Question

A disk with a c value of 1/2, a mass of 4 kg, and radius of 0.26 meters, rolls without slipping down an incline with has a length of 10 meters and angle of 30 degrees. At the top of the incline the disk is spinning at 36 rad/s. How fast is the disk spinning (the center of mass) at the bottom of the incline in rad/s?

Answer 1

here,

the mass of disk , m = 4 kg

radius , r = 0.26 m

length of incline , s = 10 m

theta = 30 degree

the angular speed of the disk at the top of incline , w0 = 36 rad/s

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

u = 0.26 * 30 m/s = 7.8 m/s

let the speed of disk at the bottom be v

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 * 10 * sin(30) + 0.75 * 7.8^2 = 0.75 * v^2

solving for v

v = 11.24 m/s

the angular speed , w = v/r

w = 11.24 /0.26 rad/s = 43.2 rad/s

the speed at the bottom of incline is 43.2 rad/s