Answer:
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
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.
[tex]\sum F = \frac{mv^2}{r} - uF_n\\\\\frac{mv^2}{r} = uF_n[/tex]....equation 1.
But, Normal force, [tex]F_n = mg[/tex]
Substituting the normal force into eqn. 1, we have:
[tex]\frac{mv^2}{r} = umg\\\\\frac{v^2}{r} = ug[/tex]....equation 2.
Also, Linear speed, [tex]v = r\omega[/tex]
Substituting Linear speed into eqn. 2, we have:
[tex]\frac{(r\omega )^2}{r} = ug\\\\r\omega ^2 = ug\\\\r = \frac{ug}{\omega ^2}[/tex]
Substituting the given parameters into the formula, we have;
[tex]r = \frac{0.5 \times 9.8}{2.5^2} \\\\r = \frac{4.9}{6.25}[/tex]
Radius, r = 0.784 meters
Read more: https://brainly.com/question/13754413
