particle of mass m start moving in a circular path of radius R=1 m with increasing speed v = ktm/s . Find the time at which its total acceleration vector will makes an angle of 45 degree with the velocity vector
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Answer:
4.4 Uniform Circular Motion
LEARNING OBJECTIVES
By the end of this section, you will be able to:
Solve for the centripetal acceleration of an object moving on a circular path.
Use the equations of circular motion to find the position, velocity, and acceleration of a particle executing circular motion.
Explain the differences between centripetal acceleration and tangential acceleration resulting from nonuniform circular motion.
Evaluate centripetal and tangential acceleration in nonuniform circular motion, and find the total acceleration vector.
Uniform circular motion is a specific type of motion in which an object travels in a circle with a constant speed. For example, any point on a propeller spinning at a constant rate is executing uniform circular motion. Other examples are the second, minute, and hour hands of a watch. It is remarkable that points on these rotating objects are actually accelerating, although the rotation rate is a constant. To see this, we must analyze the motion in terms of vectors.
Centripetal Acceleration
In one-dimensional kinematics, objects with a constant speed have zero acceleration. However, in two- and three-dimensional kinematics, even if the speed is a constant, a particle can have acceleration if it moves along a curved trajectory such as a circle. In this case the velocity vector is changing, or
d
→
v
/
d
t
≠
0.
This is shown in (Figure). As the particle moves counterclockwise in time
Δ
t
on the circular path, its position vector moves from
→
r
(
t
)
to
→
r
(
t
+
Δ
t
)
.
The velocity vector has constant magnitude and is tangent to the path as it changes from
→
v
(
t
)
to
→
v
(
t
+
Δ
t
)
,
changing its direction only. Since the velocity vector
→
v
(
t
)
is perpendicular to the position vector
→
r
(
t
)
,
the triangles formed by the position vectors and
Δ
→
r
,
and the velocity vectors and
Δ
→
v
are similar. Furthermore, since
|
→
r
(
t
)
|
=
|
→
r
(
t
+
Δ
t
)
|
and
|
→
v
(
t
)
|
=
|
→
v
(
t
+
Δ
t
)
|
,
the two triangles are isosceles. From these facts we can make the assertion
Δ
v
v
=
Δ
r
r
or
Δ
v
=
v
r
Δ
r
.