gligible friction in the groove are two metal blocks, each with a mass of 0.064 kg, and they are connected to each other by a spring.
At a particular moment in time (the “initial state”) the blocks are 0.24 m apart, with zero radial velocity (that is, they are not moving toward or away from each other). At this moment the angular speed of the disk is ?i is 17 radians/s. The spring pulls the blocks toward each other, and at a later time (the “final state”) the blocks are 0.08 m apart. Now what is the angular speed ?f? Approximate the metal blocks as point particles.
?f= radians/s
B. A hollow lightweight grooved disk whose moment of inertia is 4.0 × 10-3 kg·m2 rotates with negligible friction around a vertical axis. Free to slide with negligible friction in the groove are two metal blocks, each with a mass of 0.060 kg, and they are connected to each other by a spring.
At a particular moment in time (the “initial state”) the blocks are 0.48 m apart, with zero radial velocity (that is, they are not moving toward or away from each other). At this moment the angular speed of the disk is ?i is 15 radians/s. The spring pulls the blocks toward each other, and at a later time (the “final state”) the blocks are 0.16 m apart.
Now what is the angular speed ?f? Approximate the metal blocks as point particles.
?f= radians/s
What is the change in the potential energy of the spring, including sign?
??U= J
The correct answer and explanation is:
To solve the problem, we use the principle of conservation of angular momentum and the concept of change in the spring’s potential energy.
Conservation of Angular Momentum:
The total angular momentum of the system (disk + blocks) must remain constant because no external torque acts on it.
- Initial Angular Momentum: The angular momentum is the sum of the moment of inertia of the disk and the blocks: Li=Idiskωi+2mri2ωiL_i = I_{\text{disk}} \omega_i + 2 m r_i^2 \omega_i where:
- Idisk=4.0×10−3 kg\cdotpm2I_{\text{disk}} = 4.0 \times 10^{-3} \, \text{kg·m}^2 (moment of inertia of the disk)
- m=0.060 kgm = 0.060 \, \text{kg} (mass of each block)
- ri=0.48 mr_i = 0.48 \, \text{m} (initial distance of each block from the center)
- ωi=15 rad/s\omega_i = 15 \, \text{rad/s} (initial angular speed of the disk)
- Final Angular Momentum: Similarly: Lf=Idiskωf+2mrf2ωfL_f = I_{\text{disk}} \omega_f + 2 m r_f^2 \omega_f where:
- rf=0.16 mr_f = 0.16 \, \text{m} (final distance of each block from the center)
- ωf\omega_f (final angular speed to be found)
- Angular Momentum Conservation: Li=LfL_i = L_f Equating the two expressions: (4.0×10−3)(15)+2(0.060)(0.48)2(15)=(4.0×10−3)ωf+2(0.060)(0.16)2ωf(4.0 \times 10^{-3}) (15) + 2 (0.060) (0.48)^2 (15) = (4.0 \times 10^{-3}) \omega_f + 2 (0.060) (0.16)^2 \omega_f Simplify and solve for ωf\omega_f.
Change in Potential Energy of the Spring:
The change in potential energy (ΔU\Delta U) is related to the work done by the spring in pulling the blocks closer together. Using the formula for elastic potential energy: U=12kx2U = \frac{1}{2} k x^2
where:
- kk is the spring constant
- xx is the separation between the blocks.
The initial and final potential energies are: Ui=12k(0.48)2,Uf=12k(0.16)2U_i = \frac{1}{2} k (0.48)^2, \quad U_f = \frac{1}{2} k (0.16)^2
The change in potential energy: ΔU=Uf−Ui=12k[(0.16)2−(0.48)2]\Delta U = U_f – U_i = \frac{1}{2} k \left[ (0.16)^2 – (0.48)^2 \right]
Since kk is not given, you must express ΔU\Delta U in terms of kk. Alternatively, if kk is provided or can be deduced from other context, substitute and compute ΔU\Delta U.
Let me compute the angular speed ωf\omega_f and ΔU\Delta U numerically.
Results:
- Final Angular Speed (ωf\omega_f): ωf=67.13 rad/s\omega_f = 67.13 \, \text{rad/s}
- Change in Spring Potential Energy (ΔU\Delta U): The change in spring potential energy is: ΔU=12k(−0.1024)\Delta U = \frac{1}{2} k (-0.1024) The factor −0.1024-0.1024 represents the proportional change in potential energy, which depends on the spring constant kk. The negative sign indicates a decrease in the spring’s potential energy as the blocks move closer together.
If you know the spring constant kk, substitute its value to compute the exact change in potential energy. Let me know if further clarification or calculations are needed!