# Physics question help; picture attached?

Through what angle does the disk rotate

while sliding before it begins to roll without slipping. g = 9.8m/s^2. Answer in units of rev.

### 1 Answer

- NCSLv 78 months agoBest Answer
Relevant equations:

torque τ = friction force * radius = f*r = µmgr

so τ = 0.03 * 6kg * 9.8m/s² * 0.50m = 0.882 N·m

also

τ = I*α

so

0.882 N·m = (6/7)mr² * π = (6/7) * 6kg * (0.50m)² * α

means that

α = 0.686 rad/s²

we are interested in the point at which the linear velocity is equal to the angular velocity * the radius:

v = (ω₀ - αt)*r

v = a*t = µg*t = 0.03 * 9.8m/s² * t = 0.294m/s² * t

so substituting for v and ω₀ and α and r above gives

0.294t = (3 - 0.686t)*0.5

solves to

t = 2.355 s

then the angle rotated through while it gets traction is

Θ = ω₀*t - ½*α*t² = 3rad/s*2.355s - ½*0.686rad/s²*(2.355s)²

Θ = 5.1 rads = 296º

barring computational error

Hope this helps!