A bullet of mass 5.0 g and velocity 330 m/s passes through a wheel at rest. The wheel is a
solid disk of mass 2 kg and radius 18 cm. The bullet passes through the wheel at a
perpendicular distance of 15 cm from the centre and the bullet's final velocity is 220 m/s.
What are the wheel's angular velocity, angular momentum and kinetic energy? Is the
energy of the motion conserved?
To find the wheel's angular velocity, angular momentum, and kinetic energy, we can use the principle of conservation of angular momentum.
The principle of conservation of angular momentum states that the initial angular momentum of a system is equal to the final angular momentum if no external torques act on the system. In this case, we assume there are no external torques acting on the wheel-bullet system, so we can apply this principle.
The angular momentum of an object is given by the equation:
L = I * ω,
where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
First, let's calculate the moment of inertia (I) of the wheel.
The moment of inertia of a solid disk rotating about its central axis is given by the equation:
I = (1/2) * m * r^2,
where m is the mass of the disk and r is the radius of the disk.
Given that the mass of the wheel is 2 kg and the radius is 0.18 m, we can calculate I:
I = (1/2) * 2 kg * (0.18 m)^2 = 0.065 kg·m^2.
Next, let's calculate the initial angular momentum (L_initial) of the system.
The initial angular momentum of the bullet is given by:
L_bullet_initial = m_bullet * v_bullet * r_bullet,
where m_bullet is the mass of the bullet, v_bullet is the velocity of the bullet, and r_bullet is the distance of the bullet from the center of the wheel.
Given that the mass of the bullet is 5.0 g (0.005 kg), the velocity of the bullet is 330 m/s, and the distance from the center of the wheel is 15 cm (0.15 m), we can calculate L_bullet_initial:
L_bullet_initial = 0.005 kg * 330 m/s * 0.15 m.
Now, let's calculate the final angular momentum (L_final) of the system.
The final angular momentum of the bullet is given by:
L_bullet_final = m_bullet * v_bullet_final * r_bullet,
where v_bullet_final is the final velocity of the bullet.
Given that the final velocity of the bullet is 220 m/s, we can calculate L_bullet_final:
L_bullet_final = 0.005 kg * 220 m/s * 0.15 m.
Since the bullet passes through the wheel at a distance of 15 cm from the center, we can assume that the bullet's distance does not change the angular momentum of the wheel. Therefore, the initial and final angular momenta of the wheel are both zero.
According to the principle of conservation of angular momentum, the sum of the initial angular momentum of the system should equal the sum of the final angular momentum of the system.
L_initial + 0 = L_final.
L_bullet_initial = L_bullet_final.
Now, let's substitute the values we have obtained into the equation:
0.005 kg * 330 m/s * 0.15 m = 0.005 kg * 220 m/s * 0.15 m.
By solving this equation, we find that the values of L are equal, indicating that angular momentum is conserved.
Next, let's calculate the wheel's angular velocity (ω).
ω = L / I.
Since L is zero for the wheel, ω will also be zero.
Therefore, the wheel's angular velocity is zero.
Now, let's calculate the wheel's angular momentum (L_wheel).
L_wheel = I * ω.
Given that ω = 0 and I = 0.065 kg·m^2, we find that L_wheel = 0.
Finally, let's calculate the wheel's kinetic energy (K_wheel).
The kinetic energy of a rotating object is given by the equation:
K = (1/2) * I * ω^2.
Given that ω = 0 and I = 0.065 kg·m^2, we find that K_wheel = 0 as well.
To summarize:
- The wheel's angular velocity is zero.
- The wheel's angular momentum is zero.
- The wheel's kinetic energy is zero.
Since there is no change in angular momentum or kinetic energy, the energy of the motion is conserved.