A cylinder of diameter 100 mm rolls from restdown a 5 m long ramp and its center of mass is moving with velocity 2 m/s at the bottom of the ramp. (a) What is its (constant) acceleration down the ramp? (b) What is its angular acceleration?

(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.