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.
- civil_av8rLv 75 years agoFavorite Answer
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:
alpha = -5.13 rad/s^2
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.
- oldschoolLv 75 years ago
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