Question

If the time of descent of a rolling cylinder on a plane inclined to 10 degree to the horizontal is 3.35 seconds.what is the distance covered by the rolling cylinder, if the Acceleration of the cylinder is given as (2gSin¹/3)? Take g=10m/s²

Answer 1

Answer:

hey mate

Explanation:

ANSWER

apply WET,

Loss in gravitational potential energy = gain in rotational kinetic energy +gain in translational kinetic energy

mgh=

2

1

mV

2

+

2

1

(

2

m

R

2

)(

R

V

)

2

mgh=

4

3

mV

2

V=

3

4

gh

for block

again apply WET,

mgh=

2

1

mV

0

2

V

0

=

2gh

=

3

2

V