Question 57
Angular speed required so that the block does not slide
Given:
cylinder radius r, coefficient of friction μ, angular speed ω.
cylinder radius r, coefficient of friction μ, angular speed ω.
Key idea: At every angular position, friction must balance the tangential component of gravity, while the normal reaction changes with position.
1. Tangential direction
Let θ be measured from the lowest point. If the block does not slip relative to the cylinder, its tangential acceleration is zero.
|f| = mg|sinθ|
2. Radial direction
Towards the centre, the required centripetal acceleration is ω2r.
N − mg cosθ = mω2r
N = m(ω2r + g cosθ)
3. Static-friction condition
|f| ≤ μN
mg|sinθ| ≤ μm(ω2r + g cosθ)
g(|sinθ| − μcosθ) ≤ μω2r
4. Condition for every position
The maximum value of |sinθ| − μcosθ is
√1 + μ2
Therefore,
g√1 + μ2 ≤ μω2r
ω ≥ √g√(1 + μ2)/(μr)
ωmin = √[g√(1 + μ2)/(μr)]