GATE Mechanical Engineering (ME) 2019 Shift 1 Solved Paper

Question : 62 of 65

Marks: +1, -0

A uniform thin disk of mass

1 {kg}

and radius

0.1 {m}

is kept on a surface as shown in the figure. A spring of stiffness

k_{1}=400 {N} {/} {m}

is connected to the disk center

{A}

and another spring of stiffness

{k}_{2}=100 {N} \/ {m}

is connected at point

{B}

just above point

{A}

on the circumference of the disk. Initially, both the springs are unstretched. Assume pure rolling of the disk. For small disturbance, from the equilibrium, the natural frequency of vibration of the system is _________ rad/s (round off to one decimal place).

Your Answer:

Validate

Solution:

Consider the point at which the circle touches the bottom as ‘O’

Using the Energy method

I_{o} {θ↖{..}}+(k_{1} R θ) R+(k_{2} 2 R θ) 2 R=0

I_{o}=\;{M R^{2}}/{2}+M R^{2}=\;{3}/{2} M R^{2}

Substituting this value

\;{3}/{2} M R^{2} {θ↖{..}}+(k_{1} R^{2}+4 k_{2} R^{2}) θ=0

Or,

\;{3}/{2} M{θ↖{..}+(k_{1}+4 k_{2}) θ=0

Substituting the values of

M, k_{1}

and

_{k 2}

in above equation

\;{3}/{2} {θ↖{..}+(400+4 × 100) θ=0

Or,

\;{3}/{2} {θ↖{..}+(800) θ=0

3 {θ↖{..}+(1600) θ=0

{θ↖{..}+\;{1600}/{3} θ=0

Comparing it with standard equation,

{θ↖{..}+ω_{n}^{2} θ=0

We get,

ω_{n}=√{\;{1600}/{3}}=23.09

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