Question

A long thin rod of mass *M* and length *L* is
situated along the *y* axis with one end at the origin. A
small spherical mass *m*_{1} is placed at the
location P, which is at a distance *d* from the origin.

If *L* = 2.00 ? 10^{4} m, *d* = 18.0 ?
10^{4} m,*M* = 14.0 ? 10^{6} kg,and
*m*_{1} = 8.00 ? 10^{6} kg,what is the value
of this potential energy?

Answer #1

Gravitational potential energy between two body is given by

Since the mass are distributed so we have to use integration method to find the total potential energy

mass per unit length of the rod is

lets take a small mass "dm" at a distance "r' from the point P so

**So the value of potential energy is -4.401*10^-2 J or
-0.04401 J**

A thin, non conducting rod with length L lies along the positive
X-Axis with one end at the origin. The rod carries a charge
distributed along its length of λ(x) = bx/L. Determine the electric
potential along the X-Axis at the point x = 2 cm if L = 1 cm and b
= 50 pC/m. Answer = (0.17 V)

A thin, 1-dimensional, uniform rod of mass M and length L lies
on the x axis with one end at the origin.
(a) Find its moment of inertia tensor about the origin.
(b) Find the moment of inertia tensor if the rod’s center is
located at the origin.

6) A rod with length "l" is lied along x-axis. The
charge density of the rod is "a". Calculate the potential of the
rod for a the point p on x-axis.
7) A rod with length "l" is lied along x-axis. The
charge density of the rod is "a". Calculate the Electric field of
the rod for a point p on x-axis.

A thin rod of length l and uniform charge per unit
length λ lies along the x axis as shown figure. (a) Show that the
electric field at point P, a distance y from the rod, along the
perpendicular bisector has no x component and is given by
E=(2kλsinθ0)/y. (b) Using your result to (a), show that the field
of a rod of infinite length is given by E=2kλ/y.

A uniform rod of length L and mass M is free to swing about an
axis that is perpendicular to the rod. The axis is a distance x
from the rod's center of mass.
a) Find the period of oscillations for small angles as a
function of L and x with appropriate constants.
b) make a sketch of the period as a function of x. If you use a
spread sheet you may assume that L=1.0 m, then your graph...

A thin rod of length L has uniform linear mass density λ
(mass/length).
(a) Find the gravitational potential Φ(r) in the plane that
perpendicularly bisects the rod where r is the perpendicular
distance from the rod center. Assume the gravitational potential at
infinity is zero.
(b) Find an approximate form of your expression from part (a)
when r >> L.
(c) Find an approximate form of your expression from part (a)
when r<< L.

A thin rod sits along the xx axis with one end at x= -1.4 m and
the other end at x= 3.5 m. The rod has a non-uniform density that
increases linearly from 1.1 kg/m at the left end to 3.0 kg/m at the
right end
a What is the total mass of the rod?
answer 10 kg
b Where is the center of mass of the rod located?
find it

Consider a thin rod of length L=2.58m and mass m1=1.27 kg, and a
hollow (empty) sphere of radius R=0.16 m and mass of m2=0.82 kg.
Sphere is at one end of the rod and the other end of the rod is
fixed and oscillate like a pendulum (simple harmonic oscillations,
SHM) with small-angle oscillations. When ? ?? ?????, ???? ≈ ?. a)
Derive a second order differential equation for this pendulum to
confirm the oscillation is SHM.(b) Compare the above...

A thin, rigid, uniform rod has a mass of 1.40 kg and a length of
2.50 m. (a) Find the moment of inertia of the rod relative to an
axis that is perpendicular to the rod at one end. (b) Suppose all
the mass of the rod were located at a single point. Determine the
perpendicular distance of this point from the axis in part (a),
such that this point particle has the same moment of inertia as the
rod...

The uniform thin rod in the figure below has mass M =
2.00 kg and length L = 2.87 m and is free to rotate on a
frictionless pin. At the instant the rod is released from rest in
the horizontal position, find the magnitude of the rod's angular
acceleration, the tangential acceleration of the rod's center of
mass, and the tangential acceleration of the rod's free end.
HINT
An illustration shows the horizontal initial position and
vertical final position...

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 5 minutes ago

asked 34 minutes ago

asked 36 minutes ago

asked 39 minutes ago

asked 43 minutes ago

asked 55 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago