The figure shows an overhead view of a 3.00-kg plastic rod of length 1.20 m on a table. One end of the rod is attached to the table, and the rod is free to pivot about this point without friction. A disk of mass 37.0 g slides toward the opposite end of the rod with an initial velocity of 20.0 m/s. The disk strikes the rod and sticks to it. After the collision, the rod rotates about the pivot point. a.What is the angular velocity of the two after the collision?b) What is the kinetic energy before and after the collision?
To find the angular velocity of the system after the collision, we need to use the principle of conservation of angular momentum.
a) Conservation of angular momentum states that the total angular momentum of a system remains constant if no external torques act on it. Mathematically, this can be expressed as:
L1 = L2
Where L1 is the initial angular momentum and L2 is the final angular momentum.
Initially, the disk is moving with a linear velocity and no angular velocity, so its angular momentum is zero.
L1 = 0
After the collision, the disk sticks to the rod, and together they rotate about the pivot point. The angular momentum of this combined system can be calculated as the product of the moment of inertia (I) and the angular velocity (ω).
L2 = I * ω
The moment of inertia (I) of a rod rotating about one end is given by:
I = (1/3) * m * L^2
Where m is the mass of the rod and L is its length.
Substituting the values:
I = (1/3) * (3.00 kg) * (1.20 m)^2 = 1.44 kg⋅m^2
Using conservation of angular momentum, we can equate L1 and L2:
0 = (1.44 kg⋅m^2) * ω
Solving for angular velocity (ω):
ω = 0
After the collision, the system comes to rest, and the angular velocity is zero.
b) To find the kinetic energy before and after the collision, we need to calculate the total initial kinetic energy and the final kinetic energy of the system.
The initial kinetic energy (K1) can be calculated as:
K1 = (1/2) * m_disk * v^2
Where m_disk is the mass of the disk and v is its initial velocity.
Substituting the values:
K1 = (1/2) * (0.037 kg) * (20.0 m/s)^2 = 14.8 J
After the collision, the system is at rest, and its kinetic energy is zero.
K2 = 0
Therefore, the initial kinetic energy before the collision is 14.8 J, and the final kinetic energy after the collision is 0 J.
To find the angular velocity of the system after the collision, we can use the principle of conservation of angular momentum. The angular momentum before the collision is zero since the disk is not rotating and the rod is not moving. However, after the collision, the disk and the rod rotate together.
a) Using the conservation of angular momentum:
L(initial) = L(final)
0 = (m1 * r1 * v1) + (m2 * r2 * v2)
Here:
m1 = mass of disk = 37.0 g = 0.037 kg
r1 = distance from pivot to center of mass of disk = length of rod/2 = 1.20 m / 2 = 0.60 m
v1 = initial velocity of disk = 20.0 m/s
m2 = mass of rod = 3.00 kg
r2 = distance from pivot to center of mass of rod = length of rod/2 = 1.20 m / 2 = 0.60 m
v2 = final velocity of rod and disk
0 = (0.037 kg * 0.60 m * 20.0 m/s) + (3.00 kg * 0.60 m * v2)
Simplifying the equation:
0 = (0.444 kg⋅m^2/s) + (1.08 kg⋅m^2 * v2)
Rearranging the equation:
v2 = -0.444 kg⋅m^2 / (1.08 kg⋅m^2/s)
v2 = -0.4111 rad/s (Negative sign indicates the direction of rotation, in this case, clockwise)
Therefore, the angular velocity of the system after the collision is -0.4111 rad/s (clockwise).
b) To find the kinetic energy before and after the collision:
Kinetic energy before collision (KE_initial) = (1/2) * m1 * v1^2
KE_initial = (1/2) * 0.037 kg * (20.0 m/s)^2
KE_initial = 14.8 J
After the collision, the disk and rod stick together and rotate about the pivot point. The kinetic energy of this rotating system is given by:
KE_final = (1/2) * I * ω^2
Where:
I = moment of inertia of the rod and disk about the pivot point
ω = angular velocity of the system after the collision
The moment of inertia of the rod about its pivot point can be calculated as:
I_rod = (1/3) * m_rod * (L_rod)^2
I_rod = (1/3) * 3.00 kg * (1.20 m)^2
I_rod = 2.40 kg⋅m^2
The moment of inertia of the disk about its center of mass can be calculated as:
I_disk = (1/2) * m_disk * (r_disk)^2
I_disk = (1/2) * 0.037 kg * (0.60 m)^2
I_disk = 0.00702 kg⋅m^2
Since the disk is attached to the rod, its moment of inertia about the pivot point will be the sum of the moment of inertia of the disk and the moment of inertia of the rod about its pivot point.
I_total = I_rod + I_disk
I_total = 2.40 kg⋅m^2 + 0.00702 kg⋅m^2
I_total = 2.40702 kg⋅m^2
Substituting the values into the equation:
KE_final = (1/2) * I_total * ω^2
KE_final = (1/2) * 2.40702 kg⋅m^2 * (-0.4111 rad/s)^2
KE_final = 0.5009 J
Therefore, the kinetic energy of the system after the collision is 0.5009 J.