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? !
- 9 years agoFavorite 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
- 9 years ago
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.
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.
- welcome newsLv 69 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)