Manipal Entrance Test 2015 Physics Paper

Linear acceleration for rolling, a=

g‌sin‌θ

√1+k2∕R2

For cylinder,

k2

R2

=

1

2

∴acylinder=

2

3

g‌sin‌θ
For rotation, the torque
FR=

Iα.(MR2α)

2

(where, F = force of friction)
But Rα=a
∴F=

M

2

a
F=

M

2

.

2

3

g‌sin‌θ=

M

3

g‌sin‌θ
∴µs=

F

N

, where N is normal reaction,
µs=

M 3 g‌sin‌θ

Mg‌sin‌θ

=

tan‌θ

3

∴ For rolling without slipping of a roller down, the inclined plane, tan‌θ≤3µs