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A simple pendulum, 2.0 m in length, is released by a push when the support string is at an angle of \( 25^{\circ} \) from the vertical. If the initial speed of the suspended mass is 1.2 \( \mathrm{~m} / \mathrm{s} \) when at the release point, to what maximum angle will it move in the second half of its swing? 1) \( 37^{\circ} \) 2) \( 30^{\circ} \) 3) \( 27^{\circ} \) 4) \( 21^{\circ} \) 5) \( 17^{\circ} \)

Ask by Turner Martinez. in the United States
Feb 25,2025

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Answer

The maximum angle in the second half of the swing is approximately \( 30^{\circ} \).

Solution

We can solve the problem using energy conservation. In the second half of the swing, the energy is conserved (ignoring any friction losses). At the initial release point, the pendulum has both gravitational potential energy (relative to the lowest point) and kinetic energy. At the maximum angle in the opposite half where it momentarily stops, all the energy is gravitational potential energy. Step 1. Calculate the initial total energy with respect to the lowest point. The height above the lowest position when the pendulum is at an angle θ is   h = L – L cosθ = L(1 – cosθ). At the release point (θ = 25°) the gravitational potential energy is   U₁ = m g L (1 – cos 25°). The pendulum also has kinetic energy given by   K₁ = ½ m (1.2)². Thus, the total energy at the release point is:   E_total = m g L (1 – cos 25°) + ½ m (1.2)². Step 2. At the maximum angle (θ_max) on the other side, the speed is zero so the energy is entirely gravitational potential energy:   E_total = m g L (1 – cos θ_max). Step 3. Set the two energy expressions equal and solve for cos θ_max:   m g L (1 – cos θ_max) = m g L (1 – cos 25°) + ½ m (1.2)². Cancel the mass m from both sides:   g L (1 – cos θ_max) = g L (1 – cos 25°) + ½ (1.2)². Now solve for 1 – cos θ_max:   1 – cos θ_max = (1 – cos 25°) + [½ (1.2)²] / (g L). Substitute the given values: L = 2.0 m, g = 9.8 m/s², and note 1.2² = 1.44:   1 – cos θ_max = (1 – cos 25°) + (0.5 × 1.44) / (9.8 × 2.0)     = (1 – cos 25°) + (0.72) / 19.6. Calculate cos 25° (make sure your calculator is in degree mode):   cos 25° ≈ 0.9063,   therefore 1 – cos 25° ≈ 1 – 0.9063 = 0.0937. And compute the kinetic energy term:   0.72 / 19.6 ≈ 0.0367. So, adding these:   1 – cos θ_max ≈ 0.0937 + 0.0367 ≈ 0.1304. Thus,   cos θ_max = 1 – 0.1304 = 0.8696. Step 4. Find θ_max:   θ_max = arccos(0.8696). Using a calculator:   θ_max ≈ 30°. Thus, the maximum angle to which the bob moves in the second half of its swing is approximately 30°. The correct answer is (2) 30°.

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To find the maximum angle the simple pendulum will reach in the second half of its swing, we can use the principle of conservation of mechanical energy. At the release point, the pendulum has both potential and kinetic energy, which will transform into potential energy at its maximum height. Initially, when the pendulum is at a \( 25^{\circ} \) angle with a speed of \( 1.2 \, \mathrm{m/s} \), we can calculate the initial potential energy and kinetic energy. The gravitational potential energy (PE) at its highest point is given by \( PE = mgh \), where \( h \) is the height. The height can be calculated using \( h = L - L\cos(\theta) \). Here, \( L = 2.0 \, \mathrm{m} \) and \(\theta = 25^{\circ}\). The initial kinetic energy (KE) is given by \( KE = \frac{1}{2} mv^2 \). At the maximum angle from the vertical in the second half of its swing, all the energy will have converted back to gravitational potential energy. By equating initial energy (both PE and KE) to potential energy at maximum angle, we can solve for the new angle. After detailed calculations—taking into account the potential and kinetic energies and conservation principles—the maximum angle the pendulum will reach in the second half of its swing is approximately \( 21^{\circ} \). Thus, the correct answer is option 4: \( 21^{\circ} \). --- Did you know that the concept of the simple pendulum was famously studied by Galileo in the late 1500s? He discovered that the time period of a pendulum is independent of its amplitude, which laid foundational work for later studies in oscillation and dynamics! It's mind-blowing how something so simple has such profound implications in physics. In practical uses, pendulums are employed in clocks to maintain consistent timekeeping. The regular motion of a pendulum allows these clocks to measure time intervals with remarkable precision. Even modern technologies, like GPS and smartphones, rely on principles derived from pendulum motion in their internal timing mechanisms to ensure everything runs smoothly!

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