Question

4. A pendulum is made by letting a 2.0-kg object swing at the end of a string that has a length of 2.5 m. The maximum angle the string makes with the vertical as the pendulum swings is 30°. What is the speed of the object at the lowest point in its trajectory?

Answer 1

Answer 2

SOLUTION :

Hheight of pendulum at the extreme swing point with respect to lowest point

= L - L cos(30)

= L(1 - cos (30))

= 2.5*0.133975

= 0.3349  (m)

So, loss of height, ∆h,  at lowest point with respect to extreme point = 0.3349 (m)

So, loss of potential energy = mg ∆h = 2*9.8*0.3349 = 6.56 J

Let v m/s be the velocity at the lowest point. 

At extreme point velocity was 0 m/s.

So, K.E gained = 1/2 m v^2 - 0  = 1/2 * 2 * v^2 - 0 = v^2  (J)

As per energy conservation principle :

Gain. In K.E. = Loss of P.E.

=> v^2 = 6.56 

=> v = Velocity at the lowest point = sqrt(6.56) = 2.56 m/s (ANSWER)