GATE Mechanical Engineering (ME) 2019 Shift 2 Solved Paper

Question : 49 of 65

Marks: +1, -0

Two masses

A

and B having mass

m_{a}

and

m_{b}

, respectively, lying in the plane of the figure shown, are rigidly attached to a shaft which revolves about an axis through

{O}

perpendicular to the figure. The radii of rotation of the masses

m_{a}

and

m_{b}

are

r_{a}

and

r_{b}

, respectively. The angle between lines

{OA}

and

{OB}

90^{∘}

. If

{m}_{ {a}}=10 {kg}, {m}_{ {b}}=20 {kg}, {r}_{ {a}}=200 {mm}

and

{r}_{ {b}}=400 {mm}

, then the balance mass to be placed at a radius of

200 {mm}

is ______________ kg (round off to two decimal places).

Solution:

Let the balance mass be m

∑ F_{x}=m_{A} r_{A} \cos 0^{∘}+m_{B} r_{B} \cos 90^{∘}+m r \cos θ=0

......(1)

∑ F_{y}=m_{A} r_{A} \sin 0^{∘}+m_{B} r_{B} \sin 90^{∘}+m r \sin θ=0^{∘}=0

.....(2)

From (1),

m r \cos θ=-[m_{A} r_{A} \cos 0^{∘}+m_{B} r_{B} \cos 90^{∘}]

Or,

m r \cos θ=-[10 × 200]=-2000

From

(2), m r \sin θ=-[m_{A} r_{A} \sin 0+m_{B} r_{B} \sin 90^{∘}]

Or,

m r \sin θ=-[20 × 400]=-8000

Using

√{(m r \cos θ)^{2}+(m r \sin θ)^{2}}=m r=√{2000^{2}+8000^{2}}=8246.21

Now, since the radius of the balance mass is

200 {mm}

, therefore

m=8246.21 {/}200=41.23 {kg}

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