a particle moves along circular path such that magnitude of tangential acceleration is always equal to centripetal acceleration of initial velocity is u then find the time after which it's speed becomes 2u and radius of circle is r
Answers
Answer:
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Explanation:
For circular motion, the radial acceleration, a
r
is given by
R
v
2
, and the tangential acceleration, a
t
is given by
dt
dv
Since these accelerations are equal, so
R
v
2
=
dt
dv
Hence,(
v
2
R
)dv=dt
Upon integration
∫
v
2
R
dv=∫dt
v
R
=t+C
To evaluate integration constant C
Put v=v
0
at t=0
Therefore, C=−
v
0
R
The relation between v and t is ,
v
R
=t−
v
0
R
t=R[
v
0
.v
v−v
0
]
Now, v=
dt
ds
,
where, s id the length of the arc covered by the particle as it moves in the circle
Therefore, t=
v
0
R
−
dt
ds
R
t=
v
0
R
−R.
ds
dt
tds=
v
0
R
−Rdt
ds(
v
0
R
−t)=Rdt
R
ds
=
v
0
R
−t
dt
Upon integrating
∫
R
ds
=∫
v
0
R
−t
dt
R
s
=−ln(
v
0
R
−t)+C
putting at t=0 and s=0
C=
v
0
lnR
Therefore, the relation takes the form
R
s
=
v
0
lnR
−ln(
v
0
R
−t)=ln(
R−v
0
t
R
)
For complete revolution, putting s=2πR, t=T
R
2πR
=ln(
R−v
0
T
R
)
T=
v
0
R
(1−e
−2π
)