A large wooden wheel of radius R and the moment of inertia I is mounted on an axle so as to rotate freely. A bullet of mass m and speed v is shot tangential to the wheel and strike its edge, lodging in the wheel’s rim. If the wheel was originally at rest, what is its rotation rate just after the collision?
Initial angular momentum about axis =
Mbullet*Vbullet*R
This equals the final angular momentum,
(Iwheel + Mbullet*R^2)*w_final
w_final = (Mbullet*Vbullet)*R/(Iwheel + Mbullet*R^2)
To determine the rotation rate of the wheel just after the collision, we can use the principle of conservation of angular momentum. The total angular momentum before the collision is equal to the total angular momentum after the collision.
The initial angular momentum of the wheel is zero since it is at rest. The angular momentum of the bullet can be calculated as the product of its moment of inertia, I, and its initial angular velocity, ωi. The angular velocity of the bullet, ωi, is given by the formula v/r, where v is the linear velocity of the bullet and r is the radius of the wheel.
Initially, the total angular momentum is given by:
Linitial = Lwheel + Lbullet
= 0 + I * ωi
After the collision, the bullet lodges in the wheel's rim, becoming a part of it. This increases the moment of inertia of the wheel to I + mr^2, where m is the mass of the bullet. The final angular velocity of the wheel, ωf, can be calculated by dividing the initial angular momentum by the final moment of inertia:
Linitial = Lfinal
I * ωi = (I + mr^2) * ωf
Simplifying the equation, we have:
I * ωi = (I + mr^2) * ωf
ωf = (I * ωi) / (I + mr^2)
Therefore, the rotation rate of the wheel just after the collision is given by (I * ωi) / (I + mr^2).
To determine the rotation rate of the wheel just after the collision, we can use the principle of conservation of angular momentum.
The initial angular momentum of the system (bullet + wheel) is zero since the wheel was originally at rest. After the collision, the bullet becomes lodged in the wheel's rim and both start to rotate together. The final angular momentum of the system is given by the equation:
Anguilar Momentum (initial) = Angular Momentum (final)
Since the bullet sticks to the rim, the angular momentum of the bullet is directly transferred to the wheel. The angular momentum of a rotating object can be calculated as the product of its moment of inertia and angular velocity:
I * ω (initial) = (I + mR^2) * ω (final)
Where:
- I is the moment of inertia of the wheel initially
- ω (initial) is the initial angular velocity of the wheel
- m is the mass of the bullet
- R is the radius of the wheel
- ω (final) is the final angular velocity of the wheel after the bullet lodges in its rim
Since the initial angular velocity is zero, we can simplify the equation:
0 = (I + mR^2) * ω (final)
We can solve this equation to find the final angular velocity (ω (final)) of the wheel after the bullet lodges in its rim.
ω (final) = 0 / (I + mR^2)
Since the initial angular velocity is zero, the final angular velocity of the wheel will also be zero. Therefore, the rotation rate of the wheel just after the collision is zero.