An object is formed by attaching a uniform, thin rod with a mass of mr = 6.95 kg and length L = 4.96 m to a uniform sphere with mass ms = 34.75 kg and radius R = 1.24 m. Note ms = 5mr and L = 4R.
a. What is the moment of inertia of the object about an axis at the left end of the rod?
b. If the object is fixed at the left end of the rod, what is the angular acceleration if a force F = 492 N is exerted perpendicular to the rod at the center of the rod?
c. What is the moment of inertia of the object about an axis at the center of mass of the object? (Note: the center of mass can be calculated to be located at a point halfway between the center of the sphere and the left edge of the sphere.)
d. If the object is fixed at the center of mass, what is the angular acceleration if a force F = 492 N is exerted parallel to the rod at the end of rod?
e. What is the moment of inertia of the object about an axis at the right edge of the sphere?
f. Compare the three moments of inertia calculated above: (select one correct answer given below)
ICM < Ileft < Iright
ICM < Iright < Ileft
Iright < ICM < Ileft
ICM < Ileft = Iright
Iright = ICM < Ileft
To find the moment of inertia and angular acceleration for different scenarios in this problem, we'll need to use various formulas and principles of rotational motion. Let's go through each part of the question.
a. The moment of inertia (I) of the object about an axis at the left end of the rod can be calculated by considering the moment of inertia of the rod and the moment of inertia of the sphere. The moment of inertia of a thin rod rotating about one end is given by Irod = (1/3) * mr * L^2. The moment of inertia of a solid sphere rotating about its center is given by Isphere = (2/5) * ms * R^2. Since the rod is attached to the sphere and both rotate about the same axis, we can simply add these two moments of inertia to get the total moment of inertia (Ileft) about the left end of the rod: Ileft = Irod + Isphere.
b. To find the angular acceleration (α) when a force F is exerted perpendicular to the rod at its center, we can use the rotational analog of Newton's second law. The torque (τ) acting on the rod-sphere system is given by τ = Ileft * α, where α is the angular acceleration. The torque is also equal to the force (F) multiplied by the perpendicular distance between the point of application of the force and the axis of rotation (which is half the length of the rod in this case). So, we have τ = F * (L/2). Equating the two expressions for torque, we can solve for α: Ileft * α = F * (L/2), which gives α = (F * L) / (2 * Ileft).
c. The moment of inertia of the object about an axis at the center of mass can be calculated using the parallel axis theorem. The moment of inertia about the center of mass (ICM) can be found by subtracting the moment of inertia of the center of mass of the rod from the moment of inertia of the object about its end. Since the center of mass of a uniform rod is located at the midpoint, the moment of inertia of the rod about its center of mass is given by Icm-rod = (1/12) * mr * L^2. Using the parallel axis theorem, the moment of inertia of the object about the center of mass is given by ICM = Ileft - Icm-rod.
d. If the object is fixed at the center of mass and a force F is exerted parallel to the rod at the end of the rod, the torque acting on the system will be the force multiplied by the perpendicular distance between the axis of rotation and the point of application of the force. In this case, the perpendicular distance is L/2. So, the torque (τ) is given by τ = F * (L/2). Using the same formula τ = ICM * α as before, we can solve for α: ICM * α = F * (L/2), which gives α = (2 * F) / (L * ICM).
e. To find the moment of inertia of the object about an axis at the right edge of the sphere, we can use the parallel axis theorem again. The axis at the right edge of the sphere is at a distance of R from the center of mass. The moment of inertia about this axis (Iright) is given by Iright = ICM + ms * R^2.
f. Comparing the three moments of inertia calculated above (ICM, Ileft, Iright), the correct order is: ICM < Ileft = Iright. According to the parallel axis theorem, when considering the moment of inertia of an object about different axes, the moment of inertia about the center of mass will always be the smallest, and the moment of inertia about an axis further away from the center of mass will be larger.