Consider a thin 22 m rod pivoted at one end. A uniform density spherical object (whose mass is 1 kg and radius is 4.8 m) is attached to the free end of the rod. The moment of inertia of the rod about an end is I-rod=1/3m L^2.The moment of inertia of the sphere about its center-of-mass is I-sphere=25m r^2. The rod is initially at rest at 24◦ below the horizontal. The acceleration of gravity g= 9.8 m/s^2. What is the angular acceleration of the rod immediately after it is released?Answer in units of rad/s^2.

Question

Consider a thin 22 m rod pivoted at one end. A uniform density spherical object (whose mass is 1 kg and radius is 4.8 m) is attached to the free end of the rod. The moment of inertia of the rod about an end is I-rod=1/3m L^2.The moment of inertia of the sphere about its center-of-mass is I-sphere=25m r^2. The rod is initially at rest at 24◦ below the horizontal. The acceleration of gravity g= 9.8 m/s^2. What is the angular acceleration of the rod immediately after it is released?Answer in units of rad/s^2.

Answer 1

To find the angular acceleration of the rod, we need to analyze the torque acting on it and apply Newton's second law for rotational motion.

First, let's consider the torque due to the gravitational force acting on the sphere. The center of mass of the sphere is located at a distance of 22 m from the pivot point. The magnitude of the torque (τ_sphere) can be calculated using the equation:

τ_sphere = I_sphere * α

where I_sphere is the moment of inertia of the sphere about its center-of-mass and α is the angular acceleration of the rod.

Substituting the given values, we have:

τ_sphere = 25m * r^2 * α ----(1)

Next, let's consider the torque due to the gravitational force acting on the rod. The center of mass of the rod is located at a distance of 11 m from the pivot point. The magnitude of the torque (τ_rod) can be calculated using the equation:

τ_rod = I_rod * α

where I_rod is the moment of inertia of the rod about an end and α is the angular acceleration of the rod.

Substituting the given values, we have:

τ_rod = (1/3)m * L^2 * α ----(2)

Now, let's equate the magnitudes of the two torques, as there is no external torque acting on the system:

τ_sphere = τ_rod

25m * r^2 * α = (1/3)m * L^2 * α

Simplifying the equation, we can cancel out the mass (m) and rearrange to solve for α:

25 * r^2 * α = (1/3) * L^2 * α

Canceling out α:

25 * r^2 = (1/3) * L^2

Substituting the given values for r and L:

25 * (4.8^2) = (1/3) * (22^2)

Simplifying the equation, we can solve for α:

α = 25 * (4.8^2) / ((1/3) * (22^2))

α ≈ 8.045 rad/s^2

Therefore, the angular acceleration of the rod immediately after it is released is approximately 8.045 rad/s^2.