Question

A cylinder with radius R spins around its axis with an angular speed ω. On its inner surface, there lies a small block; the coefficient of friction between the block and the inner surface of the cylinder is µ. Find the values of ω for which the block does not slip (stays still with respect to the cylinder). Consider the cases where (a) the axis of the cylinder is horizontal; (b) the axis is inclined by angle α with respect to the horizon.

Answer 1

Answer:

A. Mrŵ² = ųMg

Ŵ = (ųg/r)^½

B.

Ŵ =[ (g /r)* tan á]^½

Explanation:

T.v.= centrepetal force = mrŵ²

Where m = mass of block,

r = radius

Ŵ = angular momentum

On a horizontal axial banking frictional force supplies the Pentecostal force is numerically equal.

So there for

Mrŵ² = ųMg

Ŵ = (ųg/r)^½

g = Gravitational pull

ų = coefficient of friction.

B. The net external force equals the horizontal centerepital force if the angle à is ideal for the speed and radius then friction becomes negligible

So therefore

N *(sin á) = mrŵ² .....equ 1

Since the car does not slide the net vertical forces must be equal and opposite so therefore

N*(cos á) = mg.....equ 2

Where N is the reaction force of the car on the surface.

Equ 2 becomes N = mg/cos á

Substituting N into equation 1

mg*(sin á /cos á) =mrŵ²

Tan á = rŵ²/g

Ŵ =[ (g /r)* tan á]^½