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a ring rolls down an inclined plane the ratio of r
Question:
A ring rolls down an inclined plane. The ratio of rotational kinetic energy to translational kinetic energy is
COMEDK UGET - 2012
COMEDK UGET
Updated On:
May 12, 2024
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1:01
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The Correct Option is
B
Solution and Explanation
Translational kinetic energy,
K
T
=
1
2
m
v
2
K_T = \frac{1}{2} mv^2
K
T
=
2
1
m
v
2
where
m
m
m
= mass of the ring,
v
v
v
= speed of the centre of mass of the ring
Rotational kinetic energy,
K
R
=
1
2
I
ω
2
K_R = \frac{1}{2} I \omega^2
K
R
=
2
1
I
ω
2
Here,
I
=
m
R
2
,
ω
=
v
R
I = mR^2 , \omega = \frac{v}{R}
I
=
m
R
2
,
ω
=
R
v
So,
K
R
=
1
2
(
m
R
2
)
(
v
R
)
2
=
1
2
m
v
2
K_R = \frac{1}{2} ( mR^2 ) \left(\frac{v}{R} \right)^2 = \frac{1}{2} mv^2
K
R
=
2
1
(
m
R
2
)
(
R
v
)
2
=
2
1
m
v
2
∴
K
R
K
T
=
1
1
\therefore\:\:\:\: \frac{K_R}{K_T} = \frac{1}{1}
∴
K
T
K
R
=
1
1
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Concepts Used:
System of Particles and Rotational Motion
The system of particles refers to the extended body which is considered a
rigid body
most of the time for simple or easy understanding. A rigid body is a body with a perfectly definite and unchangeable shape.
The distance between the pair of particles in such a body does not replace or alter. Rotational motion can be described as the motion of a rigid body originates in such a manner that all of its particles move in a circle about an axis with a common angular velocity.
The few common examples of rotational motion are the motion of the blade of a windmill and periodic motion.