at t=0 particle starts and moves in circular path of radius R of tangential acceleration is 'a' then find the time when total acceleration vector makes an angle 45° with tangential acceleration vector
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Explanation:
Correct option is
C
2
2/3
sec
using equation for a straight line (y=mx+c) where,
m=tan60
o
=
3
(slop) C=0 from graph
relation b/w a
T
and t is obtained as:-
a
T
=tan60
o
t+0
this tangential acceleration increases velocity (v) of the particless as.
a
T
=
dt
dv
⇒
dt
dv
=
3
t
⇒∫dv=∫
3
+dt⇒v=
2
3
t
2
so, a
T
a particular time, centripetal accin (a
c
) is given by :-
a
c
=
r
v
2
=
4
3
t
4
(r=1)
a
T
let a time
a
masses angle 30
o
with
a
c
. so it makes angles 60
o
with
a
T
(
a
c
K
a
T
are perpendicular)
⇒
a
T
=
3t
2
+
16
9
+8
2
1
that given,
⇒3t
2
=
11
1
(3t
2
+
16
9
+8)
⇒t=0 or t=2
2/3
.
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