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victus00 [196]
3 years ago
10

How can I show that the sphere of radius R performs a simple harmonic movement. how can i set its reference point and make the f

ree body diagram.
I have the torque sum equation which is equal to the moment of inertia by angular acceleration

Physics
1 answer:
Marysya12 [62]3 years ago
5 0

Explanation:

Draw a free body diagram of the pendulum (the combination of the sphere and the massless rod).  There are three forces on the pendulum:

Weight force mg at the center of the sphere,

Reaction force in the x direction at the pivot,

Reaction force in the y direction at the pivot.

Sum the torques about the pivot O.

∑τ = I d²θ/dt²

mg (L sin θ) = I d²θ/dt²

For small θ, sin θ ≈ θ.

mg L θ = I d²θ/dt²

Since d²θ/dt² is directly proportional to θ, this fits the definition of simple harmonic motion.

If you wish, you can use parallel axis theorem to find the moment of inertia about O:

I = Icm + md²

I = ⅖ mr² + mL²

mg L θ = (⅖ mr² + mL²) d²θ/dt²

gL θ = (⅖ r² + L²) d²θ/dt²

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Why is an absorption spectrum especially useful for astronomers?.
creativ13 [48]

Answer:

It has dark lines in it that allow astronomers to determine what elements are in the star. red-shifted (shifted toward the red end of the light spectrum).

Explanation:

4 0
2 years ago
On earth, you swing a simple pendulum in simple harmonic motion with a period of 1.6 seconds. what is the period of this same pe
polet [3.4K]
The period of a simple pendulum is given by
T= 2 \pi  \sqrt{ \frac{L}{g} }
where
L is the pendulum length
g is the acceleration of gravity

If we move the same pendulum from Earth to the Moon, its length L remains the same, while the acceleration of gravity g changes. So we can write the period of the pendulum on Earth as:
T_e= 2 \pi \sqrt{ \frac{L}{g_e} }
where g_e is the acceleration of gravity on Earth, while the period of the pendulum on the Moon is
T_m= 2 \pi \sqrt{ \frac{L}{g_m} }
where g_m is the acceleration of gravity on the Moon. 

If we do the ratio of the two periods, we get
\frac{T_m}{T_e} =  \sqrt{ \frac{g_e}{g_m} }
but the gravity acceleration on the Moon is 1/6 of the gravity acceleration on Earth, so we can write g_e = 6 g_m and we can rewrite the previous ratio as
\frac{T_m}{T_e} = \sqrt{ \frac{6 g_m}{g_m} }=  \sqrt{6}

so the period of the pendulum on the Moon is
T_m =  \sqrt{6}  T_e =  \sqrt{6} (1.6 s)=3.9 s
8 0
4 years ago
If the radius of an electron's orbit around a nucleus doubles but the wavelength remains unchanged, what happens to the number o
inessss [21]

Answer:

Number of electron wavelength will Double

Explanation:

let the radius = r  and wavelength = λ

when R doubles and  λ ( wavelength ) remains the same

The number of electron electron wavelengths will double as well

Using Bohr's angular momentum quantization to show this

attached below

3 0
3 years ago
The total amount of power that a star radiates is called its
Ede4ka [16]
Luminosity is the total amount of power a star radiate I think :)
3 0
3 years ago
Dolphins emit clicks of sound for communication and echolocation. A marine biologist is monitoring a dolphin swimming in seawate
LenKa [72]

Answer:

12.64968 Hz

Explanation:

v = Velocity of sound in seawater = 1522 m/s

u = Velocity of dolphin = 7.2 m/s

f' = Actual frequency = 2674 Hz

From Doppler effect we get the relation

f=f'\frac{v-u}{v}\\\Rightarrow f=2674\frac{1522-7.2}{1522}\\\Rightarrow f=2661.35032\ Hz

The frequency that will be received is 2661.35032 Hz

The difference in the frequency will be

2674-2661.35032=12.64968\ Hz

6 0
4 years ago
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