Question

A solid cylinder rolls up an inclined plane of angle of inclination 30°. At the bottom of the inclined plane the centre of mass of the cylinder has a speed of 5 m/s. 

(a) How far will the cylinder go up the plane? 

(b) How long will it take to return to the bottom?

Answer 1

Inclination angle (∅)= 30° 
Speed of centre of mass (v)= 5m/s 

(A) acceleration of the cylinder rolling up the inclined plane is 
Given by the 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) 
= -9.8/3 m/s²

Using equation of motion, 
V² = U² + 2aS 
0 = (5)² + 2(-9.8/3)×S 
S = 25×3/2×9.8 = 3.83 m 

Time taken to return the bottom = t 
S = ut + (1/2)at² 
3.83 = 0 + 1/2(9.8/3)t² 
t = 1.53 sec