Question

A pendulum is made by letting a 2.0-kg object swing at the end of a string that has a length of 1.5 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:

v = 2 m/s

Explanation:

Here we can use energy conservation to find the speed at the lowest point on its trajectory

As we know that by energy conservation

initial total gravitational potential energy = final total kinetic energy

now the height that is moved by the pendulum while it swing down is given as

$h = L(1 - cos30)$

$h = 1.5(1 - cos30) = 0.200 m$

now we can use energy conservation as

$mgh = \frac{1}{2}mv^2$

$v = \sqrt{2gh}$

$v = \sqrt{2(9.8)(0.200)}$

$v = 2 m/s$

Answer 2

Answer:

v = 1.978 m/s

Explanation:

Given that,

Mass of the object, m = 2 kg

Length of the string, l = 1.5 m

The maximum angle the string makes with the vertical as the pendulum swings is 30°, $\theta=30^{\circ}$

The pendulum have gravitational potential energy when the angle is maximum. The pendulum has only kinetic energy at its lowest point. Let v is the speed of the object at the lowest point in its trajectory. It can be calculated as :

$mgh=\dfrac{1}{2}mv^2$

h is the height moved by the pendulum.

$h=l(1-cos(30))$

$h=1.5(1-cos(30))$

h = 0.2 m

$v=\sqrt{2gh}$

$v=\sqrt{2\times 9.8\times 0.2}$    

v = 1.978 m/s

So, the speed of the object at the lowest point in its trajectory is 1.978 m/s.