Question

A disk with a c value of 1/2, a mass of 3 kg, and radius of 0.29 meters, rolls without slipping down an incline with has a length of 6 meters and angle of 30 degrees. At the top of the incline the disk is spinning at 28 rad/s. How fast is the disk moving (the center of mass) at the bottom of the incline in m/s?

Answer 1

m = mass of the disk = 3 kg

r = radius of disk = 0.29 m

I = moment of inertia of disk = (0.5) m r²

L = length of the incline = 6 m

= angle of incline = 30 deg

h = height of the incline

Height of the incline is given as

h = L Sin = 6 Sin30 = 3 m

w_o = initial angular speed at the top = 28 rad/s

v_o = initial speed = r w_o = (0.29) (28) = 8.12 m/s

w = final angular speed at the bottom = ?

v = final speed = r w

Using conservation of energy

Potential energy at the top + Rotational kinetic energy at the top + kinetic energy energy at the top = Rotational kinetic energy at the top + kinetic energy energy at the top

mgh + (0.5) I w_o² + (0.5) m v_o² = (0.5) I w² + (0.5) m v²

mgh + (0.5) (0.5) (mr²) (v_o/r)² + (0.5) m v_o² = (0.5) (0.5) (mr²) (v/r)² + (0.5) m v²

mgh + (0.25) m v_o² + (0.5) m v_o² = (0.25) m v² + (0.5) m v²

mgh + (0.75) m v_o² = (0.75) m v²

gh + (0.75) v_o² = (0.75) v²

(9.8 x 3) + (0.75) (8.12)² = (0.75) v²

v = 10.3 m/s