In the given setup bead of mass misconstrained to move along one of the rails this rail is rough.The bead is connected to spring the other end of spring is constrained to move along a frictionless rail such that spring is always perpendicular to the smooth rail. Relaxed length of spring is zero and setup is in gravity free space.Bead starts out at the vertex of rails with initial speed v_(0) Then select the CORRECT option(s):
Answers
Answer:
Method 1 : Observation from ground frame. Let a
r
be the acceleration of the bead relative to the rod. Then a
r
cosα is the leftward acceleration of the bead relative to the rod and a
r
sinα is downward relative acceleration of the rod. If a
y
and a
x
be the absolute leftward horizontal and downward vertical acceleration of the bead, then
(
a
bead
)
x
=(
a
bead,rod
)
x
=(
a
rod
)
x
or a
x
=a
r
cosα+a........(i)
and a
r
sinα=a
y
−0
or a
y
=a
r
sinα.........(ii)
From FBD of the bead(projecting forces vertically and horizontally)
mg−Ncos=ma
r
sinα.......(i)
and Nsinα=m(a
r
cosα+a........(ii)
Eliminating N between (i) and (ii)
mgsinα=ma
r
+macosα
or a
r
=gsinα−acosα
Method 2 : Observation from an observer moving with rod. Considering bead w.r.t. rod, i.e., from non-inertial. A pseudo force of magnitude ma will act on the bead in the magnitude ma will act on the bead in the direction opposite to acceleator of rod, i.e., in right direction.
The bead is not moving perpendicular to rod. Hence,
N=mgcosα+masinα
Also in the direction along the rod, let acceleration of the bead w.r.t. rod is a
r
.
Equation of motion of bead with rod,
mgsinα−macosα=ma
r
⇒ a
r
=gsinα−acosα
Explanation:
Method 1 : Observation from ground frame. Let a
r
be the acceleration of the bead relative to the rod. Then a
r
cosα is the leftward acceleration of the bead relative to the rod and a
r
sinα is downward relative acceleration of the rod. If a
y
and a
x
be the absolute leftward horizontal and downward vertical acceleration of the bead, then
(
a
bead
)
x
=(
a
bead,rod
)
x
=(
a
rod
)
x
or a
x
=a
r
cosα+a........(i)
and a
r
sinα=a
y
−0
or a
y
=a
r
sinα.........(ii)
From FBD of the bead(projecting forces vertically and horizontally)
mg−Ncos=ma
r
sinα.......(i)
and Nsinα=m(a
r
cosα+a........(ii)
Eliminating N between (i) and (ii)
mgsinα=ma
r
+macosα
or a
r
=gsinα−acosα
Method 2 : Observation from an observer moving with rod. Considering bead w.r.t. rod, i.e., from non-inertial. A pseudo force of magnitude ma will act on the bead in the magnitude ma will act on the bead in the direction opposite to acceleator of rod, i.e., in right direction.
The bead is not moving perpendicular to rod. Hence,
N=mgcosα+masinα
Also in the direction along the rod, let acceleration of the bead w.r.t. rod is a
r
.
Equation of motion of bead with rod,
mgsinα−macosα=ma
r
⇒ a
r
=gsinα−acosα