notesonly.in

**Example: A ball of 0.2 kg is tied to a 1 m long string and swung in a vertical circle. Find the tension in the string at the bottom if the ball’s speed at the top is 2 m/s.** *Step 1: Find the speed at the bottom using conservation of energy.* - Energy at the top: kinetic + potential = ½ m v_top² + m g (2r) (top is 2 r above the bottom). - Energy at the bottom: kinetic only = ½ m v_bottom². - Set them equal: ½ m v_top² + 2m g r = ½ m v_bottom². - Solve for v_bottom: v_bottom² = v_top² + 4gr. - Plug numbers: v_bottom² = 2² + 4·9.8·1 = 4 + 39.2 = 43.2. - v_bottom = √43.2 ≈ 6.57 m/s. *Step 2: Find centripetal acceleration at the bottom.* - a_c = v_bottom² / r = 43.2 m/s². *Step 3: Find tension at the bottom.* - T = m(g + a_c) = 0.2 (9.8 + 43.2) = 0.2 · 53 = 10.6 N. **Answer:** The tension in the string at the bottom is about 10.6 N.