Portion AB of the wedge shown in fig is rough and BC is smooth. A solid cylinder rolls without slipping from A to B.If AB = BC,
then ratio of translational kinetic energy to rotational kinetic energy, when the cylinder reaches point Cis
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Answer:
Consider that AB is a frictional surface and BC is smooth surface.
According to law of conversation of energy\
mg
2
H
=
2
1
mv
2
+
2
1
(
2
mR
2
)
R
2
v
2
2
msH
=
2
1
mv
2
+
4
1
mv
2
=
4
3
mv
2
As cylinder goes mg
2
H
downwards,
This implies that, loss of potential energy is equal to gain in Kinetic energy
2
mgH
=
4
3
mv
2
Total Kinetic energy is:
=
2
1
mv
2
+
4
3
mv
2
+
2
1
mv
2
=
4
5
mv
2
+
4
1
mv
2
KE
R
KE
T
=
4
1
mv
2
4
5
mv
2
=5
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