A pendulum bob has a speed of 3 m s$$^{-1}$$ at its lowest position. The pendulum is 50 cm long. The speed of bob, when the length makes an angle of 60$$^\circ$$ to the vertical will be ___ m s$$^{-1}$$. ($$g = 10$$ m s$$^{-2}$$)

Let the mass of the pendulum bob be $$m$$ and the length of the string be $$L = 0.50\ \text{m}$$. At its lowest position the bob has speed $$v_0 = 3\ \text{m s}^{-1}$$. We take the gravitational potential energy to be zero at this lowest point.

The principle we use is the conservation of mechanical energy, which states:

$$\text{Total mechanical energy at any position} = \text{constant.}$$

Hence,

$$\tfrac12 m v_0^{\,2} + 0 = \tfrac12 m v^{\,2} + m g h,$$

where

• $$v$$ is the speed of the bob when the string makes an angle $$\theta = 60^\circ$$ with the vertical,
• $$h$$ is the vertical height gained by the bob relative to the lowest point.

The height $$h$$ is obtained from simple geometry of the circle traced by the pendulum. When the string of length $$L$$ makes an angle $$\theta$$ with the vertical, the bob rises from the lowest point by

$$h = L - L\cos\theta = L(1 - \cos\theta).$$

Substituting $$L = 0.50\ \text{m}$$ and $$\theta = 60^\circ$$, and recalling that $$\cos 60^\circ = \tfrac12$$, we get

$$h = 0.50(1 - \tfrac12) = 0.50 \times 0.50 = 0.25\ \text{m}.$$

Now we substitute $$h$$ into the energy conservation equation:

$$\tfrac12 m (3)^{2} = \tfrac12 m v^{\,2} + m(10)(0.25).$$

Simplifying term by term, first compute the left side kinetic energy:

$$\tfrac12 m (3)^{2} = \tfrac12 m \times 9 = 4.5 m.$$

The potential energy term is

$$m g h = m \times 10 \times 0.25 = 2.5 m.$$

So the equation becomes

$$4.5 m = \tfrac12 m v^{\,2} + 2.5 m.$$

Subtract $$2.5 m$$ from both sides:

$$4.5 m - 2.5 m = \tfrac12 m v^{\,2}.$$

That gives

$$2.0 m = \tfrac12 m v^{\,2}.$$

Divide both sides by $$\tfrac12 m$$ (which is equivalent to multiplying by 2):

$$v^{\,2} = 4.$$

Finally, take the positive square root (speed is positive):

$$v = 2\ \text{m s}^{-1}.$$

So, the answer is $$2\ \text{m s}^{-1}$$.