Question

As a pendulum swings the angle theta that it makes with the vertical changes through its swing. The force of gravity pulling on the bob is given by F=mg sin theta, where g is equal to 9.8 m/s^2. If the mass of the pendulum is 0.01kg, what is the force of pulling on the pendulum when it makes a 22.5 degree angle with the vertical?

a. sqrt2-sqrt2/2 N
b.0.049sqrt-sqrt2N
c.sqrt2-sqrt2/2 N
d.0.049sqrt2-sqrt2/2 N

Answer 1

b. 0.049sqrt 2-sqrt2 N

Answer 2

Explanation:

Alright, lets get started.

Equation is : F = mg sinΘ

We have given,

`m =0.01`

`g=9.8`

Θ
`= 25`

While plugging these above values in equation:

`F= 0.01 *9.8 * sin (22.5)`

`F = 0.098 * sin (22.5)`

Now, if we look unit circle, we don't have angle 22.5, so we will use half angle identity

`F = 0.098 * sin ((45)/(2) )`

`F = 0.098 * \sqrt{(1-cos45)/(2) }`

`F = 0.098 * \sqrt{(1-(√(2) )/(2) )/(2) }`

`F = 0.098 * \sqrt{(2-√(2))/(4) }`

`F = 0.049 * \sqrt{{2-√(2) }` : Answer

Hope it will help :)