Then the tangent of angle-Θ is (Ay / Ax).
Answer:
(a): a = 0.4m/s²
(b): α = 8 radians/s²
Explanation:
First we propose an equation to determine the linear acceleration and an equation to determine the space traveled in the ramp (5m):
a= (Vf-Vi)/t = (2m/s)/t
a: linear acceleration.
Vf: speed at the end of the ramp.
Vi: speed at the beginning of the ramp (zero).
d= (1/2)×a×t² = 5m
d: distance of the ramp (5m).
We replace the first equation in the second to determine the travel time on the ramp:
d = 5m = (1/2)×( (2m/s)/t)×t² = (1m/s)×t ⇒ t = 5s
And the linear acceleration will be:
a = (2m/s)/5s = 0.4m/s²
Now we determine the perimeter of the cylinder to know the linear distance traveled on the ramp in a revolution:
perimeter = π×diameter = π×0.1m = 0.3142m
To determine the angular acceleration we divide the linear acceleration by the radius of the cylinder:
α = (0.4m/s²)/(0.05m) = 8 radians/s²
α: angular aceleration.
Gravitational potential energy, relative to some level =
(mass of the object)
times
(height above the reference level)
times
(acceleration due to gravity) .
