Question

A disk with a c value of 1/2, a mass of 9 kg, and radius of 0.26 meters, rolls without slipping down an incline with has a length of 9 meters and angle of 30 degrees. At the top of the incline the disk is spinning at 37 rad/s. What is the rotational kinetic energy of the disk at the bottom of the incline in Joules?

Answer 1

here,

the mass of disk , m = 9 kg

radius , r = 0.26 m

length of incline , s = 9 m

theta = 30 degree

the angular speed of the disk at the top of incline , w0 = 37 rad/s

the speed of the center of mass at the top of incline , u = r * w0

u = 0.26 * 37 m/s = 9.62 m/s

using conservation of mechanical energy

PEi + KEi = PEf + KEf

m * g * s * sin(theta) + (0.5 * I * w0^2 + 0.5 * m * u^2) = 0 + (0.5 * I * w^2 + 0.5 * m * v^2)

m * g * s * sin(theta) + (0.5 * (0.5 * m * r^2) * (u/r)^2 + 0.5 * m * u^2) = 0 + (0.5 * 0.5 * m * r^2 * (v/r)^2 + 0.5 * m * v^2)

m * g * s * sin(theta) + 0.75 * m * u^2 = 0.75 * m * v^2)

g * s * sin(theta) + 0.75 * u^2 = 0.75 * v^2)

9.81 * 9 * sin(30) + 0.75 * 9.62^2 = 0.75 * v^2

solving for v

v = 12.3 m/s

he rotational kinetic energy of the disk at the bottom of the incline , KEr = 0.5 * I * w^2

KEr = 0.5 * ( 0.5 * m * r^2) * ( v/r)^2

KEr = 0.25 * 9 * 12.3^2 J = 340.4 J