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Monica [59]
4 years ago
8

A simple pendulum on the surface of Earth is found to undergo 15.0 complete small‑amplitude oscillations in 8.75 s. Find the pen

dulum's length.
Physics
1 answer:
FinnZ [79.3K]4 years ago
6 0

Answer:

Length of the Pendulum is 14.4 cm

Explanation:

Time period of simple pendulum is given by

T=2\pi \sqrt{\frac{L}{g} } \\

time taken for one oscillation is

T=\frac{time taken}{oscillations} \\T=\frac{8.75}{15} \\T=0.583 s

Pendulum length on the earth surface is

L=\frac{T^2g}{4\pi^2 }\\L=\frac{0.583\times 9.8}{4\pi^2 } \\L=0.144 m

Length of the Pendulum is 14.4 cm

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Answer:

a) \omega = 50\,\frac{rad}{s}, b) \omega = 0\,\frac{rad}{s}

Explanation:

The magnitude of torque is a form of moment, that is, a product of force and lever arm (distance), and force is the product of mass and acceleration for rotating systems with constant mass. That is:

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Given that friction is of reactive nature. Frictional torque works on the unpowered flywheel until angular velocity is reduced to zero, whose instant is:

t = \frac{0\,\frac{rad}{s}-200\,\frac{rad}{s} }{\left(-3\,\frac{rad}{s^{2}} \right)}

t = 66.667\,s

Since t > 66.667\,s, then the angular velocity is equal to zero. Therefore:

\omega = 0\,\frac{rad}{s}

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