Question

The bob of a pendulum is held horizontally and released. What will be its velocity at the lowest position of its motion? The length of the pendulum is 1m and the mass of the bob is 20gm.​

Answer 1

Use the law of conservation of energy to solve this problem and then it becomes fairly easy:

Let’s name the initial position of the bob of the pendulum as ‘A’ and the position at the lowest point to be ‘B’. We take ‘B’ to be the position at which the gravitational potential energy of the bob is zero (we have the freedom to do so, we can take any distance from the ground for the gravitational potential energy to be zero).

According to this, at ‘A’, the net energy = KE + PE = 0 + mgh ————————(1)

(KE is zero because initially the bob is at rest)

then again, at ‘B’, net energy = KE + PE = (1/2)(m)(v^2) + 0 —————————(2)

(PE is zero because we took the position to be at zero gravitational potential energy) (v is the assumed velocity of the bob at ‘B’)

i.e. mgh = (1/2)(m)(v^2)

=> (2gh)^(1/2) = v

solve for v by substituting values of g=9.8 and h=5.

=> v=9.8 m/s approx.

Hope it helps you

Answer 2

amiii

ur intro plz.