A 1.4-m-long massless rod is pivoted at one end and swings around in a circle on a frictionless table. A block with a hole through the center can slide in and out along the rod. Initially, a small piece of wax holds the block 20 from the pivot. The block is spun at 50 , then the temperature of the rod is slowly increased. When the wax melts, the block slides out to the end of the rod. What is the final angular velocity?
The moment of inertia (I) for the block (the point mass, as we are not given its dimensions) is mr^2:
I1 =m•r1^2 =m •0.2^2 =0.04•m,
I2 = m•r1^2 =m •1.4^2=1.96•m,
Acording to the law of conservation of angular momentum
L1 = L2
I1•ω1 = I2•ω2,
ω2 = I1•ω1/I2 =0.04•m•50/1.96•m =
=1.02 (units?)
To solve this problem, we can use the principle of conservation of angular momentum. The initial angular momentum of the system is equal to the final angular momentum.
The initial angular momentum is given by the equation:
L_initial = I_initial * ω_initial,
where L_initial is the initial angular momentum, I_initial is the initial moment of inertia, and ω_initial is the initial angular velocity.
The final angular momentum is given by the equation:
L_final = I_final * ω_final,
where L_final is the final angular momentum, I_final is the final moment of inertia, and ω_final is the final angular velocity.
Since the rod is massless and the block initially slides along the rod, the initial moment of inertia can be approximated as:
I_initial = m * r_initial^2,
where m is the mass of the block and r_initial is the initial distance of the block from the pivot.
Furthermore, since the wax melts and the block slides out to the end of the rod, the final moment of inertia can be approximated as:
I_final = m * r_final^2,
where r_final is the length of the rod.
Given that r_initial = 0.20 m, r_final = 1.4 m, ω_initial = 50 rad/s, and solving for ω_final, we can rewrite the equation:
L_initial = L_final,
(m * r_initial^2) * ω_initial = (m * r_final^2) * ω_final.
Simplifying, we get:
(r_initial^2) * ω_initial = (r_final^2) * ω_final.
Substituting the given values, we have:
(0.20^2) * 50 = (1.4^2) * ω_final.
Solving for ω_final:
(0.04) * 50 = (1.96) * ω_final.
2 = 1.96 * ω_final.
ω_final = 2 / 1.96.
ω_final ≈ 1.02 rad/s.
Therefore, the final angular velocity is approximately 1.02 rad/s.
To determine the final angular velocity, we need to apply the principle of conservation of angular momentum. Angular momentum is a vector quantity that represents the rotational motion of an object. In this case, the angular momentum of the system will be conserved as there are no external torques acting on it.
1. First, let's calculate the initial angular momentum of the system. Angular momentum (L) is given by the product of the moment of inertia (I) and the angular velocity (ω):
L = I * ω
The moment of inertia of the block can be approximated as that of a point mass rotating about its center, given by:
I_block = m_block * r^2
where m_block is the mass of the block and r is the distance of the block from the pivot (20 cm in this case).
2. The moment of inertia of the rod rotating about its pivot can be approximated as that of a thin rod rotating about one end, given by:
I_rod = (1/3) * m_rod * L_rod^2
where m_rod is the mass of the rod and L_rod is the length of the rod (1.4 m in this case).
3. Now, let's calculate the total initial angular momentum. Initially, the block is stationary but the rod is spinning. Therefore:
L_initial = I_rod * ω_initial
4. When the wax melts and the block slides out to the end of the rod, the system has redistributed its mass. The moment of inertia of the system changes. However, the conservation of angular momentum tells us that the total angular momentum remains constant.
L_initial = L_final
Therefore:
I_rod * ω_initial = I_final * ω_final
5. At the end, the block reaches the end of the rod. This means it will have a larger moment of inertia as it is now rotating about its center. The moment of inertia of the block at the end can be approximated as:
I_block_final = m_block * r_end^2
where r_end is the distance of the block from the pivot when it reaches the end.
6. Substituting the values into the conservation equation, we get:
I_rod * ω_initial = (I_rod + I_block_final) * ω_final
Substituting the expressions for I_rod and I_block, we get:
(1/3) * m_rod * L_rod^2 * ω_initial = (1/3) * m_rod * L_rod^2 * ω_final + m_block * r_end^2 * ω_final
7. Now, let's solve for ω_final:
ω_final = [(1/3) * m_rod * L_rod^2 * ω_initial] / [(1/3) * m_rod * L_rod^2 + m_block * r_end^2]
Given the values of m_rod (mass of the rod), L_rod (length of the rod), r (initial distance of the block from the pivot), and ω_initial (initial angular velocity), you can substitute those values to find the final angular velocity ω_final.
Please note that in the actual calculation, it is essential to use consistent units (e.g., kilograms and meters for mass and distance, respectively) and double-check the calculations to ensure accurate results.