Anonymous
Anonymous asked in Science & MathematicsPhysics · 9 years ago

A uniform bar of length 50 cm is observed to balance when supported at the center of the bar.

when the support is moved 8 cm away from the center , it is found that a 150 gm mass hung 12 cm from the new pivot point will again balance the rod.

what is the mass of the rod? !

Relevance
• 9 years ago

Let M be the mass of the bar. We have, taking moments about the new pivot,

M*0.08 = 0.150*0.12 or M = (0.150*0.12)/0.08 = 0.225 kg or 225 gm

• 9 years ago

Edit:

X_cm = m_1(x)/M

Y_cm = m_2(y)/M

X_cm = Y_cm = ((m_1)(8.5cm))/M = ((150gm)(6cm))/M

M = 150gm + m_1

(25 + 8)cm = 33cm

L = 50cm - 33cm = 17cm

L/2 = 8.5cm = x

y = 12cm/2 = 6cm

m_1 = 105.8823529gm.

This is the mass of the rod.

Edit:

Note:

This more proof.

There are two torques acting in opposite direction but we would neglect the direction and focus on the magnitude.

Since the rod is balanced, the torques must be equal to each other.

τ = F_1*r_1 = F_2*r = m_1*g *r_1 = m_2*r_2*g

r_1 = 17cm

r_2 = 12cm

Solve for m_1

m_2 = 150gm

You will be obtain the same answer as

m_1 = 105.8823529gm

I hope this helps!

Source(s): Physics B brain.