Question

A small solid sphere of mass m is released
from a point A at a height h above the bottom
of a rough track as shown in the figure. If the
sphere rolls down the track without slipping,
its rotational kinetic energy when it comes to
the bottom of track is

Answer 1

Answer:

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K.E=P.E

2

1

mv

2

[1+

r

2

k

2

]=mgh

v

2

[1+

3

2

]=gh

v=

5

3

gh

Now

2

1

mv

2

+RotationalK.E=mgh

∴mgh−

2

1

m×

5

3

gh= Rotational K.E

So Rotational Kinetic Energy

=(1−

10

3

)mgh

=

10

7

mgh