- #1

Kennedy

- 70

- 2

## Homework Statement

A uniform rod (length = 2.4 m) of negligible mass has a 1.0-kg point mass attached to one end and a 2.0-kg point mass attached to the other end. The rod is mounted to rotate freely about a horizontal axis that is perpendicular to the rod and that passes through a point 1.0 m from the 2.0-kg mass. The rod is released from rest when it is horizontal. What is the angular velocity of the rod at the instant the 2-kg mass passes through its low point? (a) 1.7 rad/s (b) 2.2 rad/s (c) 2.0 rad/s (d) 1.5 rad/s (e) 3.1 rad/s

## Homework Equations

I'm thinking that using the cross product of the radius and the torque is relevant (rFsinθ), and probably the equivalent of the equations of linear motion for rotational dynamics. I would also assume that Newton's Second Law for rotational dynamics applies (τ = Iα).

## The Attempt at a Solution

First I calculated the force of gravity that would act on both point masses. So, 9.8 N for the 1.0-kg point mass, and 19.6 N for the 2.0-kg point mass. Since they're both point masses, I would assume that the torque acts at both ends of the rod. So, the torque on the 2.0-kg point mass would be 19.6 N(1 m) = 19.6 N*m, and for the 1.0-kg point mass would be 9.8 N(1.4 m) = 13.72 N*m. Because the mass of the rod is negligible, the rotational inertia is completely due to the point masses. So, I = 1(1.4^2) +2(1^2) = 3.96 kg*m^2. The net torque is 19.6 - 13.72 = 5.88 N*m. This yields the angular acceleration to be 5.88/3.96 = 1.48 rad/s^2.

Since the rod starts out horizontally, and it rotates in the direction of the 2.0-kg point mass, when the system rotates an angle of θ = pi/2, the 2.0-kg point mass is at its low point. By using the equivalent of the equations of linear motion for rotational dynamics, I solved for the final angular velocity of the 2.0-kg point mass at its low point. This gave me 2.15 rad/s, which is not the correct answer.

I was shown how to do the problem using energy conservation, but shouldn't I be able to solve the problem using this method. What's wrong with my approach?

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