a rigid oblect is rolling down an inclined plane derive expressions for the acceleration along the track and the speed after falling through a certain vertical distance.
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Answer:
Consider a circularly symmetric rigid body, like a sphere or a wheel or a disc, rolling with friction down a plane inclined at an angle to the horizontal. If the friction force on the body is large enough, the body rolls without slipping.
Let M and R be the mass and radius of the body. Let I be the moment of inertia of the body. Let I be the moment of inertia of the body for rotation about an axis through its center, Let the body start from rest at the top of the incline at a height h. Let v be the translational speed of the center of mass at the bottom of the incline. Then, its kinetic energy at the bottom of the incline is
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Systems of Particles and Rotational Motion
Rolling Motion
Answer in brief:A rigid obj...
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Answer in brief:
A rigid object is rolling down an inclined plane derive the expression for the acceleration along the track and the speed after falling through a certain vertical distance.
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ANSWER
Consider a circularly symmetric rigid body, like a sphere or a wheel or a disc, rolling with friction down a plane inclined at an angle to the horizontal. If the friction force on the body is large enough, the body rolls without slipping.
Let M and R be the mass and radius of the body. Let I be the moment of inertia of the body. Let I be the moment of inertia of the body for rotation about an axis through its center, Let the body start from rest at the top of the incline at a height h. Let v be the translational speed of the center of mass at the bottom of the incline. Then, its kinetic energy at the bottom of the incline is
E=
2
1
MV
2
[1+
MR
2
1
]=
2
1
Mv
2
(1+β) .........(1)
where β=
MR
2
1
If k is the radius of gyration of the body,
IMK
2
and β=
MR
2
1
=
R
2
k
2
From the conservation of energy,
(KE+PE)
initial
=(KE+PE)
final
...........(2)
∴0+Mgh=
2
1
Mv
2
(1+β)+0
∴Mgh=
2
1
Mv
2
(1+β) .........(3)
∴v
2
=
1+β
2gh
∴v=
1+β
2gh
=
1+(K
2
/R
2
)
2gh
......(4)
Since h=Lsinθ,
v=
1+β
2gh
............(5)
Let a be the acceleration of the center of mass of the body along the inclined plane, Since the body starts from rest.
v
2
=2aL
∴a=
2L
v
2
........(6)
∴a=
1+β
2gLsinθ
⋅
2L
1
=
1+β
gsinθ
=
1+(K
2
/R
2
)
gsinθ
.........(7)
Starting from rest, if t is the time taken to travel the distance L,
L=
2
1
at
2
∴t=
a
2L
=
gsinθ
2L
⋅(1+
R
2
K
2
)