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3 years ago

Does anyone know how to find the answer key to this?

Physics

1 answer:

Mrrafil [7]3 years ago

8 0

It’s better to donintuourself it’s better to do it yourself and learn it

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Three beads are placed along a thin rod. The first bead, of mass m1 = 28 g, is placed a distance d1 = 1.5 cm from the left end o

Vladimir79 [104]

Answer:

Part a)

Center of mass with respect to the left end is given as

$r_{cm} = 5.63 cm$

Part b)

Center of mass with respect to middle bead is

$r_{cm} = \frac{m_1(-d_2) + m_2(0) + m_3(d_3)}{m_1 + m_2 + m_3}$

Part c)

Center of mass with respect to middle bead is

$r_{cm} = 1.63 cm$

Explanation:

Part a)

As we know that the center of mass of the system of mass is given by the formula

$r_{cm} = \frac{m_1r_1 + m_2r_2 + m_3r_3}{m_1 + m_2 + m_3}$

here we have

$m_1 = 28 g$

$m_2 = 11 g$

$m_3 = 45 g$

$r_1 = 1.5 cm$

$r_2 = 1.5 + 2.5 = 4 cm$

$r_3 = 1.5 + 2.5 + 4.6 = 8.6 cm$

Now we have

$r_{cm} = \frac{28(1.5) + 11(4) + 45(8.6)}{28 + 11 + 45}$

$r_{cm} = 5.63 cm$

Part b)

As we know that the center of mass of the system of mass is given by the formula

$r_{cm} = \frac{m_1r_1 + m_2r_2 + m_3r_3}{m_1 + m_2 + m_3}$

here we have

$m_1 = 28 g$

$m_2 = 11 g$

$m_3 = 45 g$

$r_1 = -d_2 = -2.5cm$

$r_2 = 0$

$r_3 = d_3 = 4.6 cm$

$r_{cm} = \frac{m_1(-d_2) + m_2(0) + m_3(d_3)}{m_1 + m_2 + m_3}$

Part c)

Now plug in the values in above formula

$r_{cm} = \frac{m_1(-d_2) + m_2(0) + m_3(d_3)}{m_1 + m_2 + m_3}$

$r_{cm} = \frac{28(-2.5) + m_2(0) + 45(4.6)}{28 + 11 + 45}$

$r_{cm} = 1.63 cm$

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