AUS. (10 (1)
Based on the Kinematic of the Circular
Motion
22. A particle is moving on a circular path of diameter 2.0 m
with uniform speed 10 m/s. Find :
(a) Difference in distance travelled and displacement
after completing half cycle.
(b) Magnitude of change in velocity in half cycle.
(c) Time period
Ans.(a) 1.14 m (b) 20 m/s (c) 0.628 s
Answers
Answer:
a-In circular motion the direction of motion is tangential and always changing, but the speed is constant. The force required to change the velocity (and accelerate the object) is always towards the centre of the circle and perpendicular to the motion. ... The resultant force and hence the acceleration is towards the centre.
b-ANSWER
Net acceleration is the resultant acceleration of centripetal acceleration and tangential acceleration thus find both of them as follows:
A
c
− be centripetal acceleration.
A
t
− be tangential acceleration
A
net
=
(A
c
2
+A
t
2
)
Now then,
A
c
is
R
V
2
=
10
25
=2.5m/s
2
A
t
is given as 2.m/s
2
, thus
A
net
=
(2.5
2
+2
2
)
=3.20m/s
2
.
Hence,
option (A) is correct answer.
c-Distance is the total path covered by the body and is scalar, whereas displacement is the shortest distance between initial and final positive with the direction from the starting point to the ending point.
Distance covered of a moving body is always positive and can never be zero or negative
However, the displacement can be positive, negative or zero.
So for the particular problem half a revolution would imply a distance equal to half the circumference=pi*r
And displacement would be the shortest distance along the diameter =2*r
d-ANSWER
The particle starts from point P and after half rotation, it reaches at Q via path PMQ.
Distance covered by particle to reach at Q is d=πR
Time taken, t=
v
d
=
v
πR
Displacement of the particle, s=PQ=2R
∴ Average velocity, v
avg
=
t
s
=
v
πR
2R
=
π
2
v
Answer:
please give me a call from the following number please chat with you and your wattsafe number please
Explanation:
please give me a call