Question

A man stands on a merry-go-round that is rotating at 1.58 rad/s. If the coefficient of static friction between the man's shoes and the merry-go-round is μs = 0.45, how far from the axis of rotation can he stand without sliding? (Enter the maximum distance in meters.) m

Answer 1

Answer:

the maximum distance of rotation can he stand without sliding is 1.77 m

Explanation:

given information:

angular velocity , ω = 1.58 rad/s

static friction, μs = 0.45

now we calculate the vertical force

N - W = 0, N is normal force and W is weight

N = W

   = m g

next, for the horizontal force we only have frictional force, thus

F(friction) = m a

μs N = m a

μs m g = m a

a = μs g,

now we have to find the acceleration which is both translation and cantripetal.

a = $\sqrt{a_{t} ^{2}+a_{c} ^{2}  }$

$a_{t} ^{2}$ is the acceleration for translation

$a_{t} ^{2}$ = 0

$a_{c} ^{2}$ is centripetal acceleration

$a_{c} ^{2}$ = ω^2r

therefore,

a = $\sqrt{a_{c} ^{2}  }$

  = $a_{c} ^{2}$

  = ω^2r

Now, to find the radius, substitute the equation into the following formula

a = μs g

ω^2r = μs g

r = μs g / ω^2

 = (0.45 x 9.8) / (1.58)

 = 1.77 m