Physics, asked by rohanavchite7, 6 months ago

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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Answered by raginikri2007
1

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

Answered by IMystery
1

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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

)

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