Question

In the figure below, a solid cylinder of radius 18 cm and mass 12 kg startsfrom rest and rolls without slipping a distance L = 7.9 mdown a roof that is inclined at angle ? = 29

Answer 1

a) We find its angular speed as it leaves the roof using conservation of energy.

Its initial kinetic energy is K_i = 0 and its initial potential energy is U_i = Mgh where h=7.9sin 29= 3.83 m (we are using the edge of the roof as our reference level for computing U). Its final kinetic energy (as it leaves the roof) is the sum of rotational and translational kinetic energies.

Here we use v to denote the speed of its center of mass and w is its angular speed

at the moment it leaves the roof. Since (up to that moment) the ball rolls without sliding we can set v=Rw where R=0.18m and I= 1/2MR²

The mass M cancels from the equation, and we obtain

b)Now this becomes a projectile motion

v=Rw ---> v=0.18*39.3= 7.07 m/s and its components are

v_x=vcos29= 7.07cos29= 6.18 m/s

v_y=vsin29=7.07sin29= 3.43 m/s

The projectile motion equations become

and   

We find the time when y = H = 4.1 m from the second equation (using the quadratic formula, choosing the positive root):

Then we substitute this into the x equation and obtain