Subjects physics

Pendulum Tension Speed 57Db0B

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1. **Problem statement:** A small ball of mass 0.20 kg is attached to a string of length 0.80 m fixed at point S. The ball is released from rest when the string is horizontal. We analyze the situation when the string is vertical. 2. **Relevant formulas and principles:** - Conservation of mechanical energy: Total mechanical energy at the start equals total mechanical energy at the vertical position. - Gravitational potential energy (GPE) change converts into kinetic energy (KE). - Tension in the string provides centripetal force and balances weight component. 3. **Step (a)(i): Show speed of the ball is about 4 m/s when string is vertical.** - Initial height of the ball relative to vertical position is the length of the string: $h=0.80$ m. - Initial kinetic energy $KE_i=0$ (released from rest). - Initial potential energy $PE_i = mg h$. - At vertical position, height is zero, so $PE_f=0$. - By conservation of energy: $$PE_i + KE_i = PE_f + KE_f$$ $$mg h = \frac{1}{2} m v^2$$ - Cancel mass $m$: $$g h = \frac{1}{2} v^2$$ - Solve for $v$: $$v = \sqrt{2 g h}$$ - Substitute $g=9.8$ m/s$^2$, $h=0.80$ m: $$v = \sqrt{2 \times 9.8 \times 0.80} = \sqrt{15.68} \approx 3.96 \text{ m/s}$$ - Rounded to 2 significant figures, $v \approx 4.0$ m/s. 4. **Step (a)(ii): Calculate the tension in the string when vertical.** - Forces on the ball at vertical position: - Weight $W = mg = 0.20 \times 9.8 = 1.96$ N downward. - Tension $T$ upward along the string. - The ball moves in a circle of radius $r=0.80$ m with speed $v$. - Centripetal force required: $$F_c = \frac{m v^2}{r}$$ - Substitute values: $$F_c = \frac{0.20 \times (3.96)^2}{0.80} = \frac{0.20 \times 15.68}{0.80} = \frac{3.136}{0.80} = 3.92 \text{ N}$$ - Tension must provide centripetal force and balance weight: $$T - mg = F_c$$ $$T = mg + F_c = 1.96 + 3.92 = 5.88 \text{ N}$$ - Rounded to 2 significant figures, $T \approx 5.9$ N. **Final answers:** - Speed of the ball at vertical position: $\boxed{4.0 \text{ m/s}}$ - Tension in the string at vertical position: $\boxed{5.9 \text{ N}}$
0.80 mS