A rigid mass-less rod of length L is connected to a disc (pulley) of mass m and radius r = L/4 through a friction-less revolute joint. The other end of that rod is attached to a wall through a friction-less hinge. A spring of stiffness 2k is attached to the rod at its mid-span. An inextensible rope passes over half the disc periphery and is securely tied to a spring of stiffness k at point C as shown in the figure. There is no slip between the rope and the pulley. The system is in static equilibrium in the configuration shown in the figure and the rope is always taut.Neglecting the influence of gravity, the natural frequency of the system for small amplitude vibration isa)b)c)d)Correct answer is option 'C'. Can you explain this answer?

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A rigid mass-less rod of length L is connected to a disc (pulley) of mass m and radius r = L/4 through a friction-less revolute joint. The other end of that rod is attached to a wall through a friction-less hinge. A spring of stiffness 2k is attached to the rod at its mid-span. An inextensible rope passes over half the disc periphery and is securely tied to a spring of stiffness k at point C as shown in the figure. There is no slip between the rope and the pulley. The system is in static equilibrium in the configuration shown in the figure and the rope is always taut.

Neglecting the influence of gravity, the natural frequency of the system for small amplitude vibration is

Correct answer is option 'C'. Can you explain this answer?

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A rigid mass-less rod of length L is connected to a disc (pulley) of m...

MI about point O:

MI about disc centre:
I_disc = mr²/2
When rod rotates by β,
β ⋅ r = l ⋅ θ
β = l ⋅ θ/r
(If disc is also rotating about its own centre due to static friction).
Energy Method:
E = (Rotational KE)_{system about} O + (Rotational KE)_{disc about its own centre}+ (PE of spring of 2K) + (PE of spring of K)
E =

E =

dE/dt = 0

= 0

=0
ω_n =

This is exact solution.
But this solution is very close to

Because,

Therefore,

Note: If we take an approximation in moment of inertia about hinge axis.
That,

(r = l/4)

If we neglect ml²/32 because ml²/32 <<<< ml²
If we take, I = ml²
Then we get,
ω_n =

But if we take exact inertia,

Then exact answer is ω_n=

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A rigid mass-less rod of length L is connected to a disc (pulley) of mass m and radius r = L/4 through a friction-less revolute joint. The other end of that rod is attached to a wall through a friction-less hinge. A spring of stiffness 2k is attached to the rod at its mid-span. An inextensible rope passes over half the disc periphery and is securely tied to a spring of stiffness k at point C as shown in the figure. There is no slip between the rope and the pulley. The system is in static equilibrium in the configuration shown in the figure and the rope is always taut.Neglecting the influence of gravity, the natural frequency of the system for small amplitude vibration isa)b)c)d)Correct answer is option 'C'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A rigid mass-less rod of length L is connected to a disc (pulley) of mass m and radius r = L/4 through a friction-less revolute joint. The other end of that rod is attached to a wall through a friction-less hinge. A spring of stiffness 2k is attached to the rod at its mid-span. An inextensible rope passes over half the disc periphery and is securely tied to a spring of stiffness k at point C as shown in the figure. There is no slip between the rope and the pulley. The system is in static equilibrium in the configuration shown in the figure and the rope is always taut.Neglecting the influence of gravity, the natural frequency of the system for small amplitude vibration isa)b)c)d)Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A rigid mass-less rod of length L is connected to a disc (pulley) of mass m and radius r = L/4 through a friction-less revolute joint. The other end of that rod is attached to a wall through a friction-less hinge. A spring of stiffness 2k is attached to the rod at its mid-span. An inextensible rope passes over half the disc periphery and is securely tied to a spring of stiffness k at point C as shown in the figure. There is no slip between the rope and the pulley. The system is in static equilibrium in the configuration shown in the figure and the rope is always taut.Neglecting the influence of gravity, the natural frequency of the system for small amplitude vibration isa)b)c)d)Correct answer is option 'C'. Can you explain this answer?.

Solutions for A rigid mass-less rod of length L is connected to a disc (pulley) of mass m and radius r = L/4 through a friction-less revolute joint. The other end of that rod is attached to a wall through a friction-less hinge. A spring of stiffness 2k is attached to the rod at its mid-span. An inextensible rope passes over half the disc periphery and is securely tied to a spring of stiffness k at point C as shown in the figure. There is no slip between the rope and the pulley. The system is in static equilibrium in the configuration shown in the figure and the rope is always taut.Neglecting the influence of gravity, the natural frequency of the system for small amplitude vibration isa)b)c)d)Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for GATE. Download more important topics, notes, lectures and mock test series for GATE Exam by signing up for free.

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