PLEASE HELP URGENT!!!!!!!!!!!!!

Two uniform, solid cylinders of radius R and total mass M are connected along their common axis by a short, light rod and rest o

sveta [45]

Explanation:

A) To prove the motion of the center of mass of the cylinders is simple harmonic:

System diagram for given situation is shown in attached Fig. 1

We can prove the motion of the center of mass of the cylinders is simple harmonic if

$a_{x} = -\omega^{2} x$

where aₓ is acceleration when attached cylinders move in horizontal direction:

<h3>PROOF:</h3>

rotational inertia for cylinders is given as:

$I=\frac{1}{2}MR^{2}$ -----(1)

Newton's second law for angular motion is:

∑τ = Iα ------(2)

For linear motion in horizontal direction it is:

∑Fₓ = Maₓ ------ (3)

By definition of torque:

τ = RF --------(4)

Put (4) and (1) in (2)

$RF=\frac{1}{2}MR^{2}\alpha$

from Fig 3 it can be seen that fs is force by which the cylinders roll without slipping as they oscillate

So above equation becomes

$f_{s}=\frac{1}{2}MR\alpha$ ------ (5)

As angular acceleration is related to linear by:

$a= R\alpha$

Eq (5) becomes

$f_{s}=\frac{1}{2}Ma_{x}$ ---- (6)

aₓ shows displacement in horizontal direction

From (3)

∑Fₓ = Maₓ

Fₓ is sum of fs and restoring force that spring exerts:

$\sum F_{x} = f_{s} - kx$ ----(7)

Put (7) in (3)

$f_{s} - kx = Ma_{x}$ [/tex] -----(8)

Using (6) in (8)

$\frac{1}{2}Ma_{x} - kx =Ma_{x}$

$a_{x} = \frac{2k}{3M} x$ --- (9)

For spring mass system

$a= -\omega^{2} x$ ----- (10)

Equating (9) and (10)

$\omega^{2} = \frac{2k}{3M}$

$\omega = \sqrt{ \frac{2k}{3M}}$

then (9) becomes

$a_{x} = - \omega^{2}x$

(The minus sign says that x and aₓ have opposite directions as shown in fig 3)

This proves that the motion of the center of mass of the cylinders is simple harmonic.

<h3 /><h3>B) Time Period</h3>

Time period is related to angular frequency as:

$T=\frac{2\pi }{\omega}$

$T = 2\pi \sqrt{\frac{3M}{2k}$

5 0

3 years ago

QUESTION 23

Nutka1998 [239]

Answer:

E = 3600 J

Explanation:

Given that,

Voltage, V = 115 V

Power of electric bulb, P = 60 W

We need to find the electric energy used in 1 minute. The electric energy use is given by :

$E=P\times t\\\\E=60\ W\times 60\ s\\\\E = 3600\ J$

Hence, the electrical energy is 3600 J.

8 0

3 years ago

Question 8 of 10

Marrrta [24]

Answer:

C

Explanation:

If the theory were to be proved you you need to repeat the experiment over and over again so that way you can prove that it is true wuth the same results.

3 0

3 years ago

Read 2 more answers

Suppose that a person has an average heart rate of 72.0 beats/min. How many beats does he or she have in 2.0 y? In 2.00 y? In 2.

Alex17521 [72]

2.0y i think this is guess but if it right then thats good

3 0

3 years ago

What is the answer be fast

svetlana [45]

Answer:

Explanation:

Same numbers of protons but different number of neutron so i would go for A same atomic number different number of neutrons

5 0

3 years ago