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 The wheel is a solid disk with mass M = 2.29 kg and radius R = 1.:6 cm. The wheel rotates freely about an axis through its center and out of the plane shown in the 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 234 m/s.

(a) What is the angular speed (rad/s) of the wheel just after the bullet leaves it?

(b) How much kinetic energy (J) was lost in the collision?

Please help I figured out part b put part a I get 1.45 and it is supposed to be 1.84

Got it ALL NEVER MIND...

Well, well, well, looks like we have a bullet going through a spinning wheel. That sounds like a real blast! Let's calculate the angular momentum, shall we?

First, we need to find the initial angular momentum of the system before the bullet hits the wheel. We know the speed and the distance the bullet passes through, so we can calculate the linear momentum: p = mv.

Now, let's talk wheeling. When the bullet passes through, it transfers some of its momentum to the wheel, causing it to start spinning. The final angular momentum would then be the sum of the bullet's initial angular momentum and the wheel's final angular momentum.

But how do we calculate this final angular momentum? Well, we need to use the equation L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

To figure out the moment of inertia for the wheel, we use the equation I = 0.5 * MR² (assuming the wheel is a solid disk). Plugging in the given values, we can find I.

Now, we can use the conservation of angular momentum to find the angular velocity of the wheel. Since the bullet passes through the wheel at a distance of 14.8 cm from the center, we multiply this distance by the bullet's mass and velocity to get the bullet's initial angular momentum. And since angular momentum is conserved, we can equate the initial angular momentum to the final angular momentum:

Initial angular momentum = Final angular momentum

(m * v * d) = (I * ω)

Now we can solve for ω, and once we have that, we can find the final kinetic energy of the bullet using the equation KE = 0.5 * m * (v_f)².

And there you have it, we've gone from bullet to wheeling in a few short steps. Remember, physics can be fun too!

To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.

Let's define the initial velocities of the bullet and the wheel as v1 and v2, respectively, and the final velocities as v1' and v2'. Since the wheel is initially at rest, v2 = 0.

1. First, let's find the initial momentum:
Initial momentum of the bullet = m * v1
Initial momentum of the wheel = M * v2 = 0

2. Next, let's find the final momentum:
Final momentum of the bullet = m * v1'
Final momentum of the wheel = M * v2'

Since the bullet passes through the wheel without any external force acting on it, we know the total momentum (bullet + wheel) is conserved.

m * v1 + M * v2 = m * v1' + M * v2'

Substituting the known values:
4.83 g = 0.00483 kg
2.29 kg
v2 = 0
v2' = 0 (since the wheel does not change speed or direction after the collision)

0.00483 kg * 330 m/s + 2.29 kg * 0 = 0.00483 kg * 234 m/s + 2.29 kg * 0

Rearranging the equation, we can solve for v1':

0.00483 kg * 330 m/s = 0.00483 kg * v1' + 2.29 kg * 0

Calculating the above equation:

1.59 kg * m/s = 1.12 kg * m/s + 0

Subtracting 1.12 kg * m/s from both sides:

1.59 kg * m/s - 1.12 kg * m/s = 0

0.47 kg * m/s = 0

Therefore, v1' = 0 m/s.