Physics, 16.07.2020 01:01 katiebotts18

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?

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Answer from: ladyty2109

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

Given the following data:

Angular speed = 2.5 rad/s.Coefficient of static friction = 0.5

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.

$\sum F = \frac{mv^2}{r} - uF_n\\\\\frac{mv^2}{r} = uF_n$ ....equation 1.

But, Normal force, F_n = mg

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

$\frac{mv^2}{r} = umg\\\\\frac{v^2}{r} = ug$ ....equation 2.

Also, Linear speed, $v = r\omega$

Substituting Linear speed into eqn. 2, we have:

$\frac{(r\omega )^2}{r} = ug\\\\r\omega ^2 = ug\\\\r = \frac{ug}{\omega ^2}$

Substituting the given parameters into the formula, we have;

$r = \frac{0.5 \times 9.8}{2.5^2} \\\\r = \frac{4.9}{6.25}$

Radius, r = 0.784 meters

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Answer from: aweller3898

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

Answer from: Quest

explanation:

Answer from: Quest

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A man stands on a merry-go-round that is rotating at 2.5 rad/s. If the coefficient of static frictio...