Answer like this. In step 1 : Provide Given, Objective and explanation of 40 words. Then main answer in step 2 with explanation of 40 words in last and then provide final answer of maximum 30 wordsA s

Question

Answer like this. In step 1 : Provide Given, Objective and explanation of 40 words. Then main answer in step 2 with explanation of 40 words in last and then provide final answer of maximum 30 wordsA spring/pulley/cylinder system is shown above. The
motion is such that the spring always stretches in the
vertical direction, while the pulley (a thin disc) is pinned
at its center and the rope does not slip on the pulley. At
the instant shown the spring isalready stretched by a
value of s = 0.15 m, and the angular velocity of the
pulley is w = 2.9 rad/s. The other relative properties are
listed below:R = 0.56 mpulley mass = 25.6 kgcylinder
mass = 24.6 kgk = 120 N/mUsing the given
information, determine the angular velocity of the pulley
after thecylinder falls d = 1.4 m from its current location.
3 1

VIPSolutions · Accepted Answer

**Step 1:**
- **Given:**  
  Spring stretched by $ s = 0.15\,m $, angular velocity $ \omega_1 = 2.9\,rad/s $, pulley radius $ R = 0.56\,m $, pulley mass $ 25.6\,kg $, cylinder mass $ 24.6\,kg $, spring constant $ k = 120\,N/m $, drop distance $ d = 1.4\,m $.
- **Objective:**  
  Find the angular velocity of the pulley after the cylinder falls $ d = 1.4\,m $.
- **Explanation:**  
  Use conservation of energy, accounting for gravitational potential, spring potential, and rotational/translational kinetic energy, to relate the angular velocity before and after the cylinder falls.

---

**Step 2:**

**Energy Equation:**  
Initial energy (E₁):  
$$
E_1 = mgh_1 + \frac{1}{2}ks^2 + \frac{1}{2}I\omega_1^2
$$
Final energy (E₂):  
$$
E_2 = mg(h_1-d) + \frac{1}{2}k(s+d)^2 + \frac{1}{2}I\omega_2^2
$$
Where $ I = \frac{1}{2}MR^2 $, $ h_1 $ cancels, and $\omega_2 = v_2/R$ (with no slip).  
Set $ E_1 = E_2 $ and solve for $ \omega_2 $:

$$
mgd + \frac{1}{2}k[(s + d)^2 - s^2] + \frac{1}{2}I(\omega_2^2 - \omega_1^2) = 0
$$

Calculate each term:
- $ mgd = 24.6 \times 9.81 \times 1.4 = 338.1\,J $
- $ (s+d)^2 - s^2 = (0.15+1.4)^2 - (0.15)^2 = 1.55^2 - 0.0225 = 2.4025 - 0.0225 = 2.38 $
- $ \frac{1}{2}k \times 2.38 = 60 \times 2.38 = 142.8\,J $
- $ I = \frac{1}{2} \times 25.6 \times (0.56)^2 = 4.012\,kg\,m^2 $
- Initial rotational kinetic energy: $ \frac{1}{2}I\omega_1^2 = 0.5 \times 4.012 \times (2.9)^2 = 16.9\,J $

Now, combine:

$$
338.1 + 142.8 + 4.012 \frac{\omega_2^2 - (2.9)^2}{2} = 0
$$
$$
338.1 + 142.8 = \frac{4.012}{2} (\omega_2^2 - 8.41)
$$
$$
480.9 = 2.006 (\omega_2^2 - 8.41)
$$
$$
\frac{480.9}{2.006} = \omega_2^2 - 8.41
$$
$$
239.81 = \omega_2^2 - 8.41
$$
$$
\omega_2^2 = 239.81 + 8.41 = 248.22
$$
$$
\omega_2 = \sqrt{248.22} = 15.76\,rad/s
$$

**Explanation:**  
The energy lost from the spring and gain in height converts into rotational kinetic energy of the pulley, using the conservation of energy equation. All values are substituted stepwise for clarity.

---

**Final Answer:**  
The angular velocity of the pulley after the cylinder falls $ 1.4\,m $ is **15.76 rad/s**.

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