Question

A solid cylindrical 10 kg roll of roofing paper with radius of 15 cm, starting from rest rolls down a roof with a 20^o incline (Figure below). (a) If the cylinder rolls 4.0 m without slipping, what is the angular speed about its center when leaving the roof? (b) If the roof edge of the house is 6.0 m above level ground, how far from the edge of the roof does the cylindrical roll land? (Figure not to scale.)

Answer 1

The moment of inertia of a solid cylinder is
I = 1/2 m r^2

The kinetic energy of rotation of the cylinder is 1/2 I w^2, with w the angular velocity.
Because the cylinder rolls without slipping, there is a relation between angular velocity and translational velocity: v = w r, so the rotational kinetic energy can also be expressed in terms of the translation velocity:
Erot = 1/2 I w^2 = 1/4 m r^2 v^2/r^2 = 1/4 m v^2.

Hence the total kinetic energy is

Ekin = Erot + Etrans = 1/4 m v^2 + 1/2 m v^2 = 3/4 m v^2.

Now the conservation of mechanical energy dictates that the gain in kinetic energy equals loss of gravitational potential energy :

3/4 m v^2 = m g * (4.0m sin(20))

Hence at the roof edge the velocity will be
v = sqrt(4 g * 4.0 *sin(20) / 3) = 4.23 m/s

For part b: this is a standard exercise. The time of fall can be found by solving the quadratic

6.0 m = 4.23m/s* sin(20) * t + 1/2 *9.81 m/s^2 * t^2

With this (positive) time of fall t you then have for the horizontal distance:

s = 4.23 m/s * cos(20) * t