what is the average velocity and displacement of a body moving in a horizontal circle with a constant speed
Answers
Answer:
Since the particle has moved by an angle of
60
o
, if we join the initial and final positions of the particles with the center, the figure so formed will be an equilateral triangle. Since all the sides of an equilateral triangle are equal, the displacement would be equal to the radius of the circle.
Explanation:
A) Since the particle has moved by an angle of
60
o
, if we join the initial and final positions of the particles with the center, the figure so formed will be an equilateral triangle. Since all the sides of an equilateral triangle are equal, the displacement would be equal to the radius of the circle.
B)Average velocity of the particle is = total displacent
/
time
Let the radius of the circle be
R
Displacement =
R
Now,
total time taken by the particle to complete 1 rotation
i
.
e
to cover
360
o
⇒
T
=
2
π
R
v
So, time taken to cover
60
o
⇒
T
1
=
1
6
T
T
1
=
π
R
3
v
So, average velocity
=
R
π
R
v
=
v
π
C)average acceleration = change in velocity
/
time
Let us assume that the particle started with its velocity pointing upwards. ( see the image )
after it has moved by
60
o
, the angle made by its velocity at that instant is
60
o
with the vertical. ( Its just geometry going on here.)
Since we assumed that the initial velocity was in upwards direction, the angle made by the final velocity with the initial velocity is indeed
60
o
.
Since we are finding the change in the velocity,
Let final velocity be
v
2
and initial
v
1
=
Δ
V
=
√
(
v
1
)
2
+
(
v
2
)
2
−
2
v
1
v
2
(
cos
(
60
o
)
Since the particle hasn't changed its velocity,
v
1
=
v
2
Solving the above expression we get
=
Δ
V
=
v
1
Plugging in the values,
average velocity
=
3
v
2
π
R
