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?
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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
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
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