GATE Mechanical Engineering (ME) 2018 Shift 2 Solved Paper

Question : 40 of 65

Marks: +1, -0

A force of

100 {N}

is applied to the centre of a circular disc, of mass

10 {kg}

and radius

1 {m}

, resting on a floor as shown in the figure. If the disc rolls without slipping on the floor, the linear acceleration (in

m{/}{s}^{2}

) of the centre of the disc is (correct to two decimal places).

Solution:

T=I α

{T}=100 × 1=100 {Nm}

At point of contact, i.e. I-centre of rotation,

I=\;{m R^{2}}/{2}+m R^{2}

I=15

Hence,

α=100 {/}15=6.667 {rad} {/} {sec}^{2}

Acceleration at centre of

disc= {R} α=6.667 {m} {/} {sec}^{2}

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