Exercises: Rotational Inertia
A small ball of mass $0.50$ kg is attached to the end of a light rod of negligible mass, and the rod is $0.80$ m long. The system rotates about the other end of the rod.…
A small ball of mass $0.50$ kg is attached to the end of a light rod of negligible mass, and the rod is $0.80$ m long. The system rotates about the other end of the rod. Treating the ball as a point mass, calculate the rotational inertia of the system about that axis.
Define rotational inertia in your own words. Then state which two properties of an object determine its rotational inertia about a given axis, and give the SI unit of…
Define rotational inertia in your own words. Then state which two properties of an object determine its rotational inertia about a given axis, and give the SI unit of rotational inertia.
Two point masses, $2.0$ kg and $3.0$ kg, are fixed on a light rod at distances $0.40$ m and $0.20$ m from the axis of rotation, respectively. They lie on opposite sides…
Two point masses, $2.0$ kg and $3.0$ kg, are fixed on a light rod at distances $0.40$ m and $0.20$ m from the axis of rotation, respectively. They lie on opposite sides of the axis. Calculate the total rotational inertia of the two masses about the axis.
Four identical point masses of $0.25$ kg each sit at the corners of a square of side $0.60$ m, held by a frame of negligible mass. Find the rotational inertia of the…
Four identical point masses of $0.25$ kg each sit at the corners of a square of side $0.60$ m, held by a frame of negligible mass. Find the rotational inertia of the system about an axis perpendicular to the square and passing through its center. (Hint on geometry only: the distance from the center to a corner is half the diagonal.)
A uniform thin rod of mass $M = 1.2$ kg and length $L = 1.0$ m has rotational inertia $I = \tfrac{1}{12}ML^2$ about an axis through its center, perpendicular to the rod.…
A uniform thin rod of mass $M = 1.2$ kg and length $L = 1.0$ m has rotational inertia $I = \tfrac{1}{12}ML^2$ about an axis through its center, perpendicular to the rod. Use the parallel-axis theorem to find the rotational inertia about a parallel axis through one end of the rod. Then state how many times larger it is than the center value.
Sofia has two disks of the same mass of $2.0$ kg and the same radius of $0.30$ m. Disk A is a uniform solid disk with $I = \tfrac{1}{2}MR^2$. Disk B has all its mass…
Sofia has two disks of the same mass of $2.0$ kg and the same radius of $0.30$ m. Disk A is a uniform solid disk with $I = \tfrac{1}{2}MR^2$. Disk B has all its mass concentrated in a thin ring at the rim, with $I = MR^2$. (a) Calculate $I$ for each disk about its central axis. (b) Explain which one is harder to start spinning and why, in terms of mass distribution.
A student builds a dumbbell from two $1.0$ kg point-like masses at the ends of a light rod of length $0.60$ m. She considers two axes, both perpendicular to the rod:…
A student builds a dumbbell from two $1.0$ kg point-like masses at the ends of a light rod of length $0.60$ m. She considers two axes, both perpendicular to the rod: axis 1 passes through the center of the rod, and axis 2 passes through one of the masses. (a) Calculate the rotational inertia about each axis. (b) Using the same applied torque of $0.50$ N$\cdot$m, find the angular acceleration for each axis. (c) Explain, using mass distribution, why the two results differ, and verify that your axis 2 value is consistent with the parallel-axis theorem.
A figure skater spins with her arms extended. Model her as a central cylinder of mass $50$ kg and radius $0.20$ m with $I = \tfrac{1}{2}MR^2$, plus two arms, each a…
A figure skater spins with her arms extended. Model her as a central cylinder of mass $50$ kg and radius $0.20$ m with $I = \tfrac{1}{2}MR^2$, plus two arms, each a point mass of $4.0$ kg located $0.80$ m from the axis. She then pulls her arms in so that each arm is $0.20$ m from the axis. (a) Calculate her total rotational inertia in each configuration. (b) Find the ratio of the two values. (c) Explain, without using angular momentum equations, why it is easier for her to produce a large angular acceleration with arms pulled in when a given torque acts.