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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?

Asqrt2-sqrt2/2 N

b.0.049sqrt-sqrt2N

c.sqrt2-sqrt2/2 N

d.0.049sqrt2-sqrt2/2 N

Verified Answer :

B0.049sqrt 2-sqrt2 N

Answer:

Step-by-step 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

\(F = 0.098 * sin (\frac{45}{2} )\)

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

\(F = 0.098 * \sqrt{\frac{1-\frac{\sqrt{2} }{2} }{2} }\)

\(F = 0.098 * \sqrt{\frac{2-\sqrt{2}}{4} }\)

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

Hope it will help :)

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