Question

The 30-kg disk is originally spinning at v = 125 rad>s. If it is placed on the ground, for which the coefficient of kinetic friction is μC = 0.5, determine the time required for the motion to stop. What are the horizontal and vertical components of force which the member AB exerts on the pin at A during this time? Neglect the mass of AB.

Answer 1

Answer:

t = 3.82 s

Ax = 147 N  (←)

Ay = 294 N   (+↑)

Explanation:

Given

m = 30.0 Kg

ωinitial = 125 rad/s

ωfinal = 0 rad/s

μC = 0.5

R = 0.3 m

t = ?

Ax = ?

Ay = ?

For the disk, we can apply

∑ τ = I*α

where I = m*R²/2

then

⇒ R*Ffriction = (m*R²/2)*α

⇒ R*(-μC*N) = R*(-μC*m*g) = (m*R²/2)*α

⇒ α = -2*μC*g / R

⇒ α = -2*(0.5)*(9.8) / 0.3 = -32.666 rad/s²

we can use the equation to get t:

α = Δω / t      ⇒   t = Δω / α = (0 - 125) / (-32.666)

⇒   t = 3.82 s

The horizontal and vertical components of force which the member AB exerts on the pin at A during this time are

∑ Fx = 0  (+→)

Ax - Ffriction = 0

⇒  Ax = Ffriction = μC*m*g = (0.5)*(30)*(9.8) = 147 N

⇒   Ax = 147 N  (←)

∑ Fy = 0   (+↑)

⇒  Ay - m*g = 0

⇒  Ay = m*g

⇒  Ay = 30*9.8 = 294 N

⇒  Ay = 294 N   (+↑)