Anonymous
Anonymous asked in Science & MathematicsPhysics · 9 years ago

Torque question please help..!!?

Torque question please help ..!!?

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? !

3 Answers

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  • 9 years ago
    Favorite Answer

    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

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  • 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.
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  • 9 years ago

    The torque exercised by the bar = mX8

    The torque of the 150gm mass = 150X12

    Therefore mX8 = 150X12 so m = 225g

    (NB I have simplified working by not converting to Kg and by not using g to work out the force. Both these steps cancel out)

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