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
[tex]a_{x} = -\omega^{2} x[/tex]
where aₓ is acceleration when attached cylinders move in horizontal direction:
rotational inertia for cylinders is given as:
[tex]I=\frac{1}{2}MR^{2}[/tex] -----(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)
[tex]RF=\frac{1}{2}MR^{2}\alpha[/tex]
[tex]RF=\frac{1}{2}MR^{2}\alpha[/tex]
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
[tex]f_{s}=\frac{1}{2}MR\alpha[/tex]------ (5)
As angular acceleration is related to linear by:
[tex]a= R\alpha[/tex]
Eq (5) becomes
[tex]f_{s}=\frac{1}{2}Ma_{x}[/tex]---- (6)
aₓ shows displacement in horizontal direction
From (3)
∑Fₓ = Maₓ
Fₓ is sum of fs and restoring force that spring exerts:
[tex]\sum F_{x} = f_{s} - kx [/tex] ----(7)
Put (7) in (3)
[tex]f_{s} - kx = Ma_{x}[/tex][/tex] -----(8)
Using (6) in (8)
[tex]\frac{1}{2}Ma_{x} - kx =Ma_{x}[/tex]
[tex]a_{x} = \frac{2k}{3M} x[/tex] --- (9)
For spring mass system
[tex]a= -\omega^{2} x[/tex] ----- (10)
Equating (9) and (10)
[tex]\omega^{2} = \frac{2k}{3M}[/tex]
[tex]\omega = \sqrt{ \frac{2k}{3M}}[/tex]
then (9) becomes
[tex]a_{x} = - \omega^{2}x[/tex]
(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.
Time period is related to angular frequency as:
[tex]T=\frac{2\pi }{\omega}[/tex]
[tex]T = 2\pi \sqrt{\frac{3M}{2k}[/tex]
It has been shown below that the motion of the center of mass of the two uniform, solid cylinders is simple harmonic in nature.
Since all springs would obey Hooke’s law, the force acting on this spring is given by this formula:
[tex]F=kx[/tex]
Where:
In this scenario, the force exerted on this spring acts at the center of the two uniform, solid cylinders of radius (R) and total mass (M) and generates a torque with respect to the point of contact. Also, they are prevented from slipping due to the friction generated when they come in contact.
For the moment of inertia:
[tex]I=\frac{MR^2}{2}[/tex]
For the linear acceleration:
[tex]\alpha =\frac{a}{R} \\\\a=\alpha R[/tex]
For the torque:
[tex]\sum \tau =I\alpha \\\\f_s R=I\alpha\\\\f_s R=\frac{MR^2}{2} \alpha \\\\f_s =\frac{MR\alpha}{2} \\\\f_s =\frac{Ma}{2}[/tex]
From Newton's Second Law of Motion, we have:
[tex]\sum F=ma\\\\f_s-kx=-Ma\\\\\frac{Ma}{2} -kx=-Ma\\\\kx=\frac{Ma}{2}+Ma\\\\kx=\frac{3Ma}{2}\\\\a=(\frac{2k}{3M} )x[/tex]
Therefore, the motion of the center of mass of the two uniform, solid cylinders is simple harmonic in nature.
Also, angular frequency is given by:
[tex]\omega^2=\frac{2k}{3M}\\\\\omega=\sqrt{ \frac{2k}{3M}}[/tex]
Now, we can determine the period of oscillation:
[tex]T=\frac{2\pi}{\omega} \\\\T=\frac{2\pi}{\sqrt{\frac{2k}{3M} } }\\\\T= 2\pi \times \sqrt{\frac{3M}{2k} }\\\\T= 2\pi \sqrt{\frac{3M}{2k} }[/tex]
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