# Physics Question- 2 Masses on Balance?

Two blocks each of mass 1.7 kg are suspended from the ends of a rigid weightless rod of length l1 = 176.0 cm and l2 = 13.0 cm. The rod is held in the horizontal position shown and then released. Calculate the initial acceleration of the mass attached to l1. Assume up is positive and down is negative. Relevance

I think your lenghts are backwards. From the picture l1 < l2, but in the problem it says l1 > l2. I will go by the picture.

Sum of the moments about the pivot point = I*alpha

Find the moment of inertia:

Treat the blocks as point masses.

I_point mass = m*r^2

I_LHS = m*l1^2

I_RHS = m*l2^2

I_total = I_RHS + I_LHS = m*l1^2 + m*l2^2 = m*(l1^2 + l2^2)

What are the moments? Moments that produce CCW spin will be treated as positive. The mass on the left hand side produces a moment of m*g*l1, which makes the system want to spin CCW (positive). The mass on the right hand side produces a moment of m*g*l2, hich makes the system want to spin CW (negative).

Sum of the moments (CCW = +) = m*g*l1 - m*g*l2 = I_total*alpha

m*g*l1 - m*g*l2 = I_total*alpha

alpha = (m*g*l1 - m*g*l2) / I_total = m*g*(l1 - l2) / (m*(l1^2+l2^2)) = g (l1-l2) / (l1+l2)^2

Plug in numbers, remember to convert cm to m:

The negative means that the system is going to rotate in the clockwise direction. This means that the LHS is rising.

The tangential/agular acceleration relatonship

a = alpha*r

The radius in this case is l1

a = alpha*l1

Sub in alpha, but since we know it is going up, treat alpha as a positive number so our acceleration will be positive (up).

a = 5.13 rad/s^2 * 0.13 m = 0.667 m/s^2 <--- ANS

Hope I helped explain it.

• RickB
Lv 7
5 years agoReport

I am skeptical. Clearly the moments are caused by the tensions in the string, T1 and T2; that means the CCW moment is (T1)(l1) and the CW moment is (T2)(l2). But these tensions do NOT equal mg; otherwise the masses would not be accelerating! Not sure what the resolution is here.

• The moment of inertia of this rod is 1.7*0.13²+1.7*1.76² = 5.30

The net torque = 1.7g(1.76-0.13) = 27.16

We know that Net Torque = I *α => α = T/I = 27.2/5.3 = 5.13 rad/s²

a = r*α = 1.76*5.13 = 9.03m/s²

Notice that if you use 0.13m instead of 1.76m a = 0.667m/s²

which makes sense, the further from the point of rotation, the greater the tangential