Question

A solid cylinder of mass 2 kg and radius 50 cm rolls
up an inclined plane of angle inclination 30°. The
centre of mass of cylinder has speed of 4 m/s. The
distance travelled by the cylinder on the incline
surface will be : (Take g = 10 m/s2)
(1) 2.2 m
(2) 1.6 m
(3) 1.2 m
(4) 2.4 m

explain it properly​

Answer 1

$\huge{\underline{\mathfrak{Solution:-}}}$

$\boxed{\sf{→1.6m}}$

$\bigstar{\mathfrak{\underline{Given:-}}}$

• Inclination angle (∅) = 30°

• Speed of centre of mass (v)= 5m/s

$\huge{\underline{\mathfrak{Explaination:-}}}$

$\bigstar{\underline{\sf{Acceleration\:of \:the\:cylinder\:rolling\:up\:the\:inclined\:plane\:is:-}}}$

$\bigstar{\underline{\sf{According\:to\: formula:-}}}$

a = - gsin∅ / (1 + k²/r²)

Where K = radius of gyration

Moment of inertia of cylinder = (1/2)mr²

K² = r²/2

a = -gsin30°/(1 + 1/2)

$\boxed{\sf{a = {\frac{-9.8}{3}m/s^{2}}}}$

$\bigstar{\underline{\sf{Using\:equation\:of\:motion:-}}}$

V² = U² + 2aS

0 = (5)² + 2(-9.8/3) × S

S = 25 × 3/2 × 9.8 = 3.83 m

$\bigstar{\underline{\sf{Time\:taken\:to\:return\:the\: bottom = t}}}$

S = ut + (1/2)at²

3.83 = 0 + 1/2(9.8/3)t²

$\boxed{\sf{t = 1.53 sec}}$

______________________________________________________________________________

Answer 2

Answer:2.4m

Explanation:

At the top of the inclined plane there is only potential energy.

Thus, tthe potential energy is equal to the kinetic energy of rolling. So plz see the attached image for full answer