A horizontal vinyl record of mass 0.10 kg and radius 0.10 m rotates freely about a vertical axis through its center with an angular speed of 4.7 rad/s and a rotational inertia of 5.0 *10-4 kg m2. Putty of mass 0.020 kg drops vertically onto the record from above and sticks to the edge of the record. What is the angular speed of the record immediately afterwards?
The Correct Answer and Explanation is:
Let’s analyze the problem step-by-step:
Given:
- Mass of record, mr=0.10 kgm_r = 0.10 \, \text{kg}
- Radius of record, R=0.10 mR = 0.10 \, \text{m}
- Initial angular speed, ωi=4.7 rad/s\omega_i = 4.7 \, \text{rad/s}
- Rotational inertia of record, Ir=5.0×10−4 kg⋅m2I_r = 5.0 \times 10^{-4} \, \text{kg} \cdot \text{m}^2
- Mass of putty, mp=0.020 kgm_p = 0.020 \, \text{kg}
- Putty sticks at the edge of the record (radius R=0.10 mR = 0.10 \, m)
- Find the final angular speed ωf\omega_f after the putty sticks.
Step 1: Understand the principle involved
This is a problem involving conservation of angular momentum. The system is isolated in terms of external torques (assuming no friction on the axis), so the total angular momentum before the putty lands equals the total angular momentum after.
Step 2: Calculate initial angular momentum
The initial angular momentum LiL_i is given by: Li=IrωiL_i = I_r \omega_i
Substitute the values: Li=(5.0×10−4 kg⋅m2)(4.7 rad/s)=2.35×10−3 kg⋅m2/sL_i = (5.0 \times 10^{-4} \, \text{kg} \cdot \text{m}^2)(4.7 \, \text{rad/s}) = 2.35 \times 10^{-3} \, \text{kg} \cdot \text{m}^2/\text{s}
Step 3: Calculate the rotational inertia of the putty after it sticks
Since the putty sticks at the edge, it acts like a point mass at radius RR: Ip=mpR2=0.020×(0.10)2=0.020×0.01=2.0×10−4 kg⋅m2I_p = m_p R^2 = 0.020 \times (0.10)^2 = 0.020 \times 0.01 = 2.0 \times 10^{-4} \, \text{kg} \cdot \text{m}^2
Step 4: Total rotational inertia after the putty sticks
If=Ir+Ip=5.0×10−4+2.0×10−4=7.0×10−4 kg⋅m2I_f = I_r + I_p = 5.0 \times 10^{-4} + 2.0 \times 10^{-4} = 7.0 \times 10^{-4} \, \text{kg} \cdot \text{m}^2
Step 5: Use conservation of angular momentum to find ωf\omega_f
Li=Lf ⟹ Irωi=IfωfL_i = L_f \implies I_r \omega_i = I_f \omega_f ωf=IrωiIf=5.0×10−4×4.77.0×10−4=2.35×10−37.0×10−4≈3.36 rad/s\omega_f = \frac{I_r \omega_i}{I_f} = \frac{5.0 \times 10^{-4} \times 4.7}{7.0 \times 10^{-4}} = \frac{2.35 \times 10^{-3}}{7.0 \times 10^{-4}} \approx 3.36 \, \text{rad/s}
Final answer:
ωf≈3.36 rad/s\boxed{ \omega_f \approx 3.36 \, \text{rad/s} }
Explanation (300 words):
This problem is a classic example of angular momentum conservation during a sudden change in a rotating system. Initially, the vinyl record is spinning freely at an angular speed of 4.7 rad/s, with a known rotational inertia. When a putty mass drops vertically and sticks to the edge, it adds extra mass at the outer radius, thereby increasing the system’s rotational inertia.
The key to solving this problem is recognizing that no external torque acts on the system about its rotation axis during the putty’s impact and sticking. This means the system’s angular momentum just before and immediately after the event remains constant.
The initial angular momentum is the product of the record’s rotational inertia and its angular speed. The rotational inertia is a measure of how mass is distributed relative to the axis of rotation; more mass further from the axis increases inertia.
When the putty sticks to the edge, it acts like a point mass at radius 0.1 m. We calculate its rotational inertia as Ip=mpR2I_p = m_p R^2, which adds to the original inertia of the record. The total inertia after the putty sticks is the sum of the record’s original inertia and the putty’s inertia.
Because angular momentum is conserved, the product of rotational inertia and angular velocity before the event equals that after. We solve for the new angular velocity ωf\omega_f by dividing the initial angular momentum by the increased inertia.
As expected, the angular velocity decreases after the putty sticks because the system’s inertia increases, making it harder to rotate at the original speed. This decrease demonstrates the inverse relationship between rotational inertia and angular velocity when angular momentum is conserved.
In summary, this problem shows the importance of conservation laws in rotational dynamics and how adding mass at the periphery of a rotating object affects its angular speed.