Results given on page 300: TABLE 12.2 Moments of inertia of objects with uniform density Object...
Results given on page 300 TABLE 12.2 Moments of inertia of objects with uniform density Object and axis Picture Object and axis Picture Thin rod, about center | Cylinder or disk, about center MR ML2 Thin rod, about end ML Cylindrical hoop. MR2 about center | Solid sphere, about diameter Маг Plane or slab, about center Ma2 MR Plane or slab, about edge Ma2 Spherical shell, about diameter MR2 1. b. A very thin, straight, uniform rod has a length...
Results given on page 300 TABLE 12.2 Moments of inertia of objects with uniform density Object and axis Picture Object and axis Picture Thin rod, about center MCylinder or disk, MR 2 about center Thin rod about end ML Cylindrical hoop, MR2 about center Plane or slab, about center Маг | Solid sphere, about RMR2 diameter Plane or slab, about edge 1Ma2 I spherical shell, about diameter MR2 5. Again, use the table of integration results on page 300 of...
Results given on page 300: TABLE 12.2 Moments of inertia of objects with uniform density Object and axis Picture Object and axis Picture Thin rod about center Cylinder or disk, ML MR2 about center Thin rod about end ML Cylindrical hoop, MR2 about center Plane or slab, about center Solid sphere, about diameter 3MR2 Plane or slab, about edge Ma Spherical shell, about diameter MR2 4. Use the results on page 300 of the textbook to do the following: A...
2. A very thin, flat, uniform slab has a width (W) of 2.00 m, a height (H) of 30.0 cm, and a total mass of 16.0 kg. Treating the slab as essentially a sheet of mass- distributed uniformly over its area- do the following (i) Use integration to prove that the slab's center of mass is located at its center point. a. (Reminders: dm - ndA dA can be written here as either Hdx or Wdy What is the value...
2. A very thin, flat, uniform slab has a width (W) of 2.00 m, a height (H) of 30.0 cm, and a total mass of 16.0 kg. Treating the slab as essentially a sheet of mass- distributed uniformly over its area- do the following (i) Use integration to prove that the slab's center of mass is located at its center point. a. (Reminders: dm - ndA dA can be written here as either Hdx or Wdy What is the value...
5*) Find the angular velocity of the Earth due to its daily
rotation and express it in radians per second. Then use it, and a
model of the Earth as a solid sphere of mass M=
5.97 × 1024 kg and radius R
= 6.37 × 106 m, to estimate the angular momentum of the Earth due
to its rotation around its axis. (The result should be of the order
of 1033 kg m2/s. This is called the Earth’s “intrinsic”...
The Parallel-Axis Theorem allows one to find the moment of inertia of an object if the moment of inertia through the center of mass (c.o.m.) is known and the second axis is parallel to the axis through the c.o.m.. The equation is given by I= Icom +md2, where Icom is the moment of inertia about an axis through the c.o.m., m is the mass of the object, d is the perpendicular distance from the axis through the c.o.m. to the...
An object is formed by attaching a uniform, thin rod with a mass of mr = 7.31 kg and length L = 5.68 m to a uniform sphere with mass ms = 36.55 kg and radius R = 1.42 m. Note ms = 5mr and L = 4R. *What is the moment of inertia of the object about an axis at the left end of the rod? *If the object is fixed at the left end of the rod, what...
10*) The Sun has approximate radius 7×108 m, and rotates around
its axis once every 27 days. a) Find its angular velocity in rad/s,
and (assuming it is a uniform sphere) write a formula for its
angular momentum expressing it in terms of its mass M (you do not
need to substitute a value for M). b) Suppose the Sun were to
collapse to a neutron star, which is a much denser state, without
losing any mass and without being...
A very thin, straight, uniform rod has a length of 3.00 m and a total mass of 7.00 kg. Treating the rod as essentially a line segment of mass (distributed uniformly), do the following: (i) Use integration to prove that the rod's center of mass is located at its center point. (Reminders: dmnds mass (and that axis is perpendicular to the rod). with the previous result-to calculate lemr the moment of inertia of the rod about an axis through one...