.29 kg) * (0.186 m)² = 0.040547 kg·m²
Using the principle of conservation of angular momentum, we have:
L_bullet = L_wheel
2.2634 kg·m²/s = (0.040547 kg·m²) * ω
ω = 2.2634 kg·m²/s / 0.040547 kg·m²
ω ≈ 55.84 rad/s
Therefore, the angular speed of the wheel just after the bullet leaves it is approximately 55.84 rad/s.
2) The kinetic energy before the collision is equal to the kinetic energy after the collision plus the kinetic energy of the wheel. The kinetic energy before the collision is given by:
KE_before = (1/2) * m_bullet * v_bullet²
Substituting the given values:
KE_before = (1/2) * (4.83 g) * (330 m/s)² = 26.9594 J
The kinetic energy after the collision is given by:
KE_after = (1/2) * m_bullet * v_bullet² + (1/2) * I * ω²
Substituting the given values:
KE_after = (1/2) * (4.83 g) * (201 m/s)² + (1/2) * (0.040547 kg·m²) * (55.84 rad/s)²
KE_after ≈ 2.045 J
The kinetic energy lost in the collision is the difference between the kinetic energy before and after the collision:
KE_lost = KE_before - KE_after
KE_lost = 26.9594 J - 2.045 J
KE_lost ≈ 24.914 J
Therefore, the kinetic energy lost in the collision is approximately 24.914 J.