A man stands on a merry-go-round that is rotating at 2.5 rad/s. If the coefficient of static friction between the man’s shoes and the merry-go-round is µs = 0.5, how far from the axis of rotation can he stand without sliding?

The distance (radius) from the axis of rotation which the man can stand without sliding is 0.784 meters.

Given the following data:

To determine how far (radius) from the axis of rotation can the man stand without sliding:

We would apply Newton's Second Law of Motion, to express the centripetal and force of static friction acting on the man.

....equation 1.

But, Normal force,  

Substituting the normal force into eqn. 1, we have:

....equation 2.

Also, Linear speed,

Substituting Linear speed into eqn. 2, we have:

Substituting the given parameters into the formula, we have;

Radius, r = 0.784 meters

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0.8 m

Explanation:

Draw a free body diagram.  There are three forces:

Weight force mg pulling down,

Normal force N pushing up,

and friction force Nμ pushing towards the center.

Sum of forces in the y direction:

∑F = ma

N − mg = 0

N = mg

Sum of forces in the centripetal direction:

∑F = ma

Nμ = m v²/r

Substitute and simplify:

mgμ = m v²/r

gμ = v²/r

Write v in terms of ω and solve for r:

gμ = ω²r

r = gμ/ω²

Plug in values:

r = (10 m/s²) (0.5) / (2.5 rad/s)²

r = 0.8 m

explanation: magnetic pole, region at each end of a magnet where the external magnetic field is strongest. the south-seeking pole, or any pole similar to it, is called a south magnetic pole. unlike poles of different magnets attract each other; like poles repel each other.

the answer is b i think