A thin uniform rod of mass m=2.4 kg and length L=1.1 m can rotate on a frictionless pin that passes through one of its ends. The other end has a small ball of mass M=1.4 kg attached to it. The rod is released from rest while in the horizontal position. What will be the speed of the ball when it reaches its lowest position?
To determine the speed of the ball when it reaches its lowest position, we can use the principle of conservation of mechanical energy.
1. Let's denote the angular velocity of the rod when the ball reaches its lowest position as ω (omega).
Since the rod is released from rest, its initial angular velocity is 0.
2. At the lowest position, the potential energy of the system is at its minimum value, while the kinetic energy is at its maximum value.
The potential energy is given by the gravitational potential energy (mgh), where h is the vertical distance from the lowest position to the initial position of the center of mass of the ball-rod system.
3. Since the rod starts from the horizontal position and rotates freely, at the lowest position, the center of mass of the ball-rod system has descended a distance of L/2 (half the length of the rod).
4. Therefore, h = L/2.
5. The potential energy of the system at the highest position is equal to the kinetic energy at the lowest position, so we can equate them:
mgh = (1/2)Iω^2
6. The moment of inertia of the rod about its pivot point is I = (1/3)mL^2, using the parallel axis theorem.
7. Applying these values to the equation:
mgh = (1/2)(1/3)mL^2 ω^2
8. Simplifying,
gh = (1/6)L^2 ω^2
9. Since g is the acceleration due to gravity (approximately 9.8 m/s^2), we can substitute its value into the equation:
9.8(L/2) = (1/6)L^2 ω^2
10. Solving for ω^2:
4.9 = (1/6)Lω^2
11. Rearranging the equation,
30ω^2 = 1.1
12. Solving for ω^2:
ω^2 = 1.1/30
13. Taking the square root of both sides,
ω = √(1.1/30)
14. The speed of the ball when it reaches its lowest position is the tangential speed at the end of the rod due to this angular velocity.
v = ωL/2
15. Substituting the given values:
v = (√(1.1/30))(1.1)/2
16. Evaluating the expression, we find:
v ≈ 0.179 m/s
Therefore, the speed of the ball when it reaches its lowest position is approximately 0.179 m/s.