Consider a thin 34 m rod pivoted at one end. A uniform density spherical object (whose mass is 5 kg and radius is 3.6 m) is attached to the free end of the rod. The moment of inertia of the rod about an end is I rod =1/3 mL^2 and the moment of inertia of the sphere about its center of mass is I sphere =(2/5) mr^2.
What is the angular acceleration of the rod immediately after it is released from its initial position of 61◦ from the vertical? The acceleration of gravity g = 9.8 m/s2.
Answer in units of rad/s2
To find the angular acceleration of the rod, we first need to calculate the net torque acting on the system.
The torque acting on the rod is given by the equation:
τ_rod = I_rod * α_rod
where τ_rod represents the torque, I_rod is the moment of inertia of the rod, and α_rod is the angular acceleration of the rod.
Since the rod is released from its initial position, the torque acting on it is due to the gravitational force acting on the attached sphere. The gravitational force exerts a torque that tends to rotate the rod about the pivot point.
The torque due to the gravitational force acting on the sphere is given by:
τ_sphere = m * g * d
where m is the mass of the sphere, g is the acceleration due to gravity, and d is the perpendicular distance from the pivot point to the center of mass of the sphere.
In this case, the distance d is equal to the length of the rod, L. Thus, τ_sphere = m * g * L.
Since the torque acting on the rod is caused by the torque from the sphere, we can equate the two torques and solve for the angular acceleration of the rod:
I_rod * α_rod = m * g * L
Substituting the given values:
(1/3) * m * L^2 * α_rod = m * g * L
Simplifying:
(1/3) * L * α_rod = g
α_rod = 3g/L
Now, we can substitute the given values into the formula to calculate the angular acceleration:
α_rod = 3 * 9.8 m/s^2 / 34 m
α_rod ≈ 0.862 rad/s^2
Therefore, the angular acceleration of the rod immediately after it is released is approximately 0.862 rad/s^2.