To tackle this problem, we need to analyze the system consisting of a uniform rod and a ball, both of which are rotating about the bottom end of the rod after being pushed. We'll break this down into three parts: calculating the rotational kinetic energy, determining the angular speed, and finding the linear speed of the ball's center of mass.
Rotational Kinetic Energy Calculation
The rotational kinetic energy (KE) of a rotating object can be calculated using the formula:
KE = (1/2) I ω²
where I is the moment of inertia and ω is the angular speed. To find the moment of inertia of the system, we need to consider both the rod and the ball.
Moment of Inertia of the Rod
The moment of inertia of a uniform rod rotating about one end is given by:
I_rod = (1/3) m l²
Substituting the values:
- m = 1.20 kg
- l = 0.24 m (converted from cm)
I_rod = (1/3) * 1.20 kg * (0.24 m)² = 0.02304 kg·m²
Moment of Inertia of the Ball
The moment of inertia of a solid sphere rotating about its diameter is:
I_ball = (2/5) M r²
However, since the ball is attached to the end of the rod, we need to use the parallel axis theorem to find its moment of inertia about the pivot point:
I_ball = (2/5) M r² + M d²
Here, r is the radius of the ball (0.04 m) and d is the distance from the pivot to the center of the ball (0.24 m):
- M = 2.00 kg
- r = 0.04 m
- d = 0.24 m
I_ball = (2/5) * 2.00 kg * (0.04 m)² + 2.00 kg * (0.24 m)²
I_ball = (2/5) * 2.00 * 0.0016 + 2.00 * 0.0576 = 0.00064 + 0.1152 = 0.11584 kg·m²
Total Moment of Inertia
The total moment of inertia of the system is:
I_total = I_rod + I_ball = 0.02304 kg·m² + 0.11584 kg·m² = 0.13888 kg·m²
Finding Angular Speed
Next, we need to find the angular speed (ω) after the rotation. When the rod and ball fall through 90°, the potential energy at the top is converted into rotational kinetic energy at the bottom.
The potential energy (PE) at the top is given by:
PE = mgh + Mgh
Where h is the height of the center of mass of each object from the pivot point:
- For the rod, the center of mass is at l/2 = 0.12 m.
- For the ball, the center of mass is at l = 0.24 m.
PE = (1.20 kg * 9.81 m/s² * 0.12 m) + (2.00 kg * 9.81 m/s² * 0.24 m)
PE = (1.20 * 9.81 * 0.12) + (2.00 * 9.81 * 0.24) = 1.176 + 4.7136 = 5.8896 J
Setting the potential energy equal to the rotational kinetic energy:
5.8896 J = (1/2) * I_total * ω²
Now, substituting the total moment of inertia:
5.8896 = (1/2) * 0.13888 * ω²
ω² = (5.8896 * 2) / 0.13888 = 84.66
ω = √84.66 ≈ 9.2 rad/s
Linear Speed of the Ball's Center of Mass
The linear speed (v) of the center of mass of the ball can be calculated using the relationship:
v = ω * r
Where r is the distance from the pivot to the center of the ball (0.24 m):
v = 9.2 rad/s * 0.24 m ≈ 2.21 m/s
Summary of Results
- Rotational Kinetic Energy: 5.89 J
- Angular Speed: 9.2 rad/s
- Linear Speed of the Ball's Center of Mass: 2.21 m/s
This analysis illustrates how energy conservation principles apply in rotational motion, allowing us to derive key quantities like kinetic energy and speeds effectively. If you have any further questions or need clarification on any step, feel free to ask!
