A pendulum is suspended by a string of length 250 cm. The mass of the bob of the pendulum is 200 g. The bob is pulled aside until the string is at 60° with vertical as shown in the figure. After releasing the bob, the maximum velocity attained by the bob will be ____________ ms⁻¹. (if g = 10 m/s²)

Solution

Physics sir · 4 steps

At 60° this is not small-angle SHM, so use energy: the height the bob falls sets its top speed.

1. Use energy. The tension is always perpendicular to the motion, so it does no work. Only gravity does work, and the bob is fastest at the lowest point. The mass of 200 g cancels and is not needed.
2. Height dropped. With l = 250 cm = 2.5 m, the bob starts at a height h = l(1 - cos 60°) = 2.5 × 0.5 = 1.25 m above the lowest point.

   1/2mvₘₐₓ² = mgl(1-cos θ) ⇒ vₘₐₓ = √2gl(1-cos θ)

3. Put in numbers. vₘₐₓ = √(2 × 10 × 1.25) = √25 = 5 m s⁻¹.
4. So. The maximum velocity of the bob is 5 m s⁻¹.

The common slip: 50. Leaving the length as 250 cm gives h = 125 and v = √2500 = 50. Convert l to 2.5 m before using g = 10 m s⁻².

The idea to keep. For a pendulum released at any angle, vₘₐₓ = √2gl(1-cos θ). Use Aω only for small angles, and always take l in metres.

Answer from the NTA final answer key. Solution by PrepWiser.