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Standard XII

Physics

Centripetal Acceleration

Two masses m ...

Question

Two masses m are attached to opposite sides of a rigid rotating shaft in the vertical plane. Another pair of equal masses $m_{1}$ is attached to the opposite sides of the shaft in the vertical plane as shown in figgure. Consider m = 1 kgm e = 50 mm, $e_{1}$ =20 mm, b=0.3 m, a = 2 m and $a_{1}$ =2.5 m.
For the system to be dynamically balanced. $m_{1}$ should be kg

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Solution

Method I:

Balance mement of all forces
$ω^{2} (m_{1} \times 0.02 \times 2.5) + (1 \times 0.05 \times 0.3) ω^{2} = (m_{1} \times 0.02 \times 0) ω^{2} + (1 \times 0.05 \times 2.3) ω^{2}$
$m_{1} = 2 k g$

Method II:

Let referance PQ, passes through $m_{1}$ mass on left side,
Taking moment about PQ.
$m_{1} e_{1} \times 0 \times c o s 180^{o} + m e b c o s)^{o} + (a + b) c o s 180^{o} + m_{1} e_{1} a_{1} c o s 0^{o} = 0$
$0 + m e b - m e (a + b) + m_{1} e_{1} a_{1} = 0$
$m_{1} e_{1} a_{1} = m e a$
$m_{1} = \frac{m e a}{e_{1} a_{1}} = \frac{1 \times 50 \times 2}{20 \times 2.5} = 2$ kg

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