A uniform stick 1.5 m long with a total mass of 210 g is pivoted at its center. A 3.9-{\rm g} bullet is shot through the stick midway between the pivot and one end The bullet approaches at 250 {\rm{ m}}/{\rm{s}} and leaves at 140 {\rm{ m}}/{\rm{s}}. With what angular speed is the stick spinning after the collision?
no, it's 5
To find the angular speed at which the stick is spinning after the collision, we can use the principle of conservation of angular momentum.
Angular momentum is defined as the product of the moment of inertia and the angular velocity. In this case, the moment of inertia of the stick can be calculated by considering it as a thin rod rotating about its center.
The moment of inertia (I) of a thin rod about its center can be expressed as:
I = (1/12) * M * L^2
where M is the mass of the rod and L is the length of the rod.
In this problem, the total mass of the stick is given as 210 g, which is equal to 0.21 kg. The length of the stick is given as 1.5 m.
So, the moment of inertia of the stick is:
I = (1/12) * 0.21 * (1.5)^2 = 0.00525 kg.m^2
The initial angular momentum of the system (before the collision) is zero because the stick is at rest.
After the bullet is shot through the stick, it imparts an angular momentum to the stick. We can find this angular momentum using the equation:
L = I * ω
where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
The angular momentum imparted by the bullet can be calculated as the product of the mass and the change in velocity of the bullet:
L_bullet = m_bullet * (v_f - v_i)
where m_bullet is the mass of the bullet, v_f is the final velocity of the bullet, and v_i is the initial velocity of the bullet.
In this problem, the mass of the bullet is given as 3.9 g, which is equal to 0.0039 kg. The initial velocity of the bullet is 250 m/s, and the final velocity of the bullet is 140 m/s.
So, the angular momentum imparted by the bullet is:
L_bullet = 0.0039 * (140 - 250) = -0.4293 kg.m^2/s
According to the conservation of angular momentum, the angular momentum imparted by the bullet is equal to the angular momentum acquired by the stick:
L_bullet = L_stick
Therefore,
I * ω = L_stick
Solving for ω gives:
ω = L_stick / I
= L_bullet / I
= -0.4293 / 0.00525
= -81.9429 rad/s
The negative sign indicates that the stick is spinning in the opposite direction to the bullet's motion.
Hence, the angular speed at which the stick is spinning after the collision is approximately 81.94 rad/s.